Rd Sharma Xi 2020 2021 _volume 2 for Class 11 Science Maths Chapter 30 - Derivatives

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Page No 30.25:

Question 1:

(i) $\frac{2}{x}$

(ii) $\frac{1}{\sqrt{x}}$

(iii) $\frac{1}{x^{3}}$

(iv) $\frac{x^{2} + 1}{x}$

(v) $\frac{x^{2} - 1}{x}$

(vi) $\frac{x + 1}{x + 2}$

(vii) $\frac{x + 2}{3 x + 5}$

(viii) k xⁿ

(ix) $\frac{1}{\sqrt{3 - x}}$

(x) x² + x + 3

(xi) (x + 2)³

(xii) x³ + 4x² + 3x + 2

(xiii) (x² + 1) (x − 5)

(xiv) $\sqrt{2 x^{2} + 1}$

(xv) $\frac{2 x + 3}{x - 2}$

Answer:

$(i) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\frac{2}{x + h} - \frac{2}{x}}{h} = \lim_{h \to 0} \frac{2 x - 2 x - 2 h}{h x (x + h)} = \lim_{h \to 0} \frac{- 2 h}{h x (x + h)} = \lim_{h \to 0} \frac{- 2}{x (x + h)} = \frac{- 2}{x^{2}}$
$(ii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h} = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x + h}}{h \sqrt{x} \sqrt{x + h}} \times \frac{\sqrt{x} + \sqrt{x + h}}{\sqrt{x} + \sqrt{x + h}} = \lim_{h \to 0} \frac{x - x - h}{h \sqrt{x} \sqrt{x + h} (\sqrt{x} + \sqrt{x + h})} = \lim_{h \to 0} \frac{- h}{h \sqrt{x} \sqrt{x + h} (\sqrt{x} + \sqrt{x + h})} = \lim_{h \to 0} \frac{- 1}{\sqrt{x} \sqrt{x + h} (\sqrt{x} + \sqrt{x + h})} = \frac{- 1}{\sqrt{x} \sqrt{x} (\sqrt{x} + \sqrt{x})} = \frac{- 1}{x \times 2 \sqrt{x}} = \frac{- 1}{2 x^{\frac{3}{2}}} = - \frac{1}{2} x^{\frac{- 3}{2}}$
$(iii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\frac{1}{(x + h)^{3}} - \frac{1}{x^{3}}}{h} = \lim_{h \to 0} \frac{x^{3} - (x + h)^{3}}{h (x + h)^{3} x^{3}} = \lim_{h \to 0} \frac{x^{3} - x^{3} - 3 x^{2} h - 3 x h^{2} - h^{3}}{h (x + h)^{3} x^{3}} = \lim_{h \to 0} \frac{- 3 x^{2} h - 3 x h^{2} - h^{3}}{h (x + h)^{3} x^{3}} = \lim_{h \to 0} \frac{h (- 3 x^{2} - 3 x h - h^{2})}{h (x + h)^{3} x^{3}} = \lim_{h \to 0} \frac{(- 3 x^{2} - 3 x h - h^{2})}{(x + h)^{3} x^{3}} = \frac{- 3 x^{2}}{x^{6}} = \frac{- 3}{x^{4}} = - 3 x^{- 4}$
$(iv) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\frac{(x + h)^{2} + 1}{x + h} - \frac{x^{2} + 1}{x}}{h} = \lim_{h \to 0} \frac{\frac{x^{2} + 2 x h + h^{2} + 1}{x + h} - \frac{x^{2} + 1}{x}}{h} = \lim_{h \to 0} \frac{x^{3} + 2 x^{2} h + h^{2} x + x - x^{3} - x^{2} h - x - h}{x h (x + h)} = \lim_{h \to 0} \frac{x^{2} h + h^{2} x - h}{x (x + h)} = \lim_{h \to 0} \frac{h (x^{2} + h x - 1)}{x h (x + h)} = \lim_{h \to 0} \frac{x^{2} + h x - 1}{x (x + h)} = \frac{x^{2} - 1}{x^{2}}$

$(v) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\frac{(x + h)^{2} - 1}{x + h} - \frac{x^{2} - 1}{x}}{h} = \lim_{h \to 0} \frac{\frac{x^{2} + 2 x h + h^{2} - 1}{x + h} - \frac{x^{2} - 1}{x}}{h} = \lim_{h \to 0} \frac{x^{3} + 2 x^{2} h + h^{2} x - x - x^{3} - x^{2} h + x + h}{x h (x + h)} = \lim_{h \to 0} \frac{x^{2} h + h^{2} x + h}{x (x + h)} = \lim_{h \to 0} \frac{h (x^{2} + h x + 1)}{x h (x + h)} = \lim_{h \to 0} \frac{x^{2} + h x + 1}{x (x + h)} = \frac{x^{2} + 1}{x^{2}}$

$(vi) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\frac{x + h + 1}{x + h + 2} - \frac{x + 1}{x + 2}}{h} = \lim_{h \to 0} \frac{(x + h + 1) (x + 2) - (x + h + 2) (x + 1)}{h (x + h + 2) (x + 2)} = \lim_{h \to 0} \frac{x^{2} + 2 x + h x + 2 h + x + 2 - x^{2} - x - h x - h - 2 x - 2}{h (x + h + 2) (x + 2)} = \lim_{h \to 0} \frac{h}{h (x + h + 2) (x + 2)} = \lim_{h \to 0} \frac{1}{(x + h + 2) (x + 2)} = \frac{1}{(x + 0 + 2) (x + 2)} = \frac{1}{{(x + 2)}^{2}}$

$(vii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\frac{x + h + 2}{3 (x + h) + 5} - \frac{x + 2}{3 x + 5}}{h} = \lim_{h \to 0} \frac{\frac{x + h + 2}{3 x + 3 h + 5} - \frac{x + 2}{3 x + 5}}{h} = \lim_{h \to 0} \frac{(x + h + 2) (3 x + 5) - (3 x + 3 h + 5) (x + 2)}{h (3 x + 3 h + 5) (3 x + 5)} = \lim_{h \to 0} \frac{3 x^{2} + 3 x h + 6 x + 5 x + 5 h + 10 - 3 x^{2} - 3 x h - 5 x - 6 x - 6 h - 10}{h (3 x + 3 h + 5) (3 x + 5)} = \lim_{h \to 0} \frac{- h}{h (3 x + 3 h + 5) (3 x + 5)} = \lim_{h \to 0} \frac{- 1}{(3 x + 3 h + 5) (3 x + 5)} = \frac{- 1}{{(3 x + 5)}^{2}}$

$(viii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{k {(x + h)}^{n} - k x^{n}}{h} = \lim_{(x + h) - x \to 0} \frac{k [{(x + h)}^{n} - x^{n}]}{(x + h) - x} Here, we have: \lim_{x \to a} \frac{x^{m} - a^{m}}{x - a} = m a^{m - 1} = k n x^{n - 1}$

$(ix) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\frac{1}{\sqrt{3 - x - h}} - \frac{1}{\sqrt{3 - x}}}{h} = \lim_{h \to 0} \frac{(\sqrt{3 - x} - \sqrt{3 - x - h})}{h \sqrt{3 - x} \sqrt{3 - x - h}} = \lim_{h \to 0} \frac{(\sqrt{3 - x} - \sqrt{3 - x - h})}{h \sqrt{3 - x} \sqrt{3 - x - h}} \times \frac{(\sqrt{3 - x} + \sqrt{3 - x - h})}{(\sqrt{3 - x} + \sqrt{3 - x - h})} = \lim_{h \to 0} \frac{(3 - x - 3 + x + h)}{h \sqrt{3 - x} \sqrt{3 - x - h} (\sqrt{3 - x} + \sqrt{3 - x - h})} = \lim_{h \to 0} \frac{h}{h \sqrt{3 - x} \sqrt{3 - x - h} (\sqrt{3 - x} + \sqrt{3 - x - h})} = \lim_{h \to 0} \frac{1}{\sqrt{3 - x} \sqrt{3 - x - h} (\sqrt{3 - x} + \sqrt{3 - x - h})} = \frac{1}{\sqrt{3 - x} \sqrt{3 - x - 0} (\sqrt{3 - x} + \sqrt{3 - x - 0})} = \frac{1}{(3 - x) (2 \sqrt{3 - x})} = \frac{1}{2 {(3 - x)}^{\frac{3}{2}}}$

$(x) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{{(x + h)}^{2} + x + h + 3 - (x^{2} + x + 3)}{h} = \lim_{h \to 0} \frac{x^{2} + h^{2} + 2 x h + x + h + 3 - x^{2} - x - 3}{h} = \lim_{h \to 0} \frac{h^{2} + 2 x h + h}{h} = \lim_{h \to 0} \frac{h (h + 2 x + 1)}{h} = \lim_{h \to 0} h + 2 x + 1 = 0 + 2 x + 1 = 2 x + 1$

$(xi) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{{(x + h + 2)}^{3} - {(x + 2)}^{3}}{h} = \lim_{h \to 0} \frac{(x + h + 2 - x - 2) [{(x + h + 2)}^{2} + (x + h + 2) (x + 2) + {(x + 2)}^{2}]}{h} = \lim_{h \to 0} \frac{h [{(x + h + 2)}^{2} + (x + h + 2) (x + 2) + {(x + 2)}^{2}]}{h} = \lim_{h \to 0} [{(x + h + 2)}^{2} + (x + h + 2) (x + 2) + {(x + 2)}^{2}] = [{(x + 0 + 2)}^{2} + (x + 0 + 2) (x + 2) + {(x + 2)}^{2}] = {(x + 2)}^{2} + {(x + 2)}^{2} + {(x + 2)}^{2} = 3 {(x + 2)}^{2}$

$(xii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{{(x + h)}^{3} + 4 {(x + h)}^{2} + 3 (x + h) + 2 - (x^{3} + 4 x^{2} + 3 x + 2)}{h} = \lim_{h \to 0} \frac{x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3} + 4 x^{2} + 4 h^{2} + 8 x h + 3 x + 3 h + 2 - x^{3} - 4 x^{2} - 3 x - 2}{h} = \lim_{h \to 0} \frac{3 x^{2} h + 3 x h^{2} + h^{3} + 4 h^{2} + 8 x h + 3 h + 2}{h} = \lim_{h \to 0} \frac{h (3 x^{2} + 3 x h + h^{2} + 4 h + 8 x + 3)}{h} = \lim_{h \to 0} (3 x^{2} + 3 x h + h^{2} + 4 h + 8 x + 3) = 3 x^{2} + 8 x + 3$

$(xiii) (x^{2} + 1) (x - 5) = x^{3} - 5 x^{2} + x - 5 \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{{(x + h)}^{3} - 5 {(x + h)}^{2} + (x + h) - 5 - (x^{3} - 5 x^{2} + x - 5)}{h} = \lim_{h \to 0} \frac{x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3} - 5 x^{2} - 5 h^{2} - 10 x h + x + h - 5 - x^{3} + 5 x^{2} - x + 5}{h} = \lim_{h \to 0} \frac{3 x^{2} h + 3 x h^{2} + h^{3} - 5 h^{2} - 10 x h + h}{h} = \lim_{h \to 0} \frac{h (3 x^{2} + 3 x h + h^{2} - 5 h - 10 x + 1)}{h} = \lim_{h \to 0} (3 x^{2} + 3 x h + h^{2} - 5 h - 10 x + 1) = 3 x^{2} - 10 x + 1$

$(xiv) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\sqrt{2 {(x + h)}^{2} + 1} - \sqrt{2 x^{2} + 1}}{h} = \lim_{h \to 0} \frac{\sqrt{2 x^{2} + 2 h^{2} + 4 x h + 1} - \sqrt{2 x^{2} + 1}}{h} = \lim_{h \to 0} \frac{\sqrt{2 x^{2} + 2 h^{2} + 4 x h + 1} - \sqrt{2 x^{2} + 1}}{h} \times \frac{\sqrt{2 x^{2} + 2 h^{2} + 4 x h + 1} + \sqrt{2 x^{2} + 1}}{\sqrt{2 x^{2} + 2 h^{2} + 4 x h + 1} + \sqrt{2 x^{2} + 1}} = \lim_{h \to 0} \frac{2 x^{2} + 2 h^{2} + 4 x h + 1 - 2 x^{2} - 1}{h (\sqrt{2 x^{2} + 2 h^{2} + 4 x h + 1} + \sqrt{2 x^{2} + 1})} = \lim_{h \to 0} \frac{h (2 h + 4 x)}{h (\sqrt{2 x^{2} + 2 h^{2} + 4 x h + 1} + \sqrt{2 x^{2} + 1})} = \lim_{h \to 0} \frac{(2 h + 4 x)}{(\sqrt{2 x^{2} + 2 h^{2} + 4 x h + 1} + \sqrt{2 x^{2} + 1})} = \frac{4 x}{\sqrt{2 x^{2} + 1} + \sqrt{2 x^{2} + 1}} = \frac{4 x}{2 \sqrt{2 x^{2} + 1}} = \frac{2 x}{\sqrt{2 x^{2} + 1}}$

$(xv) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\frac{2 (x + h) + 3}{x + h - 2} - \frac{2 x + 3}{x - 2}}{h} = \lim_{h \to 0} \frac{(2 x + 2 h + 3) (x - 2) - (x + h - 2) (2 x + 3)}{h (x + h - 2) (x - 2)} = \lim_{h \to 0} \frac{2 x^{2} + 2 x h + 3 x - 4 x - 4 h - 6 - 2 x^{2} - 2 x h + 4 x - 3 x - 3 h + 6}{h (x + h - 2) (x - 2)} = \lim_{h \to 0} \frac{- 7 h}{h (x + h - 2) (x - 2)} = \lim_{h \to 0} \frac{- 7}{(x + h - 2) (x - 2)} = \frac{- 7}{(x - 2) (x - 2)} = \frac{- 7}{{(x - 2)}^{2}}$

Page No 30.25:

Question 2:

Differentiate each of the following from first principles:

(i) e⁻^x

(ii) e^3x

(iii) e^ax^+ b

(iv) x e^x

(v) − x

(vi) (−x)⁻¹

(vii) sin (x + 1)

(viii) $\cos (x - \frac{π}{8})$

(ix) x sin x

(x) x cos x

(xi) sin (2x − 3)

Answer:

$(i) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (e^{x}) = \lim_{h \to 0} \frac{e^{- (x + h)} - e^{- x}}{h} = \lim_{h \to 0} \frac{e^{- x} e^{- h} - e^{- x}}{h} = \lim_{h \to 0} \frac{e^{- x} (e^{- h} - 1)}{h} = - e^{- x} \lim_{h \to 0} \frac{e^{- h} - 1}{- h} = - e^{- x} (1) = - e^{- x}$

$(ii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (e^{3 x}) = \lim_{h \to 0} \frac{e^{3 (x + h)} - e^{3 x}}{h} = \lim_{h \to 0} \frac{e^{3 x} e^{3 h} - e^{3 x}}{h} = \lim_{h \to 0} \frac{e^{3 x} (e^{3 h} - 1)}{3 h} = 3 e^{3 x} \lim_{h \to 0} \frac{e^{3 h} - 1}{3 h} = 3 e^{3 x} (1) = 3 e^{3 x}$

$(iii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (e^{a x + b}) = \lim_{h \to 0} \frac{e^{a (x + h) + b} - e^{a x + b}}{h} = \lim_{h \to 0} \frac{e^{a x + b} e^{a h} - e^{a x + b}}{h} = \lim_{h \to 0} \frac{e^{a x + b} (e^{a h} - 1)}{h} = a e^{a x + b} \lim_{h \to 0} \frac{e^{a h} - 1}{a h} = a e^{a x + b} (1) = a e^{a x + b}$

$(iv) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (x e^{x}) = \lim_{h \to 0} \frac{(x + h) e^{(x + h)} - x e^{x}}{h} = \lim_{h \to 0} \frac{(x + h) e^{x} e^{h} - x e^{x}}{h} = \lim_{h \to 0} \frac{x e^{x} e^{h} + h e^{x} e^{h} - x e^{x}}{h} = \lim_{h \to 0} \frac{x e^{x} e^{h} - x e^{x}}{h} + \lim_{h \to 0} \frac{h e^{x} e^{h}}{h} = \lim_{h \to 0} \frac{x e^{x} (e^{h} - 1)}{h} + \lim_{h \to 0} e^{x} e^{h} = x e^{x} (1) + e^{x} (e^{0}) = x e^{x} + e^{x}$

$(v) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (- x) = \lim_{h \to 0} \frac{- (x + h) - (- x)}{h} = \lim_{h \to 0} \frac{- x - h + x}{h} = \lim_{h \to 0} \frac{- h}{h} = \lim_{h \to 0} - 1 = - 1$

$(vi) {(- x)}^{- 1} = \frac{1}{- x} \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (\frac{1}{- x}) = \lim_{h \to 0} \frac{\frac{1}{- (x + h)} - \frac{1}{- x}}{h} = \lim_{h \to 0} \frac{\frac{- 1}{x + h} + \frac{1}{x}}{h} = \lim_{h \to 0} \frac{- x + x + h}{h x (x + h)} = \lim_{h \to 0} \frac{h}{h x (x + h)} = \lim_{h \to 0} \frac{1}{x (x + h)} = \frac{1}{x . x} = \frac{1}{x^{2}}$

$(vii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (\sin (x + 1)) = \lim_{h \to 0} \frac{\sin (x + h + 1) - \sin (x + 1)}{h} We know: \sin C - \sin D = 2 \cos (\frac{C + D}{2}) \sin (\frac{C - D}{2}) = \lim_{h \to 0} \frac{2 \cos (\frac{x + h + 1 + x + 1}{2}) \sin (\frac{x + h + 1 - x - 1}{2})}{h} = \lim_{h \to 0} \frac{2 \cos (\frac{2 x + h + 2}{2}) \sin (\frac{h}{2})}{h} = 2 \lim_{h \to 0} \cos (\frac{2 x + h + 2}{2}) \lim_{h \to 0} \frac{\sin (\frac{h}{2})}{\frac{h}{2}} \times \frac{1}{2} = 2 \cos (x + 1) \times \frac{1}{2} = \cos (x + 1)$

$(viii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (\cos (x - \frac{π}{8})) = \lim_{h \to 0} \frac{\cos (x + h - \frac{π}{8}) - \cos (x - \frac{π}{8})}{h} We know: \cos C - \cos D = - 2 \sin (\frac{C + D}{2}) \sin (\frac{C - D}{2}) = \lim_{h \to 0} \frac{- 2 \sin (\frac{x + h - \frac{π}{8} + x - \frac{π}{8}}{2}) \sin (\frac{x + h - \frac{π}{8} - x + \frac{π}{8}}{2})}{h} = \lim_{h \to 0} \frac{- 2 \sin (\frac{2 x + h - \frac{π}{4}}{2}) \sin (\frac{h}{2})}{h} = - 2 \lim_{h \to 0} \sin (\frac{2 x + h - \frac{π}{4}}{2}) \lim_{h \to 0} \frac{\sin (\frac{h}{2})}{\frac{h}{2}} \times \frac{1}{2} = - 2 \sin (x - \frac{π}{8}) \times \frac{1}{2} = - \sin (x - \frac{π}{8})$

$(ix) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{(x + h) \sin (x + h) - x \sin x}{h} = \lim_{h \to 0} \frac{(x + h) (\sin x \cos h + \cos x \sin h) - x \sin x}{h} = \lim_{h \to 0} \frac{x \sin x \cos h + x \cos x \sin h + h \sin x \cos h + h \cos x \sin h}{h} = \lim_{h \to 0} \frac{x \sin x \cos h - x \sin x + x \cos x \sin h + h \sin x \cos h + h \cos x \sin h - x \sin x}{h} = x \sin x \lim_{h \to 0} \frac{(\cos h - 1)}{h} + x \cos x \lim_{h \to 0} \frac{\sin h}{h} + \sin x \lim_{h \to 0} \cos h + \cos x \lim_{h \to 0} \sin h = x \sin x \lim_{h \to 0} \frac{- 2 \sin^{2} \frac{h}{2}}{\frac{h^{2}}{4}} \times \frac{h}{4} + x \cos x (1) + \sin x (1) + \cos x (0) = x \sin x \times \frac{- h}{2} + x \cos x (1) + \sin x (1) + \cos x (0) = - 2 x \sin x (\frac{1}{2}) (0) + x \cos x + \sin x = x \cos x + \sin x$

$(x) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{(x + h) \cos (x + h) - x \cos x}{h} = \lim_{h \to 0} \frac{(x + h) (\cos x \cos h - \sin x \sin h) - x \cos x}{h} = \lim_{h \to 0} \frac{x \cos x \cos h - x \sin x \sin h + h \cos x \cos h - h \sin x \sin h - x \cos x}{h} = \lim_{h \to 0} \frac{x \cos x \cos h - x \cos x - x \sin x \sin h + h \cos x \cos h - h \sin x \sin h}{h} = x \cos x \lim_{h \to 0} \frac{(\cos h - 1)}{h} - x \sin x \lim_{h \to 0} \frac{\sin h}{h} + \cos x \lim_{h \to 0} \cos h + \sin x \lim_{h \to 0} \sin h = x \cos x \lim_{h \to 0} \frac{- 2 \sin^{2} \frac{h}{2}}{\frac{h^{2}}{4}} \times \frac{h}{4} - x \sin x (1) + \cos x (1) + \sin x (0) = x \cos x \lim_{h \to 0} \frac{- h}{2} - x \sin x (1) + \cos x (1) + \sin x (0) = - x \cos x (0) - x \sin x + \cos x = - x \sin x + \cos x$

$(xi) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\sin (2 x + 2 h - 3) - \sin (2 x - 3)}{h} We know: sin C-sin D=2 cos (\frac{C + D}{2}) sin (\frac{C - D}{2}) = \lim_{h \to 0} \frac{2 \cos (\frac{2 x + 2 h - 3 + 2 x - 3}{2}) \sin (\frac{2 x + 2 h - 3 + 2 x - 3}{2})}{h} = \lim_{h \to 0} \frac{2 \cos (\frac{4 x + 2 h - 6}{2}) \sin (h)}{h} = \lim_{h \to 0} 2 \cos (\frac{4 x + 2 h - 6}{2}) \lim_{h \to 0} \frac{\sin h}{h} = 2 \cos (\frac{4 x - 6}{2}) (1) = 2 \cos (2 x - 3)$

Page No 30.26:

Question 3:

Differentiate each of the following from first principles:

(i) $\sqrt{\sin 2 x}$

(ii) $\frac{\sin x}{x}$

(iii) $\frac{\cos x}{x}$

(iv) x² sin x

(v) $\sqrt{\sin (3 x + 1)}$

(vi) sin x + cos x

(vii) x² e^x

(viii) $e^{x^{2} + 1}$

(ix) $e^{\sqrt{2 x}}$

(x) $e^{\sqrt{a x + b}}$

(xi) $a^{\sqrt{x}}$

(x) $3^{x^{2}}$

Answer:

$(i) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\sqrt{\sin (2 x + 2 h)} - \sqrt{\sin 2 x}}{h} \times \frac{\sqrt{\sin (2 x + 2 h)} + \sqrt{\sin 2 x}}{\sqrt{\sin (2 x + 2 h)} + \sqrt{\sin 2 x}} = \lim_{h \to 0} \frac{\sin (2 x + 2 h) - \sin 2 x}{h (\sqrt{\sin (2 x + 2 h)} + \sqrt{\sin 2 x})} We have: sin C-sin D= 2 cos (\frac{C + D}{2}) sin (\frac{C - D}{2}) = \lim_{h \to 0} \frac{2 \cos (\frac{2 x + 2 h + 2 x}{2}) \sin (\frac{2 x + 2 h - 2 x}{2})}{h (\sqrt{\sin (2 x + 2 h)} + \sqrt{\sin 2 x})} = \lim_{h \to 0} \frac{2 \cos (2 x + h) \sin h}{h (\sqrt{\sin (2 x + 2 h)} + \sqrt{\sin 2 x})} = \lim_{h \to 0} 2 \cos (2 x + h) \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{(\sqrt{\sin (2 x + 2 h)} + \sqrt{\sin 2 x})} = 2 \cos 2 x (1) \frac{1}{\sqrt{\sin 2 x} + \sqrt{\sin 2 x}} = \frac{2 \cos 2 x}{2 \sqrt{\sin 2 x}} = \frac{\cos 2 x}{\sqrt{\sin 2 x}}$

$(i i) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\frac{\sin (x + h)}{x + h} - \frac{\sin x}{x}}{h} = \lim_{h \to 0} \frac{x \sin (x + h) - (x + h) \sin x}{h x (x + h)} = \lim_{h \to 0} \frac{x (\sin x \cos h + \cos x \sin h) - x \sin x - h \sin x}{h x (x + h)} = \lim_{h \to 0} \frac{x \sin x \cos h + x \cos x \sin h - x \sin x - h \sin x}{h x (x + h)} = \lim_{h \to 0} \frac{x \sin x \cos h - x \sin x + x \cos x \sin h - h \sin x}{h x (x + h)} = x \sin x \lim_{h \to 0} \frac{\cos h - 1}{h} + \frac{x \cos x}{x} \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{x + h} - \frac{\sin x}{x} \lim_{h \to 0} \frac{1}{x + h} = x \sin x \lim_{h \to 0} \frac{- 2 \sin^{2} \frac{h}{2}}{h} + \frac{x \cos x}{x} \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{x + h} - \frac{\sin x}{x} \lim_{h \to 0} \frac{1}{x + h} = x \sin x \lim_{h \to 0} \frac{- 2 \sin^{2} \frac{h}{2}}{\frac{h^{2}}{4}} \times \frac{h}{4} + \frac{x \cos x}{x} \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{x + h} - \frac{\sin x}{x} \lim_{h \to 0} \frac{1}{x + h} = - x \sin x \times \lim_{h \to 0} \frac{h}{2} + \frac{x \cos x}{x} \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{x + h} - \frac{\sin x}{x} \lim_{h \to 0} \frac{1}{x + h} = - x \sin x (\frac{1}{2}) (0) + \frac{c o s x}{x} - \frac{s i n x}{x^{2}} = \frac{\cos x}{x} - \frac{\sin x}{x^{2}} = \frac{x \cos x - \sin x}{x^{2}}$

$(i i i) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\frac{\cos (x + h)}{x + h} - \frac{\cos x}{x}}{h} = \lim_{h \to 0} \frac{x \cos (x + h) - (x + h) \cos x}{h x (x + h)} = \lim_{h \to 0} \frac{x (\cos x \cos h - \sin x \sin h) - x \cos x - h \cos x}{h x (x + h)} = \lim_{h \to 0} \frac{x \cos x \cos h - x \sin x \sin h - x \cos x - h \cos x}{h x (x + h)} = \lim_{h \to 0} \frac{x \cos x \cos h - x \cos x - x \sin x \sin h - h \cos x}{h x (x + h)} = x \cos x \lim_{h \to 0} \frac{\cos h - 1}{h} - \frac{x \sin x}{x} \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{x + h} - \frac{\cos x}{x} \lim_{h \to 0} \frac{1}{x + h} = x \cos x \lim_{h \to 0} \frac{- 2 \sin^{2} \frac{h}{2}}{\frac{h^{2}}{4}} \times \frac{h}{4} - \frac{x \sin x}{x} \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{x + h} - \frac{\cos x}{x} \lim_{h \to 0} \frac{1}{x + h} [∵ \lim_{h \to 0} \frac{\sin^{2} \frac{h}{2}}{\frac{h^{2}}{4}} = \lim_{h \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \times \lim_{h \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} = 1 \times 1, i . e . 1] = - x \cos x \lim_{h \to 0} \frac{h}{2} - \frac{x \sin x}{x} \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{x + h} - \frac{\cos x}{x} \lim_{h \to 0} \frac{1}{x + h} = - x \cos x \times 0 - \sin x (1) \frac{1}{x} - \frac{\cos x}{x} \frac{1}{x} = 0 - \frac{\sin x}{x} - \frac{\cos x}{x^{2}} = - \frac{\sin x}{x} - \frac{\cos x}{x^{2}} = \frac{- x \sin x - \cos x}{x^{2}}$

$(i v) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{{(x + h)}^{2} \sin (x + h) - x^{2} \sin x}{h} = \lim_{h \to 0} \frac{(x^{2} + h^{2} + 2 x h) (\sin x \cos h + \cos x \sin h) - x^{2} \sin x}{h} = \lim_{h \to 0} \frac{x^{2} \sin x \cos h + x^{2} \cos x \sin h + h^{2} \sin x \cos h + h^{2} \cos x \sin h + 2 x h \sin x \cos h + 2 x h \cos x \sin h - x^{2} \sin x}{h} = \lim_{h \to 0} \frac{x^{2} \sin x \cos h - x^{2} \sin x + x^{2} \cos x \sin h + h^{2} \sin x \cos h + h^{2} \cos x \sin h + 2 x h \sin x \cos h + 2 x h \cos x \sin h}{h} = x^{2} \sin x \lim_{h \to 0} \frac{\cos h - 1}{h} + x^{2} \cos x \lim_{h \to 0} \frac{\sin h}{h} + \sin x \lim_{h \to 0} h \cos h + \cos x \lim_{h \to 0} h \sin h + 2 x \sin x \lim_{h \to 0} \cos h + 2 x \cos x \lim_{h \to 0} \sin h = x^{2} \sin x \lim_{h \to 0} \frac{- 2 \sin^{2} \frac{h}{2}}{\frac{h^{2}}{4}} \times \frac{h}{4} + x^{2} \cos x \lim_{h \to 0} \frac{\sin h}{h} + \sin x \lim_{h \to 0} h \cos h + \cos x \lim_{h \to 0} h \sin h + 2 x \sin x \lim_{h \to 0} \cos h + 2 x \cos x \lim_{h \to 0} \sin h [∵ \lim_{h \to 0} \frac{\sin^{2} \frac{h}{2}}{\frac{h^{2}}{4}} = \lim_{h \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \times \lim_{h \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} = 1 \times 1, i . e . 1] = - x^{2} \sin x \times \lim_{h \to 0} \frac{h}{2} + x^{2} \cos x \lim_{h \to 0} \frac{\sin h}{h} + \sin x \lim_{h \to 0} h \cos h + \cos x \lim_{h \to 0} h \sin h + 2 x \sin x \lim_{h \to 0} \cos h + 2 x \cos x \lim_{h \to 0} \sin h = - x^{2} \sin x \times 0 + x^{2} \cos x (1) + \sin x (0) + \cos x (0) + 2 x \sin x (1) + 2 x \cos x (0) = 0 + x^{2} \cos x + 2 x \sin x = 0 + x^{2} \cos x + 2 x \sin x = x^{2} \cos x + 2 x \sin x$

$(v) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\sqrt{\sin (3 (x + h) + 1)} - \sqrt{\sin (3 x + 1)}}{h} = \lim_{h \to 0} \frac{\sqrt{\sin (3 x + 3 h + 1)} - \sqrt{\sin (3 x + 1)}}{h} \times \frac{\sqrt{\sin (3 x + 3 h + 1)} + \sqrt{\sin (3 x + 1)}}{\sqrt{\sin (3 x + 3 h + 1)} + \sqrt{\sin (3 x + 1)}} = \lim_{h \to 0} \frac{\sin (3 x + 3 h + 1) - \sin (3 x + 1)}{h (\sqrt{\sin (3 x + 3 h + 1)} + \sqrt{\sin (3 x + 1)})} We have: sin C-sin D= 2 cos (\frac{C + D}{2}) sin (\frac{C - D}{2}) = \lim_{h \to 0} \frac{2 \cos (\frac{3 x + 3 h + 1 + 3 x + 1}{2}) \sin (\frac{3 x + 3 h + 1 - 3 x - 1}{2})}{h (\sqrt{\sin (3 x + 3 h + 1)} + \sqrt{\sin (3 x + 1)})} = \lim_{h \to 0} \frac{2 \cos (\frac{6 x + 3 h + 2}{2}) \sin \frac{3 h}{2}}{h (\sqrt{\sin (3 x + 3 h + 1)} + \sqrt{\sin (3 x + 1)})} = \lim_{h \to 0} 2 \cos (\frac{6 x + 3 h + 2}{2}) \lim_{h \to 0} \frac{\sin \frac{3 h}{2}}{h \times \frac{3}{2}} \times \frac{3}{2} \times \lim_{h \to 0} \frac{1}{(\sqrt{\sin (3 x + 3 h + 1)} + \sqrt{\sin (3 x + 1)})} = 2 \cos (3 x + 1) \times (\frac{3}{2}) \times \frac{1}{\sqrt{\sin (3 x + 1)} + \sqrt{\sin (3 x + 1)}} = \frac{3 \cos (3 x + 1)}{2 \sqrt{\sin (3 x + 1)}}$

$(v i) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\sin (x + h) + co s (x + h) - \sin x - \cos x}{h} = \lim_{h \to 0} \frac{\sin (x + h) - \sin x}{h} + \lim_{h \to 0} \frac{\cos (x + h) - \cos x}{h} We have: sin C-sin D= 2 cos (\frac{C + D}{2}) sin (\frac{C - D}{2}) And, cos C- \cos D = - 2 sin (\frac{C + D}{2}) sin (\frac{C - D}{2}) = \lim_{h \to 0} \frac{2 \cos (\frac{2 x + h}{2}) \sin \frac{h}{2}}{h} + \lim_{h \to 0} \frac{- 2 \sin (\frac{2 x + h}{2}) \sin \frac{h}{2}}{h} = 2 \lim_{h \to 0} \cos (\frac{2 x + h}{2}) \lim_{h \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \times \frac{1}{2} - 2 \lim_{h \to 0} \sin (\frac{2 x + h}{2}) \lim_{h \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \times \frac{1}{2} = 2 \cos x \times \frac{1}{2} - 2 \sin x \times \frac{1}{2} = \cos x - \sin x$

$(vii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (x^{2} e^{x}) = \lim_{h \to 0} \frac{(x + h)^{2} e^{(x + h)} - x^{2} e^{x}}{h} = \lim_{h \to 0} \frac{(x^{2} + 2 x h + h^{2}) e^{x} e^{h} - x^{2} e^{x}}{h} = \lim_{h \to 0} \frac{x^{2} e^{x} e^{h} + 2 x h e^{x} e^{h} + h^{2} e^{x} e^{h} - x^{2} e^{x}}{h} = \lim_{h \to 0} \frac{x^{2} e^{x} e^{h} - x^{2} e^{x}}{h} + \lim_{h \to 0} \frac{2 x h e^{x} e^{h}}{h} + \lim_{h \to 0} \frac{h^{2} e^{x} e^{h}}{h} = \lim_{h \to 0} \frac{x^{2} e^{x} (e^{h} - 1)}{h} + \lim_{h \to 0} 2 x e^{x} e^{h} + \lim_{h \to 0} h e^{x} e^{h} = x^{2} e^{x} (1) + 2 x e^{x} (1) + 0 = x^{2} e^{x} + 2 x e^{x} = (x^{2} + 2 x) e^{x}$

$(viii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (e^{x^{2} + 1}) = \lim_{h \to 0} \frac{e^{(x + h)^{2} + 1} - e^{x^{2} + 1}}{h} = \lim_{h \to 0} \frac{e^{x^{2} + h^{2} + 2 x h + 1} - e^{x^{2} + 1}}{h} = \lim_{h \to 0} \frac{e^{x^{2} + 1} e^{h^{2} + 2 x h} - e^{x^{2} + 1}}{h} = \lim_{h \to 0} \frac{e^{x^{2} + 1} (e^{h (h + 2 x)} - 1)}{h} \times \frac{(h + 2 x)}{(h + 2 x)} = e^{x^{2} + 1} \lim_{h \to 0} \frac{e^{h (h + 2 x)} - 1}{h (h + 2 x)} \lim_{h \to 0} (h + 2 x) = e^{x^{2} + 1} (1) (2 x) = 2 x e^{x^{2} + 1}$

$(ix) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (e^{\sqrt{2 x}}) = \lim_{h \to 0} \frac{e^{\sqrt{2 (x + h)}} - e^{\sqrt{2 x}}}{h} = 2 \lim_{h \to 0} \frac{e^{\sqrt{2 x + 2 h}} - e^{\sqrt{2 x}}}{2 x + 2 h - 2 x} = 2 \lim_{h \to 0} \frac{e^{\sqrt{2 x}} (e^{\sqrt{2 x + 2 h} - \sqrt{2 x}} - 1)}{{(\sqrt{2 x + 2 h})}^{2} - {(\sqrt{2 x})}^{2}} = 2 e^{\sqrt{2 x}} \lim_{h \to 0} \frac{e^{\sqrt{2 x + 2 h} - \sqrt{2 x}} - 1}{(\sqrt{2 x + 2 h} - \sqrt{2 x}) (\sqrt{2 x + 2 h} + \sqrt{2 x})} = 2 e^{\sqrt{2 x}} \lim_{h \to 0} \frac{e^{\sqrt{2 x + 2 h} - \sqrt{2 x}} - 1}{(\sqrt{2 x + 2 h} - \sqrt{2 x})} \lim_{h \to 0} \frac{1}{(\sqrt{2 x + 2 h} + \sqrt{2 x})} = 2 e^{\sqrt{2 x}} (1) \frac{1}{2 \sqrt{2 x}} = \frac{e^{\sqrt{2 x}}}{\sqrt{2 x}}$

$(x) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (e^{\sqrt{a x + b}}) = \lim_{h \to 0} \frac{e^{\sqrt{a x + a h + b}} - e^{\sqrt{a x + b}}}{h} = a \lim_{h \to 0} \frac{e^{\sqrt{a x + a h + b}} - e^{\sqrt{a x + b}}}{(a x + a h + b) - (a x + b)} = a \lim_{h \to 0} \frac{e^{\sqrt{a x + b}} (e^{\sqrt{a x + a h + b} - \sqrt{a x + b}} - 1)}{{(\sqrt{(a x + a h + b)})}^{2} - {(\sqrt{(a x + b)})}^{2}} = a e^{\sqrt{a x + b}} \lim_{h \to 0} \frac{e^{\sqrt{a x + a h + b} - \sqrt{a x + b}} - 1}{(\sqrt{(a x + a h + b)} - \sqrt{(a x + b)}) (\sqrt{(a x + a h + b)} + \sqrt{(a x + b)})} = a e^{\sqrt{a x + b}} \lim_{h \to 0} \frac{e^{\sqrt{a x + a h + b} - \sqrt{a x + b}} - 1}{\sqrt{(a x + a h + b)} - \sqrt{(a x + b)}} \lim_{h \to 0} \frac{1}{\sqrt{(a x + a h + b)} + \sqrt{(a x + b)}} = a e^{^{\sqrt{a x + b}}} (1) \frac{1}{2 \sqrt{a x + b}} = \frac{a e^{^{\sqrt{a x + b}}}}{2 \sqrt{a x + b}}$

$(xi) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (a^{\sqrt{x}}) = \lim_{h \to 0} \frac{a^{\sqrt{x + h}} - a^{\sqrt{x}}}{h} = \lim_{h \to 0} \frac{a^{\sqrt{x}} (a^{\sqrt{x + h} - \sqrt{x}} - 1)}{(x + h) - (x)} = a^{\sqrt{x}} \lim_{h \to 0} \frac{(a^{\sqrt{x + h} - \sqrt{x}} - 1)}{{(\sqrt{x + h})}^{2} - {(\sqrt{x})}^{2}} = a^{\sqrt{x}} \lim_{h \to 0} \frac{(a^{\sqrt{x + h} - \sqrt{x}} - 1)}{(\sqrt{x + h} - \sqrt{x}) (\sqrt{x + h} + \sqrt{x})} = a^{\sqrt{x}} \lim_{h \to 0} \frac{(a^{\sqrt{x + h} - \sqrt{x}} - 1)}{(\sqrt{x + h} - \sqrt{x})} \lim_{h \to 0} \frac{1}{(\sqrt{x + h} + \sqrt{x})} = a^{\sqrt{x}} \log_{e} a \frac{1}{2 \sqrt{x}} = \frac{1}{2 \sqrt{x}} a^{\sqrt{x}} \log_{e} a$

$(xii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (3^{x^{2}}) = \lim_{h \to 0} \frac{3^{{(x + h)}^{2}} - 3^{x^{2}}}{h} = \lim_{h \to 0} \frac{3^{x^{2} + 2 x h + h^{2}} - 3^{x^{2}}}{h} = \lim_{h \to 0} \frac{3^{x^{2}} (3^{x^{2} + 2 x h + h^{2} - x^{2}} - 1)}{h} \times \frac{(h + 2 x)}{(h + 2 x)} = 3^{x^{2}} \lim_{h \to 0} \frac{3^{h (h + 2 x)} - 1}{h (h + 2 x)} \lim_{h \to 0} (h + 2 x) = 3^{x^{2}} \log 3 (2 x) = 2 x 3^{x^{2}} \log 3$

Page No 30.26:

Question 4:

(i) tan² x

(ii) tan (2x + 1)

(iii) tan 2x

(iv) $\sqrt{\tan x}$

Answer:

$(i) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\tan^{2} (x + h) - \tan^{2} x}{h} = \lim_{h \to 0} \frac{[\tan (x + h) + \tan x] [\tan (x + h) - \tan x]}{h} = \lim_{h \to 0} \frac{[\frac{\sin (x + h)}{\cos (x + h)} + \frac{\sin x}{\cos x}] [\frac{\sin (x + h)}{\cos (x + h)} - \frac{\sin x}{\cos x}]}{h} = \lim_{h \to 0} \frac{[\sin (x + h) \cos x + \cos (x + h) \sin x] [\sin (x + h) \cos x - \cos (x + h) \sin x]}{h \cos^{2} x \cos^{2} (x + h)} = \lim_{h \to 0} \frac{[\sin (2 x + h)] [\sin h]}{h \cos^{2} x \cos^{2} (x + h)} = \frac{1}{\cos^{2} x} \lim_{h \to 0} \sin (2 x + h) \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{\cos^{2} (x + h)} = \frac{1}{\cos^{2} x} \sin (2 x) (1) \frac{1}{\cos^{2} x} = \frac{1}{\cos^{2} x} 2 \sin x \cos x \frac{1}{\cos^{2} x} = 2 \times \frac{\sin x}{\cos x} \times \frac{1}{\cos^{2} x} = 2 \tan x \sec^{2} x$

$(ii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\tan (2 x + 2 h + 1) - \tan (2 x + 1)}{h} = \lim_{h \to 0} \frac{\frac{si n (2 x + 2 h + 1)}{\cos (2 x + 2 h + 1)} - \frac{\sin (2 x + 1)}{\cos (2 x + 1)}}{h} = \lim_{h \to 0} \frac{si n (2 x + 2 h + 1) \cos (2 x + 1) - \cos (2 x + 2 h + 1) \sin (2 x + 1)}{h \cos (2 x + 2 h + 1) \cos (2 x + 1)} = \lim_{h \to 0} \frac{\sin (2 x + 2 h + 1 - 2 x - 1)}{h \cos (2 x + 2 h + 1) \cos (2 x + 1)} = \frac{1}{\cos (2 x + 1)} \lim_{h \to 0} \frac{\sin (2 h)}{2 h} \times 2 \lim_{h \to 0} \frac{1}{\cos (2 x + 2 h + 1)} = \frac{1}{\cos (2 x + 1)} \times 2 \times \frac{1}{\cos (2 x + 1)} = \frac{2}{\cos^{2} (2 x + 1)} = 2 \sec^{2} (2 x + 1)$

$(iii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\tan (2 x + 2 h) - \tan (2 x)}{h} = \lim_{h \to 0} \frac{\frac{si n (2 x + 2 h)}{\cos (2 x + 2 h)} - \frac{\sin (2 x)}{\cos (2 x)}}{h} = \lim_{h \to 0} \frac{si n (2 x + 2 h) \cos (2 x) - \cos (2 x + 2 h) \sin (2 x)}{h \cos (2 x + 2 h) \cos (2 x)} = \lim_{h \to 0} \frac{\sin (2 x + 2 h - 2 x)}{h \cos (2 x + 2 h) \cos (2 x)} = \frac{1}{\cos 2 x} \lim_{h \to 0} \frac{\sin (2 h)}{2 h} \times 2 \times \lim_{h \to 0} \frac{1}{\cos (2 x + 2 h)} = \frac{1}{\cos 2 x} \times 2 \times \frac{1}{\cos 2 x} = \frac{2}{\cos^{2} (2 x)} = 2 \sec^{2} (2 x)$

$(iv) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\sqrt{\tan (x + h)} - \sqrt{\tan x}}{h} \times \frac{\sqrt{\tan (x + h)} + \sqrt{\tan x}}{\sqrt{\tan (x + h)} + \sqrt{\tan x}} = \lim_{h \to 0} \frac{\tan (x + h) - \tan x}{h (\sqrt{\tan (x + h)} + \sqrt{\tan x})} = \lim_{h \to 0} \frac{\frac{\sin (x + h)}{\cos (x + h)} - \frac{\sin x}{\cos x}}{h (\sqrt{\tan (x + h)} + \sqrt{\tan x})} = \lim_{h \to 0} \frac{\sin (x + h) \cos x - \cos (x + h) \sin x}{h (\sqrt{\tan (x + h)} + \sqrt{\tan x}) \cos (x + h) \cos x} = \lim_{h \to 0} \frac{\sin h}{h (\sqrt{\tan (x + h)} + \sqrt{\tan x}) \cos (x + h) \cos x} = \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{(\sqrt{\tan (x + h)} + \sqrt{\tan x}) \cos (x + h) \cos x} = (1) \frac{1}{2 \sqrt{\tan x} \cos^{2} x} = \frac{\sec^{2} x}{2 \sqrt{\tan x}}$

Page No 30.26:

Question 5:

(i) $\sin \sqrt{2 x}$
(ii) $\cos \sqrt{x}$
(iii) $\tan \sqrt{x}$
(iv) tan x²

Answer:

$(i) Let f (x) = \sin \sqrt{2 x} Thus, we have : f (x + h) = \sin \sqrt{2 (x + h)} \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\sin \sqrt{2 x + 2 h} - \sin \sqrt{2 x}}{h} We know: sin C- sin D=2 sin (\frac{C - D}{2}) cos (\frac{C + D}{2}) = \lim_{h \to 0} \frac{2 \sin (\sqrt{2 x + 2 h} - \sqrt{2 x}) \cos (\sqrt{2 x + 2 h} - \sqrt{2 x})}{h} = \lim_{h \to 0} \frac{2 \times 2 \sin (\frac{\sqrt{2 x + 2 h} - \sqrt{2 x}}{2}) \cos (\frac{\sqrt{2 x + 2 h} + \sqrt{2 x}}{2})}{2 h + 2 x - 2 x} = \lim_{h \to 0} \frac{2 \times 2 \sin (\frac{\sqrt{2 x + 2 h} - \sqrt{2 x}}{2}) \cos (\frac{\sqrt{2 x + 2 h} - \sqrt{2 x}}{2})}{(\sqrt{2 x + 2 h} - \sqrt{2 x}) \sqrt{2 x + 2 h} + \sqrt{2 x}} = \lim_{h \to 0} \frac{2 \times 2 \sin (\frac{\sqrt{2 x + 2 h} - \sqrt{2 x}}{2}) \cos (\frac{\sqrt{2 x + 2 h} - \sqrt{2 x}}{2})}{2 \times (\frac{\sqrt{2 x + 2 h} - \sqrt{2 x}}{2}) (\sqrt{2 x + 2 h} + \sqrt{2 x})} = \lim_{h \to 0} \frac{\sin (\frac{\sqrt{2 x + 2 h} - \sqrt{2 x}}{2})}{(\frac{\sqrt{2 x + 2 h} - \sqrt{2 x}}{2})} \lim_{h \to 0} \frac{2 \cos (\frac{\sqrt{2 x + 2 h} - \sqrt{2 x}}{2})}{\sqrt{2 x + 2 h} + \sqrt{2 x}} = 1 \times \frac{2 \cos \sqrt{2 x}}{2 \sqrt{2 x}} [∵ \lim_{h \to 0} \frac{\sin (\frac{\sqrt{2 x + 2 h} - \sqrt{2 x}}{2})}{(\frac{\sqrt{2 x + 2 h} - \sqrt{2 x}}{2})} = 1] = \frac{\cos \sqrt{2 x}}{\sqrt{2 x}}$

$(ii) L e t f (x) = \cos \sqrt{x} Thus, we have : f (x + h) = \cos \sqrt{x + h} \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\cos \sqrt{x + h} - \cos \sqrt{x}}{h} We know : \cos C - \cos D = - 2 \sin (\frac{C + D}{2}) \sin (\frac{C - D}{2}) = \lim_{h \to 0} \frac{- 2 \sin (\frac{\sqrt{x + h} + \sqrt{x}}{2}) \sin (\frac{\sqrt{x + h} - \sqrt{x}}{2})}{h} = \lim_{h \to 0} \frac{- 2 \sin (\frac{\sqrt{x + h} + \sqrt{x}}{2}) \sin (\frac{\sqrt{x + h} - \sqrt{x}}{2})}{x + h - x} = \lim_{h \to 0} \frac{- 2 \sin (\frac{\sqrt{x + h} + \sqrt{x}}{2}) \sin (\frac{\sqrt{x + h} - \sqrt{x}}{2})}{2 \times (\sqrt{x + h} + \sqrt{x}) \frac{(\sqrt{x + h} - \sqrt{x})}{2}} = \lim_{h \to 0} \frac{\sin (\frac{\sqrt{x + h} - \sqrt{x}}{2})}{\frac{\sqrt{x + h} - \sqrt{x}}{2}} \lim_{h \to 0} \frac{- \sin (\frac{\sqrt{x + h} + \sqrt{x}}{2})}{\sqrt{x + h} + \sqrt{x}} = 1 \times \frac{- \sin \sqrt{x}}{2 \sqrt{x}} [∵ \lim_{h \to 0} \frac{\sin (\frac{\sqrt{x + h} - \sqrt{x}}{2})}{\frac{\sqrt{x + h} - \sqrt{x}}{2}} = 1] = \frac{- \sin \sqrt{x}}{2 \sqrt{x}}$

$(iii) Let f (x) = \tan \sqrt{x} Thus, we have : (x + h) = \tan \sqrt{x + h} \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\tan \sqrt{x + h} - \tan \sqrt{x}}{h} = \lim_{h \to 0} \frac{\sin (\sqrt{x + h} - \sqrt{x})}{h \cos \sqrt{x + h} \cos \sqrt{x}} [∵ \tan A - \tan B = \frac{\sin (A - B)}{\cos A \cos B}] = \lim_{h \to 0} \frac{\sin (\sqrt{x + h} - \sqrt{x})}{(x + h - x) \cos \sqrt{x + h} \cos \sqrt{x}} = \lim_{h \to 0} \frac{\sin (\sqrt{x + h} - \sqrt{x})}{(\sqrt{x + h} - \sqrt{x}) (\sqrt{x + h} - \sqrt{x}) \cos \sqrt{x + h} \cos \sqrt{x}} = \lim_{h \to 0} \frac{\sin (\sqrt{x + h} - \sqrt{x})}{(\sqrt{x + h} - \sqrt{x})} . \lim_{h \to 0} \frac{1}{(\sqrt{x + h} + \sqrt{x}) \cos \sqrt{x + h} \cos \sqrt{x}} [∵ \lim_{h \to 0} \frac{\sin (\sqrt{x + h} - \sqrt{x})}{\sqrt{x + h} - \sqrt{x}} = 1] = 1 \times \frac{1}{2 \sqrt{x} \cos \sqrt{x} \cos \sqrt{x}} = \frac{1}{2 \sqrt{x}} \sec^{2} \sqrt{x}$

$(i v) Let f (x) = \tan x^{2} Thus, we have : f (x + h) = \tan (x + h)^{2} \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\tan (x + h)^{2} - \tan x^{2}}{h} = \lim_{h \to 0} \frac{\sin ({(x + h)}^{2} - x^{2})}{h \cos {(x + h)}^{2} \cos x^{2}} [∵ \tan A - \tan B = \frac{\sin (A - B)}{\cos A \cos B}] = \lim_{h \to 0} \frac{\sin (x^{2} + h^{2} + 2 h x - x^{2})}{h \cos {(x + h)}^{2} \cos x^{2}} = \lim_{h \to 0} \frac{\sin (h (h + 2 x))}{h (h + 2 x) \cos {(x + h)}^{2} \cos x^{2}} \times (h + 2 x) = \lim_{h \to 0} \frac{\sin (h (h + 2 x))}{(h (h + 2 x))} \lim_{h \to 0} \frac{h + 2 x}{\cos (x + h)^{2} \cos x^{2}} [As \lim_{h \to 0} \frac{\sin (h (h + 2 x))}{(h (h + 2 x))} = 1] = 1 \times \frac{2 x}{\cos^{2} x^{2}} = 2 x \sec^{2} x^{2}$

Page No 30.3:

Question 1:

Find the derivative of f (x) = 3x at x = 2

Answer:

We have:

$\underset{h \to 0}{f' (2) = \lim} \frac{f (2 + h) - f (2)}{h} = \lim_{h \to 0} \frac{3 (2 + h) - 3 (2)}{h} = \lim_{h \to 0} \frac{6 + 3 h - 6}{h} = \lim_{h \to 0} \frac{3 h}{h} = 3$

Page No 30.3:

Question 2:

Find the derivative of f (x) = x² − 2 at x = 10

Answer:

We have:
$f' (x) = \lim_{h \to 0} \frac{f (10 + h) - f (10)}{h} = \lim_{h \to 0} \frac{(10 + h)^{2} - 2 - (10^{2} - 2)}{h} = \lim_{h \to 0} \frac{100 + h^{2} + 20 h - 2 - 100 + 2}{h} = \lim_{h \to 0} \frac{h^{2} + 20 h}{h} = \lim_{h \to 0} \frac{h (h + 20)}{h} = \lim_{h \to 0} h + 20 = 0 + 20 = 20$

Page No 30.3:

Question 3:

Find the derivative of f (x) = 99x at x = 100

Answer:

We have:
$f' (100) = \lim_{h \to 0} \frac{f (100 + h) - f (100)}{h} = \lim_{h \to 0} \frac{99 (100 + h) - 99 (100)}{h} = \lim_{h \to 0} \frac{9900 + 99 h - 9900}{h} = \lim_{h \to 0} \frac{99 h}{h} = \lim_{h \to 0} 99 = 99$

Page No 30.3:

Question 4:

Find the derivative of f (x) x at x = 1

Answer:

We have:
$f' (x) = \lim_{h \to 0} \frac{f (1 + h) - f (1)}{h} = \lim_{h \to 0} \frac{1 + h - 1}{h} = \lim_{h \to 0} 1 = 1$

Page No 30.3:

Question 5:

Find the derivative of f (x) = cos x at x = 0

Answer:

We have:
$f' (x) = \lim_{h \to 0} \frac{f (0 + h) - f (0)}{h} = \lim_{h \to 0} \frac{f (h) - f (0)}{h} = \lim_{h \to 0} \frac{\cos h - \cos 0}{h} = \lim_{h \to 0} \frac{\cos h - 1}{h} \underset{h \to 0}{= \lim} \frac{- 2 \sin^{2} \frac{h}{2}}{h} \underset{}{= \lim_{h \to 0} \frac{- 2 \sin^{2} \frac{h}{2}}{\frac{h^{2}}{4}}} \times \frac{h}{4} = \underset{}{= \lim_{h \to 0} - 1} \times \frac{h}{2} = 0$

Page No 30.3:

Question 6:

Find the derivative of f (x) = tan x at x = 0

Answer:

We have:
$f' (x) = \lim_{h \to 0} \frac{f (0 + h) - f (0)}{h} = \lim_{h \to 0} \frac{f (h) - f (0)}{h} = \lim_{h \to 0} \frac{\tan h - \tan 0}{h} = \lim_{h \to 0} \frac{\tan h}{h} = 1$

Page No 30.3:

Question 7:

Find the derivatives of the following functions at the indicated points:

(i) sin x at x = $\frac{π}{2}$

(ii) x at x = 1

(iii) 2 cos x at x = $\frac{π}{2}$

(iv) sin 2x at x = $\frac{π}{2}$

Answer:

$(i) We have : f' (\frac{π}{2}) = \lim_{h \to 0} \frac{f (\frac{π}{2} + h) - f (\frac{π}{2})}{h} = \lim_{h \to 0} \frac{s i n (\frac{π}{2} + h) - s i n (\frac{π}{2})}{h} = \lim_{h \to 0} \frac{c o s h - 1}{h} \underset{h \to 0}{= \lim} \frac{- 2 \sin^{2} \frac{h}{2}}{h} \underset{}{= \lim_{h \to 0} \frac{- 2 \sin^{2} \frac{h}{2}}{\frac{h^{2}}{4}}} \times \frac{h}{4} \underset{}{= \lim_{h \to 0} - 1} \times \frac{h}{2} = 0$
$(ii) We have : f' (x) = \lim_{h \to 0} \frac{f (1 + h) - f (1)}{h} = \lim_{h \to 0} \frac{1 + h - 1}{h} = \lim_{h \to 0} 1 = 1$

$(iii) We have : f' (\frac{π}{2}) = \lim_{h \to 0} \frac{f (\frac{π}{2} + h) - f (\frac{π}{2})}{h} = \lim_{h \to 0} \frac{2 c o s (\frac{π}{2} + h) - c o s (\frac{π}{2})}{h} = \lim_{h \to 0} \frac{- 2 s i n h - 0}{h} = - 2 \lim_{h \to 0} \frac{\sin h}{h} = - 2 (1) = - 2$
$(iv) We have : f' (\frac{π}{2}) = \lim_{h \to 0} \frac{f (\frac{π}{2} + h) - f (\frac{π}{2})}{h} = \lim_{h \to 0} \frac{\sin 2 (\frac{π}{2} + h) - \sin 2 (\frac{π}{2})}{h} = \lim_{h \to 0} \frac{\sin (π + 2 h) - 0}{h} = \lim_{h \to 0} \frac{- \sin 2 h}{h} \times \frac{2}{2} = \lim_{h \to 0} - \frac{\sin 2 h}{2 h} \times 2 = - 2$

Page No 30.33:

Question 1:

x⁴ − 2 sin x + 3 cos x

Answer:

$\frac{d}{d x} (x^{4} - 2 \sin x + 3 \cos x) = \frac{d}{d x} (x^{4}) - 2 \frac{d}{d x} (\sin x) + 3 \frac{d}{d x} (\cos x) = 4 x^{4} - 2 \cos x - 3 \sin x$

Page No 30.33:

Question 2:

3^x + x³ + 3³

Answer:

$\frac{d}{d x} (3^{x} + x^{3} + 3^{3}) = \frac{d}{d x} (3^{x}) + \frac{d}{d x} (x^{3}) + \frac{d}{d x} (3^{3}) = 3^{x} \log 3 + 3 x^{2} + 0 = 3^{x} \log 3 + 3 x^{2}$

Page No 30.33:

Question 3:

$\frac{x^{3}}{3} - 2 \sqrt{x} + \frac{5}{x^{2}}$

Answer:

$\frac{d}{d x} (\frac{x^{3}}{3} - 2 \sqrt{x} + \frac{5}{x^{2}}) = \frac{1}{3} \frac{d}{d x} (x^{3}) - 2 \frac{d}{d x} (x^{\frac{1}{2}}) + 5 \frac{d}{d x} (x^{- 2}) = \frac{1}{3} (3 x^{2}) - 2 . \frac{1}{2} . x^{\frac{- 1}{2}} + 5 (- 2) x^{- 3} = x^{2} - x^{\frac{- 1}{2}} - 10 x^{- 3}$

Page No 30.33:

Question 4:

e^x^{log a} + e^a^{long x} + e^a^{log a}

Answer:

$\frac{d}{d x} (e^{x \log a} + e^{a \log x} + e^{a \log a}) = \frac{d}{d x} (e^{x \log a}) + \frac{d}{d x} (e^{a \log x}) + \frac{d}{d x} (e^{a \log a}) = \frac{d}{d x} (e^{\log a^{x}}) + \frac{d}{d x} ({e^{\log x}}^{a}) + \frac{d}{d x} (e^{\log a^{a}}) = \frac{d}{d x} (a^{x}) + \frac{d}{d x} (x^{a}) + \frac{d}{d x} (a^{a}) = a^{x} \log a + a x^{a - 1} + 0 = a^{x} \log a + a x^{a - 1}$

Page No 30.33:

Question 5:

(2x² + 1) (3x + 2)

Answer:

$\frac{d}{d x} ((2 x^{2} + 1) (3 x + 2)) = \frac{d}{d x} (6 x^{3} + 4 x^{2} + 3 x + 2) = 6 \frac{d}{d x} (x^{3}) + 4 \frac{d}{d x} (x^{2}) + 3 \frac{d}{d x} (x) + \frac{d}{d x} (2) = 6 (3 x^{2}) + 4 (2 x) + 3 (1) + 0 = 18 x^{2} + 8 x + 3$

Page No 30.33:

Question 6:

log₃x + 3 log_e x + 2 tan x

Answer:

$\frac{d}{d x} (\log_{3} x + 3 \log_{e} x + 2 \tan x) = \frac{d}{d x} (\frac{\log x}{\log 3}) + 3 \frac{d}{d x} (\log_{e} x) + 2 \frac{d}{d x} (\tan x) = \frac{1}{\log 3} . \frac{1}{x} + 3 . \frac{1}{x} + 2 \sec^{2} x = \frac{1}{x \log 3} + \frac{3}{x} + 2 \sec^{2} x$

Page No 30.34:

Question 7:

$(x + \frac{1}{x}) (\sqrt{x} + \frac{1}{\sqrt{x}})$

Answer:

$\frac{d}{d x} [(x + \frac{1}{x}) (\sqrt{x} + \frac{1}{\sqrt{x}})] = \frac{d}{d x} [(x + x^{- 1}) (x^{\frac{1}{2}} + x^{\frac{- 1}{2}})] = \frac{d}{d x} (x^{\frac{3}{2}} + x^{\frac{1}{2}} + x^{\frac{- 1}{2}} + x^{\frac{- 3}{2}}) = \frac{d}{d x} (x^{\frac{3}{2}}) + \frac{d}{d x} (x^{\frac{1}{2}}) + \frac{d}{d x} (x^{\frac{- 1}{2}}) + \frac{d}{d x} (x^{\frac{- 3}{2}}) = \frac{3}{2} x^{\frac{1}{2}} + \frac{1}{2} x^{\frac{- 1}{2}} - \frac{1}{2} x^{\frac{- 3}{2}} - \frac{3}{2} x^{\frac{- 5}{2}}$

Page No 30.34:

Question 8:

${(\sqrt{x} + \frac{1}{\sqrt{x}})}^{3}$

Answer:

$\frac{d}{d x} {(\sqrt{x} + \frac{1}{\sqrt{x}})}^{3} = \frac{d}{d x} [{(\sqrt{x})}^{3} + 3 {(\sqrt{x})}^{2} (\frac{1}{\sqrt{x}}) + 3 (\sqrt{x}) {(\frac{1}{\sqrt{x}})}^{2} + {(\frac{1}{\sqrt{x}})}^{3}] = \frac{d}{d x} (x^{\frac{3}{2}}) + 3 \frac{d}{d x} (x^{\frac{1}{2}}) + 3 \frac{d}{d x} (x^{\frac{- 1}{2}}) + \frac{d}{d x} (x^{\frac{- 3}{2}}) = \frac{3}{2} x^{\frac{3}{2} - 1} + 3 . \frac{1}{2} x^{\frac{1}{2} - 1} + 3 (\frac{- 1}{2}) x^{\frac{- 1}{2} - 1} + (\frac{- 3}{2}) x^{\frac{- 3}{2} - 1} = \frac{3}{2} x^{\frac{1}{2}} + \frac{3}{2} x^{\frac{- 1}{2}} - \frac{3}{2} x^{\frac{- 3}{2}} - \frac{3}{2} x^{\frac{- 5}{2}}$

Page No 30.34:

Question 9:

$\frac{2 x^{2} + 3 x + 4}{x}$

Answer:

$\frac{d}{d x} (\frac{2 x^{2} + 3 x + 4}{x}) = \frac{d}{d x} (\frac{2 x^{2}}{x}) + \frac{d}{d x} (\frac{3 x}{x}) + \frac{d}{d x} (\frac{4}{x}) = 2 \frac{d}{d x} (x) + 3 \frac{d}{d x} (1) + 4 \frac{d}{d x} (x^{- 1}) = 2 (1) + 3 (0) + 4 (- 1) x^{- 2} = 2 - \frac{4}{x^{2}}$

Page No 30.34:

Question 10:

$\frac{(x^{3} + 1) (x - 2)}{x^{2}}$

Answer:

$\frac{d}{d x} (\frac{(x^{3} + 1) (x - 2)}{x^{2}}) = \frac{d}{d x} (\frac{x^{4} - 2 x^{3} + x - 2}{x^{2}}) = \frac{d}{d x} (\frac{x^{4}}{x^{2}}) - 2 \frac{d}{d x} (\frac{x^{3}}{x^{2}}) + \frac{d}{d x} (\frac{x}{x^{2}}) - \frac{d}{d x} (\frac{2}{x^{2}}) = \frac{d}{d x} (x^{2}) - 2 \frac{d}{d x} (x) + \frac{d}{d x} (x^{- 1}) - 2 \frac{d}{d x} (x^{- 2}) = 2 x - 2 - \frac{1}{x^{2}} - 2 (- 2) x^{- 3} = 2 x - 2 - \frac{1}{x^{2}} + \frac{4}{x^{3}}$

Page No 30.34:

Question 11:

$\frac{a \cos x + b \sin x + c}{\sin x}$

Answer:

$\frac{d}{d x} (\frac{a \cos x + b \sin x + c}{\sin x}) = \frac{d}{d x} (\frac{a \cos x}{\sin x}) + \frac{d}{d x} (\frac{b \sin x}{\sin x}) + \frac{d}{d x} (\frac{c}{\sin x}) = a \frac{d}{d x} (c o t x) + \frac{d}{d x} (b) + c \frac{d}{d x} (\cos e c x) = - a \cos e c^{2} x + 0 - c \cos e c x c o t x = - a \cos e c^{2} x - c \cos e c x c o t x$

Page No 30.34:

Question 12:

2 sec x + 3 cot x − 4 tan x

Answer:

$\frac{d}{d x} (2 s e c x + 3 c o t x - 4 \tan x) = 2 \frac{d}{d x} (\sec x) + 3 \frac{d}{d x} (\cot x) - 4 \frac{d}{d x} (\tan x) = 2 \sec x \tan x - 3 \cos e c^{2} x - 4 \sec^{2} x$

Page No 30.34:

Question 13:

a₀ xⁿ + a₁ xⁿ⁻¹ + a₂ xⁿ⁻² + ... + a_n₋₁ x + a_n.

Answer:

$\frac{d}{d x} [(a_{0} x^{n}) + \frac{d}{d x} (a_{1} x^{n - 1}) + \frac{d}{d x} (a_{2} x^{n - 2}) + . . . + a_{n - 1} x + a_{n}] = a_{0} \frac{d}{d x} (x^{n}) + a_{1} \frac{d}{d x} (x^{n - 1}) + a_{2} \frac{d}{d x} (x^{n - 2}) + . . . + a_{n - 1} \frac{d}{d x} (x) + \frac{d}{d x} (a_{n}) = n a_{0} x^{n - 1} + (n - 1) a_{1} x^{n - 2} + (n - 2) a_{2} x^{n - 3} + . . . . + a_{n - 1} (1) + 0 = n a_{0} x^{n - 1} + (n - 1) a_{1} x^{n - 2} + (n - 2) a_{2} x^{n - 3} + . . . . + a_{n - 1}$

Page No 30.34:

Question 14:

$\frac{1}{\sin x} + 2^{x + 3} + \frac{4}{\log_{x} 3}$

Answer:

$\frac{d}{d x} (\frac{1}{\sin x} + 2^{x + 3} + \frac{4}{\log_{x} 3}) = \frac{d}{d x} (\cos e c x + 2^{x} . 2^{3} + \frac{4}{\frac{\log 3}{\log x}}) = \frac{d}{d x} (\cos e c x) + 2^{3} \frac{d}{d x} (2^{x}) + \frac{4}{\log 3} \frac{d}{d x} (\log x) = - \cos e c x c o t x + 2^{3} . 2^{x} . \log 2 + \frac{4}{\log 3} . \frac{1}{x} = - \cos e c x c o t x + 2^{x + 3} . \log 2 + \frac{4}{x \log 3}$

Page No 30.34:

Question 15:

$\frac{(x + 5) (2 x^{2} - 1)}{x}$

Answer:

$\frac{d}{d x} (\frac{(x + 5) (2 x^{2} - 1)}{x}) = \frac{d}{d x} (\frac{2 x^{3} + 10 x^{2} - x - 5}{x}) = \frac{d}{d x} (\frac{2 x^{3}}{x}) + \frac{d}{d x} (\frac{10 x^{2}}{x}) - \frac{d}{d x} (\frac{x}{x}) - \frac{d}{d x} (\frac{5}{x}) = 2 \frac{d}{d x} (x^{2}) + 10 \frac{d}{d x} (x) - \frac{d}{d x} (1) - 5 \frac{d}{d x} (x^{- 1}) = 2 (2 x) + 10 (1) - 0 - 5 (- 1) x^{- 2} = 4 x + 10 + \frac{5}{x^{2}}$

Page No 30.34:

Question 16:

$\log (\frac{1}{\sqrt{x}}) + 5 x^{a} - 3 a^{x} + \sqrt[3]{x^{2}} + 6 \sqrt[4]{x^{- 3}}$

Answer:

$\frac{d}{d x} (\log (\frac{1}{\sqrt{x}}) + 5 x^{a} - 3 a^{x} + \sqrt[3]{x^{2}} + 6 \sqrt[4]{x^{- 3}}) = \frac{d}{d x} [l og (x^{\frac{- 1}{2}})] + 5 \frac{d}{d x} (x^{a}) - 3 \frac{d}{d x} (a^{x}) + \frac{d}{d x} (x^{\frac{2}{3}}) + 6 \frac{d}{d x} (x^{\frac{- 3}{4}}) = \frac{d}{d x} (\frac{- 1}{2} \log x) + 5 \frac{d}{d x} (x^{a}) - 3 \frac{d}{d x} (a^{x}) + \frac{d}{d x} (x^{\frac{2}{3}}) + 6 \frac{d}{d x} (x^{\frac{- 3}{4}}) = \frac{- 1}{2} . \frac{1}{x} + 5 a x^{a - 1} - 3 a^{x} \log a + \frac{2}{3} x^{\frac{- 1}{3}} + 6 (\frac{- 3}{4}) x^{\frac{- 7}{4}} = \frac{- 1}{2 x} + 5 a x^{a - 1} - 3 a^{x} \log a + \frac{2}{3} x^{\frac{- 1}{3}} - \frac{9}{2} x^{\frac{- 7}{4}}$

Page No 30.34:

Question 17:

cos (x + a)

Answer:

$\frac{d}{d x} [\cos (x + a)] = \frac{d}{d x} (\cos x \cos a - \sin x \sin a) = \cos a \frac{d}{d x} (\cos x) - \sin a \frac{d}{d x} (\sin x) = - \cos a \sin x - \sin a \cos x = - (\sin x \cos a + \cos x \sin a) = - \sin (x + a)$

Page No 30.34:

Question 18:

$\frac{\cos (x - 2)}{\sin x}$

Answer:

$\frac{d}{d x} [\frac{\cos (x - 2)}{\sin x}] = \frac{d}{d x} (\frac{\cos x \cos 2 + \sin x \sin 2}{\sin x}) = \frac{d}{d x} (\frac{\cos x \cos 2}{\sin x}) + \frac{d}{d x} (\frac{\sin x \sin 2}{\sin x}) = \cos 2 \frac{d}{d x} (c o t x) + \sin 2 \frac{d}{d x} (1) = \cos 2 (- \cos e c^{2} x) + \sin 2 (0) = - \cos e c^{2} x \cos 2$

Page No 30.34:

Question 19:

$If y = {(\sin \frac{x}{2} + \cos \frac{x}{2})}^{2}, find \frac{d y}{d x} at x = \frac{π}{6} .$

Answer:

$\frac{d y}{d x} = \frac{d}{d x} {(\sin \frac{x}{2} + \cos \frac{x}{2})}^{2} = \frac{d}{d x} (\sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2}) = \frac{d}{d x} (1 + \sin x) = \frac{d}{d x} (1) + \frac{d}{d x} (\sin x) = 0 + \cos x = \cos x \frac{d y}{d x} at x = \frac{π}{6} = cos \frac{π}{6} = \frac{\sqrt{3}}{2}$

Page No 30.34:

Question 20:

$If y = (\frac{2 - 3 \cos x}{\sin x}), find \frac{d y}{d x} at x = \frac{π}{4}$

Answer:

$\frac{d y}{d x} = \frac{d}{d x} (\frac{2 - 3 \cos x}{\sin x}) = \frac{d}{d x} (\frac{2}{\sin x}) - \frac{d}{d x} (\frac{3 \cos x}{\sin x}) = 2 \frac{d}{d x} (\cos e c x) - 3 \frac{d}{d x} (\cot x) = - 2 \cos e c x \cot x + 3 \cos e c^{2} x \frac{d y}{d x} at x= \frac{π}{4} = - 2 \cos e c \frac{π}{4} \cot \frac{π}{4} + 3 \cos e c^{2} \frac{π}{4} = - 2 (\sqrt{2}) (1) + 3 {(\sqrt{2})}^{2} = - 2 \sqrt{2} + 6 = 6 - 2 \sqrt{2}$

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Question 21:

Find the slope of the tangent to the curve f (x) = 2x⁶ + x⁴ − 1 at x = 1.

Answer:

$Slope of the tangent = f' (x) = \frac{d}{d x} (2 x^{6} + x^{4} - 1) = 2 \frac{d}{d x} (x^{6}) + \frac{d}{d x} (x^{4}) - \frac{d}{d x} (1) = 12 x^{5} + 4 x^{3} ∴ Slope of the tangent at x =1: 12 {(1)}^{5} + 4 {(1)}^{3} = 12 + 4 = 16$

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Question 22:

$If y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}, prove that 2 x y \frac{d y}{d x} = (\frac{x}{a} - \frac{a}{x})$

Answer:

$y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}} = \frac{1}{\sqrt{a}} x^{\frac{1}{2}} + \sqrt{a} x^{\frac{- 1}{2}} \frac{d y}{d x} = \frac{1}{\sqrt{a}} \frac{1}{2} x^{\frac{- 1}{2}} + \sqrt{a} (\frac{- 1}{2}) x^{\frac{- 3}{2}} LHS = 2 x y \frac{d y}{d x} = 2 x (\frac{1}{\sqrt{a}} x^{\frac{1}{2}} + \sqrt{a} x^{\frac{- 1}{2}}) (\frac{1}{\sqrt{a}} \frac{1}{2} x^{\frac{- 1}{2}} + \sqrt{a} (\frac{- 1}{2}) x^{\frac{- 3}{2}}) = 2 x (\frac{1}{2 a} - \frac{1}{2 x} + \frac{1}{2 x} - \frac{a}{2 x^{2}}) = 2 x (\frac{1}{2 a} - \frac{a}{2 x^{2}}) = (\frac{x}{a} - \frac{a}{x}) = RHS Hence, proved .$

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Question 23:

Find the rate at which the function f (x) = x⁴ − 2x³ + 3x² + x + 5 changes with respect to x.

Answer:

$Rate = f' (x) = \frac{d}{d x} (x^{4} - 2 x^{3} + 3 x^{2} + x + 5) = \frac{d}{d x} (x^{4}) - 2 \frac{d}{d x} (x^{3}) + 3 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (x) + \frac{d}{d x} (5) = 4 x^{3} - 2 (3 x^{2}) + 3 (2 x) + 1 + 0 = 4 x^{3} - 6 x^{2} + 6 x + 1$

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Question 24:

$If y = \frac{2 x^{9}}{3} - \frac{5}{7} x^{7} + 6 x^{3} - x, find \frac{d y}{d x} at x = 1 .$

Answer:

$\frac{d y}{d x} = \frac{d}{d x} (\frac{2 x^{9}}{3} - \frac{5}{7} x^{7} + 6 x^{3} - x) = \frac{2}{3} \frac{d}{d x} (x^{9}) - \frac{5}{7} \frac{d}{d x} (x^{7}) + 6 \frac{d}{d x} (x^{3}) - \frac{d}{d x} (x) = \frac{2}{3} (9 x^{8}) - \frac{5}{7} (7 x^{6}) + 6 (3 x^{2}) - 1 = 6 x^{8} - 5 x^{6} + 18 x^{2} - 1 \frac{d y}{d x} at x = 1: 6 {(1)}^{8} - 5 {(1)}^{6} + 18 {(1)}^{2} - 1 = 6 - 5 + 18 - 1 = 18$

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Question 25:

If for f (x) = λ x² + μ x + 12, f' (4) = 15 and f' (2) = 11, then find λ and μ.

Answer:

$f' (x) = λ \frac{d}{d x} (x^{2}) + μ \frac{d}{d x} (x) + \frac{d}{d x} (12) f' (x) = 2 λ x + μ (1) Given: f' (4) = 15 2 λ (4) + μ = 15 (From (1)) \Rightarrow 8 λ + μ = 15 (2) Also, given: f' (2) = 11 2 λ (2) + μ = 11 (From (1)) 4 λ + μ = 11 (3) Subtracting equation (3) from equation (2): 4 λ = 4 λ = 1 Substituting this in equation (3): 4 (1) + μ = 11 μ = 7 ∴ λ=1 and μ=7$

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Question 26:

For the function $f (x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + . . . + \frac{x^{2}}{2} + x + 1 .$ Prove that f' (1) = 100 f' (0).

Answer:

$f' (x) = \frac{d}{d x} (\frac{x^{100}}{100} + \frac{x^{99}}{99} + . . . + \frac{x^{2}}{2} + x + 1) = \frac{1}{100} (100 x^{99}) + \frac{1}{99} (99 x^{98}) + . . . + \frac{1}{2} (2 x) + 1 + 0 = x^{99} + x^{98} + . . . + x + 1 f' (1) = 1^{99} + 1^{98} + . . . + 1 + 1 = 99 + 1 = 100 f' (0) = 0 + 0 + . . . + 0 + 1 = 1 RHS = 100 f' (0) = 100 (1) = 100 = f' (1) = LHS ∴ f' (1) = 100 f' (0)$

Page No 30.39:

Question 1:

x³ sin x

Answer:

$Let u = x^{3}; v = \sin x Then, u' = 3 x^{2}; v' = \cos x Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} (x^{3} \sin x) = x^{3} \cos x + \sin x (3 x^{2}) = x^{2} (x \cos x + 3 \sin x)$

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Question 2:

x³ e^x

Answer:

$Let u = x^{3}; v = e^{x} Then, u' = 3 x^{2}; v' = e^{x} Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} (x^{3} e^{x}) = x^{3} e^{x} + e^{x} (3 x^{2}) = x^{2} e^{x} (x + 3)$

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Question 3:

x² e^x log x

Answer:

$Let u = x^{2}; v = e^{x}; w = \log x Then, u' = 2 x; v' = e^{x}, w = \frac{1}{x} Using the product rule: \frac{d}{d x} (u v w) = u' v w + + u v' w + u v w' \frac{d}{d x} (x^{2} e^{x} \log x) = 2 x e^{x} \log x + x^{2} e^{x} \log x + x^{2} e^{x} \frac{1}{x} = 2 x e^{x} \log x + x^{2} e^{x} \log x + x e^{x} = x e^{x} (2 \log x + x \log x + 1)$

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Question 4:

xⁿ tan x

Answer:

$Let u = x^{n}; v = \tan x Then, u' = n x^{n - 1}; v' = \sec^{2} x Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} (x^{n} \tan x) = x^{n} \sec^{2} x + \tan x (n x^{n - 1}) = x^{n - 1} (x \sec^{2} x + n \tan x)$

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Question 5:

xⁿ log_a x

Answer:

$Let u = x^{n}; v = \log_{a} x = \frac{\log x}{\log a} Then, u' = n x^{n - 1}; v' = \frac{1}{x \log a} Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} (x^{n} \log_{a} x) = x^{n} . \frac{1}{x \log a} + \log_{a} x (n x^{n - 1}) = x^{n - 1} \frac{1}{\log a} + \log_{a} x (n x^{n - 1}) = x^{n - 1} (\frac{1}{\log a} + n \log_{a} x)$

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Question 6:

(x³ + x² + 1) sin x

Answer:

$Let u = x^{3} + x^{2} + 1; v = \sin x Then, u' = 3 x^{2} + 2 x; v' = \cos x By product rule, \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} [(x^{3} + x^{2} + 1) \sin x] = (x^{3} + x^{2} + 1) \cos x + (3 x^{2} + 2 x) \sin x$

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Question 7:

sin x cos x

Answer:

$Let u = \sin x; v = \cos x Then, u' = \cos x; v' = - \sin x Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} (\sin x \cos x) = \sin x (- \sin x) + \cos x . \cos x = - \sin^{2} x + \cos^{2} x = \cos 2 x$

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Question 8:

$\frac{2^{x} \cot x}{\sqrt{x}}$

Answer:

$\frac{2^{x} \cot x}{\sqrt{x}} = 2^{x} \cot x (x^{\frac{- 1}{2}}) Let u = 2^{x}; v = \cot x; w = x^{\frac{- 1}{2}} Then, u' = 2^{x} \log 2; v' = - {cosec}^{2} x; w' = \frac{- 1}{2} x^{\frac{- 3}{2}} Using the product rule: \frac{d}{d x} (u v w) = u' v w + u v' w + u v w' \frac{d}{d x} [2^{x} \cot x (x^{\frac{- 1}{2}})] = 2^{x} \log 2 . \cot x . x^{\frac{- 1}{2}} + 2^{x} (- {cosec}^{2} x) x^{\frac{- 1}{2}} + 2^{x} \cot x (\frac{- 1}{2} x^{\frac{- 3}{2}}) = 2^{x} \log 2 . \cot x . \frac{1}{\sqrt{x}} + 2^{x} (- {cosec}^{2} x) \frac{1}{\sqrt{x}} + 2^{x} \cot x (\frac{- 1}{2 x \sqrt{x}}) = \frac{2^{x}}{\sqrt{x}} (\log 2 . \cot x - {cosec}^{2} x - \frac{\cot x}{2 x})$

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Question 9:

x² sin x log x

Answer:

$Let u = x^{2}; v = \sin x; w = \log x Then, u' = 2 x; v' = \cos x; w' = \frac{1}{x} Using the product rule: \frac{d}{d x} (u v w) = u' v w + u v' w + u v w' \frac{d}{d x} (x^{2} \sin x \log x) = 2 x \sin x \log x + x^{2} \cos x \log x + x^{2} \sin x . \frac{1}{x} = 2 x \sin x \log x + x^{2} \cos x \log x + x \sin x$

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Question 10:

x⁵ e^x + x⁶ log x

Answer:

$\frac{d}{d x} (x^{5} e^{x} + x^{6} \log x) = x^{5} \frac{d}{d x} (e^{x}) + e^{x} \frac{d}{d x} (x^{5}) + x^{6} \frac{d}{d x} (\log x) + \log x \frac{d}{d x} (x^{6}) = x^{5} e^{x} + e^{x} (5 x^{4}) + x^{6} . \frac{1}{x} + \log x (6 x^{5}) = x^{5} e^{x} + e^{x} (5 x^{4}) + x^{5} + \log x (6 x^{5}) = x^{4} (x e^{x} + 5 e^{x} + x + 6 x \log x)$

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Question 11:

(x sin x + cos x) (x cos x − sin x)

Answer:

$u = x \sin x + \cos x; v = x \cos x - \sin x u' = x \cos x + \sin x - \sin x = x \cos x; v' = - x \sin x + \cos x - \cos x = - x \sin x Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} [(x \sin x + \cos x) (x \cos x - \sin x)] = (x \sin x + \cos x) (- x \sin x) + (x \cos x - \sin x) (x \cos x) = - x^{2} \sin^{2} x - x \cos x \sin x + x^{2} \cos^{2} x - x \cos x \sin x = x^{2} (\cos^{2} x - \sin^{2} x) - x (2 \sin x \cos x) = x^{2} \cos (2 x) - x (\sin (2 x)) = x [x \cos (2 x) - \sin (2 x)]$

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Question 12:

(x sin x + cos x ) (e^x + x² log x)

Answer:

$Let u = x \sin x + \cos x; v = e^{x} + x^{2} \log x Then, u' = [x \frac{d}{d x} (\sin x) + \sin x \frac{d}{d x} (x)] - \sin x = x \cos x + \sin x - \sin x = x \cos x v' = e^{x} + [x^{2} \frac{d}{d x} (\log x) + \log x \frac{d}{d x} (x^{2})] = e^{x} + x + 2 x \log x Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} [(x \sin x + \cos x) (e^{x} + x^{2} \cos x)] = (x \sin x + \cos x) (e^{x} + x + 2 x \log x) + (e^{x} + x^{2} \log x) (x \cos x)$

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Question 13:

(1 − 2 tan x) (5 + 4 sin x)

Answer:

$Let u = 1 - 2 \tan x; v = 5 + 4 \sin x Then, u' = - 2 \sec^{2} x; v' = 4 \cos x Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} [(1 - 2 \tan x) (5 + 4 \sin x)] = (1 - 2 \tan x) (4 \cos x) + (5 + 4 \sin x) (- 2 \sec^{2} x) = 4 \cos x - 8 \times \frac{\sin x}{\cos x} \cos x - 10 \sec^{2} x - 8 \times \frac{\sin x}{\cos^{2} x} = 4 \cos x - 8 \sin x - 10 \sec^{2} x - 8 \sec x \tan x = 4 (\cos x - 2 \sin x - \frac{5}{2} \sec^{2} x - 2 \sec x \tan x)$

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Question 14:

(1 +x²) cos x

Answer:

$Let u = 1 + x^{2}; v = \cos x Then, u' = 2 x; v' = - \sin x Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} [(1 + x^{2}) (\cos x)] = (1 + x^{2}) (- \sin x) + (\cos x) (2 x) = - \sin x - x^{2} \sin x + 2 x \cos x$

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Question 15:

sin² x

Answer:

$\frac{d}{d x} (\sin^{2} x) = 2 \sin x \frac{d}{d x} (\sin x) (Using the chain rule) = 2 \sin x \cos x = \sin 2 x$

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Question 16:

log_x² x

Answer:

$\log_{x^{2}} x = \frac{\log x}{\log x^{2}} (by change of base property) = \frac{\log x}{2 \log x} [\log x^{2} = 2 \log x] = \frac{1}{2} Now \frac{d}{d x} (\log_{x^{2}} x) = \frac{d}{d x} (\frac{1}{2}) = 0 (∵ \frac{1}{2} is a constant)$

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Question 17:

$e^{x} \log \sqrt{x} \tan x$

Answer:

$Let u = e^{x}; v = \log \sqrt{x}; w = \tan x Then, u' = e^{x}; v' = \frac{1}{\sqrt{x}} \times \frac{1}{2 \sqrt{x}} = \frac{1}{2 x}; w' = \sec^{2} x Using the product rule: \frac{d}{d x} (u v w) = u' v w + u v' w + u v w' = e^{x} \log \sqrt{x} \tan x + e^{x} \times \frac{1}{2 x} \tan x + e^{x} \log \sqrt{x} \sec^{2} x = e^{x} (\log x^{\frac{1}{2}} . \tan x + \frac{\tan x}{2 x} + \log x^{\frac{1}{2}} . \sec^{2} x) = e^{x} (\frac{1}{2} \log x . \tan x + \frac{\tan x}{2 x} + \frac{1}{2} \log x . \sec^{2} x) = \frac{e^{x}}{2} (\log x . \tan x + \frac{\tan x}{x} + \log x . \sec^{2} x)$

Page No 30.39:

Question 18:

x³ e^x cos x

Answer:

$Let u = x^{3}; v = e^{x}; w = \cos x Then, u' = 3 x^{2}; v' = e^{x}; w' = - \sin x Using the product rule: \frac{d}{d x} (u v w) = u' v w + u v' w + u v w' \frac{d}{d x} (x^{3} e^{x} \cos x) = 3 x^{2} e^{x} \cos x + x^{3} e^{x} \cos x + x^{3} e^{x} (- \sin x) = x^{2} e^{x} (3 \cos x + x \cos x - x \sin x)$

Page No 30.39:

Question 19:

$\frac{x^{2} \cos \frac{π}{4}}{\sin x}$

Answer:

$\frac{x^{2} \cos \frac{π}{4}}{\sin x} = x^{2} \cos \frac{π}{4} cosec x Let u = x^{2}; v = \cos \frac{π}{4}; w = cosec x Then, u' = 2 x; v' = 0; w' = - cosec x \cot x Using the product rule: \frac{d}{d x} (u v w) = u' v w + u v' w + u v w' \frac{d}{d x} (x^{2} \cos \frac{π}{4} cosec x) = 2 x \cos \frac{π}{4} cosec x + x^{2} . 0 . cosec x + x^{2} \cos \frac{π}{4} (- \cos e c x \cot x) = \cos \frac{π}{4} (2 x cosec x - x^{2} cosec x \cot x) = \cos \frac{π}{4} (\frac{2 x}{\sin x} - x^{2} \frac{\cot x}{\sin x})$

Page No 30.39:

Question 20:

x⁴ (5 sin x − 3 cos x)

Answer:

$Let u = x^{4}; v = 5 \sin x - 3 \cos x Then, u' = 4 x^{3}; v' = 5 \cos x - 3 (- \sin x) = 5 \cos x + 3 \sin x According to the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} (x^{4} (5 \sin x - 3 \cos x)) = x^{4} (5 \cos x + 3 \sin x) + (5 \sin x - 3 \cos x) 4 x^{3} = x^{3} (5 x \cos x + 3 x \sin x + 20 \sin x - 12 \cos x) = x^{3} ((3 x + 20) \sin x + (5 x - 12) \cos x) = 3 x^{4} \sin x + 20 x^{3} \sin x + 5 x \cos x - 12 \cos x$

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Question 21:

(2x² − 3) sin x

Answer:

$Let u = 2 x^{2} - 3; v = \sin x Then, u' = 4 x; v' = \cos x Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} [(2 x^{2} - 3) \sin x] = (2 x^{2} - 3) \cos x + 4 x \sin x$

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Question 22:

x⁵ (3 − 6x⁻⁹)

Answer:

$Let u = x^{5}; v = (3 - 6 x^{- 9}) Then, u' = 5 x^{4}; v' = 54 x^{- 10} Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} [x^{5} (3 - 6 x^{- 9})] = x^{5} (54 x^{- 10}) + (3 - 6 x^{- 9}) (5 x^{4}) = 54 x^{- 5} + 15 x^{4} - 30 x^{- 5} = 15 x^{4} + 24 x^{- 5}$

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Question 23:

x⁻⁴ (3 − 4x⁻⁵)

Answer:

$Let u = x^{- 4}; v = 3 - 4 x^{- 5} Then, u' = - 4 x^{- 5}; v' = 20 x^{- 6} Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} [x^{- 4} (3 - 4 x^{- 5})] = x^{- 4} (20 x^{- 6}) + (3 - 4 x^{- 5}) (- 4 x^{- 5}) = 20 x^{- 10} - 12 x^{- 5} + 16 x^{- 10} = - 12 x^{- 5} + 36 x^{- 10}$

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Question 24:

x⁻³ (5 + 3x)

Answer:

$Let u = x^{- 3}; v = (5 + 3 x) Then, u = - 3 x^{- 4}; v' = 3 Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} [x^{3} (5 + 3 x)] = x^{- 3} . 3 + (5 + 3 x) (- 3 x^{- 4}) = 3 x^{- 3} - 15 x^{- 4} - 9 x^{- 3} = - 15 x^{- 4} - 6 x^{- 3}$

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Question 25:

Differentiate in two ways, using product rule and otherwise, the function (1 + 2 tan x) (5 + 4 cos x). Verify that the answers are the same.

Answer:

${Product rule (1}^{s t} method): Let u = 1 + 2 \tan x; v = 5 + 4 \cos x Then, u' = 2 \sec^{2} x; v' = - 4 \sin x Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} [(1 + 2 \tan x) (5 + 4 \cos x)] = (1 + 2 \tan x) (- 4 \sin x) + (5 + 4 \cos x) (2 \sec^{2} x) = - 4 \sin x - 8 \tan x \sin x + 10 \sec^{2} x + 8 \sec x = - 4 \sin x + 10 \sec^{2} x + (\frac{8}{\cos x} - \frac{8 \sin^{2} x}{\cos x}) = - 4 \sin x + 10 \sec^{2} x + 8 (\frac{1 - \sin^{2} x}{\cos x}) = - 4 \sin x + 10 \sec^{2} x + 8 (\frac{\cos^{2} x}{\cos x}) = - 4 \sin x + 10 \sec^{2} x + 8 \cos x 2^{nd} method: (1 + 2 \tan x) (5 + 4 \cos x) = 5 + 4 \cos x + 10 \tan x + 8 \sin x Now, we have: \frac{d}{d x} [(1 + 2 \tan x) (5 + 4 \cos x)] = \frac{d}{d x} (5 + 4 \cos x + 10 \tan x + 8 \sin x) = - 4 \sin x + 10 \sec^{2} x + 8 \cos x Using both the methods, we get the same answer.$

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Question 26:

Differentiate each of the following functions by the product rule and the other method and verify that answer from both the methods is the same.
(i) (3x² + 2)²
(ii) (x + 2) (x + 3)
(iii) (3 sec x − 4 cosec x) (−2 sin x + 5 cos x)

Answer:

$(i) {Product rule (1}^{st} method): Let u = 3 x^{2} + 2; v = 3 x^{2} + 2 Then, u' = 6 x; v' = 6 x Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} [(3 x^{2} + 2) (3 x^{2} + 2)] = (3 x^{2} + 2) (6 x) + (3 x^{2} + 2) (6 x) = 18 x^{3} + 12 x + 18 x^{3} + 12 x = 36 x^{3} + 24 x 2^{nd} method: \frac{d}{d x} [{(3 x^{2} + 2)}^{2}] = \frac{d}{d x} (9 x^{4} + 12 x^{2} + 4) = 36 x^{3} + 24 x Using both the methods, we get the same answer.$

$(ii) {Product rule (1}^{st} method): Let u = x + 2; v = x + 3 Then, u' = 1; v' = 1 Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} [(x + 2) (x + 3)] = (x + 2) 1 + (x + 3) 1 = x + 2 + x + 3 = 2 x + 5 2^{nd} method: \frac{d}{d x} [(x + 2) (x + 3)] = \frac{d}{d x} (x^{2} + 5 x + 6) = 2 x + 5 Using both the methods, we get the same answer.$

$(iii) {Product rule (1}^{st} method): Let u = 3 \sec x - 4 \cos e c x; v = - 2 \sin x + 5 \cos x Then, u' = 3 \sec x \tan x + 4 cosec x \cot x; v' = - 2 \cos x - 5 \sin x Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} [(3 s e c x - 4 \cos e c x) (- 2 \sin x + 5 \cos x)] = (3 s e c x - 4 \cos e c x) (- 2 \cos x - 5 \sin x) + (- 2 \sin x + 5 \cos x) (3 \sec x \tan x + 4 \cos e c x co t x) = - 6 + 15 \tan x + 8 \cot x + 20 - 6 \tan^{2} x - 8 c o t x - 15 \tan x + 20 \cot^{2} x = - 6 + 20 - 6 (\sec^{2} x - 1) + 20 ({cosec}^{2} x - 1) = - 6 + 20 - 6 \sec^{2} x + 6 + 20 {cosec}^{2} x - 20 = - 6 \sec^{2} x + 20 \cos e c^{2} x 2^{nd} method: \frac{d}{d x} [(3 s e c x - 4 \cos e c x) (- 2 \sin x + 5 \cos x)] = \frac{d}{d x} (- 6 \sec x \sin x + 15 \sec x \cos x + 8 \cos e c x \sin x - 20 \cos e c x \cos x) = \frac{d}{d x} (- 6 \frac{\sin x}{\cos x} + 15 \frac{\cos x}{\cos x} + 8 \frac{\sin x}{\sin x} - 20 \frac{\cos x}{\sin x}) = \frac{d}{d x} (- 6 \tan x + 15 + 8 - 20 \cot x) = \frac{d}{d x} (- 6 \tan x - 20 \cot x + 23) = - 6 \sec^{2} x + 20 \cos e c^{2} x Using both the methods, we get the same answer.$

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Question 27:

(ax + b) (a + d)²

Answer:

$(a x + b) (a + d)^{2} Let u = a x + b, v = {(a + d)}^{2} Then, u' = a, v' = 0 Using the product rule: \frac{d}{d x} (u v) = u v' + v u' \frac{d}{d x} ((a x + b) (a + d)^{2}) = (a x + b) \times 0 + {(a + d)}^{2} \times a ∴ \frac{d}{d x} ((a x + b) (a + d)^{2}) = a {(a + d)}^{2}$

Page No 30.39:

Question 28:

(ax + b)ⁿ (cx + d)ⁿ

Answer:

${(a x + b)}^{n} {(c x + d)}^{n} Let u = {(a x + b)}^{n}, v = {(c x + d)}^{n} Then, u' = n a {(a x + b)}^{n - 1}, v' = n c {(c x + d)}^{n - 1} Using the product rule : \frac{d}{d x} (u v) = u v' + u' v \frac{d}{d x} [{(a x + b)}^{n} {(c x + d)}^{n}] = {(a x + b)}^{n} \times n c {(c x + d)}^{n - 1} + n a {(a x + b)}^{n - 1} \times {(c x + d)}^{n} = n {(a x + b)}^{n - 1} {(c x + d)}^{n - 1} (a c x + c b + a c x + a d) = n {(a x + b)}^{n - 1} {(c x + d)}^{n - 1} (2 a c x + c b + a d)$

Page No 30.44:

Question 1:

$\frac{x^{2} + 1}{x + 1}$

Answer:

$Let u = x^{2} + 1; v = x + 1 Then, u' = 2 x; v' = 1 Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{x^{2} + 1}{x + 1}) = \frac{(x + 1) 2 x - (x^{2} + 1) 1}{(x + 1)^{2}} = \frac{2 x^{2} + 2 x - x^{2} - 1}{(x + 1)^{2}} = \frac{x^{2} + 2 x - 1}{(x + 1)^{2}}$

Page No 30.44:

Question 2:

$\frac{2 x - 1}{x^{2} + 1}$

Answer:

$Let u = 2 x - 1; v = x^{2} + 1; Then, u' = 2; v' = 2 x Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{2 x - 1}{x^{2} + 1}) = \frac{(x^{2} + 1) 2 - (2 x - 1) 2 x}{(x^{2} + 1)^{2}} = \frac{2 x^{2} + 2 - 4 x^{2} + 2 x}{(x^{2} + 1)^{2}} = \frac{- 2 x^{2} + 2 x + 2}{(x^{2} + 1)^{2}} = \frac{2 (1 + x - x^{2})}{(x^{2} + 1)^{2}}$

Page No 30.44:

Question 3:

$\frac{x + e^{x}}{1 + \log x}$

Answer:

$Let u = x + e^{x}; v = 1 + \log x Then, u' = 1 + e^{x}; v' = \frac{1}{x} Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{x + e^{x}}{1 + \log x}) = \frac{(1 + \log x) (1 + e^{x}) - (x + e^{x}) (\frac{1}{x})}{(1 + \log x)^{2}} = \frac{x + x e^{x} + x \log x + x \log x e^{x} - x - e^{x}}{x (1 + \log x)^{2}} = \frac{x \log x + x \log x e^{x} - e^{x} + x e^{x}}{x (1 + \log x)^{2}} = \frac{x \log x (1 + e^{x}) - e^{x} (1 - x)}{x (1 + \log x)^{2}}$

Page No 30.44:

Question 4:

$\frac{e^{x} - \tan x}{\cot x - x^{n}}$

Answer:

$Let u = e^{x} - \tan x; v = \cot x - x^{n} Then, u' = e^{x} - \sec^{2} x; v' = - \cos e c^{2} x - n x^{n - 1} Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{e^{x} - \tan x}{c o t x - x^{n}}) = \frac{(\cot x - x^{n}) (e^{x} - \sec^{2} x) - (e^{x} - \tan x) (- \cos e c^{2} x - n x^{n - 1})}{{(\cot x - x^{n})}^{2}} = \frac{(\cot x - x^{n}) (e^{x} - \sec^{2} x) + (e^{x} - \tan x) (\cos e c^{2} x + n x^{n - 1})}{{(\cot x - x^{n})}^{2}}$

Page No 30.44:

Question 5:

$\frac{a x^{2} + b x + c}{p x^{2} + q x + r}$

Answer:

$Let u = a x^{2} + b x + c; v = p x^{2} + q x + r Then, u' = 2 a x + b; v' = 2 p x + q Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{a x^{2} + b x + c}{p x^{2} + q x + r}) = \frac{(p x^{2} + q x + r) (2 a x + b) - (a x^{2} + b x + c) (2 p x + q)}{{(p x^{2} + q x + r)}^{2}} = \frac{2 a p x^{3} + 2 a q x^{2} + 2 a r x + b p x^{2} + b q x + b r - 2 a p x^{3} - 2 b p x^{2} - 2 p c x - a q x^{2} - b q x - c q}{{(p x^{2} + q x + r)}^{2}} = \frac{(a q - b p) x^{2} + 2 (a r - x p) x + b r - c q}{{(p x^{2} + q x + r)}^{2}}$

Page No 30.44:

Question 6:

$\frac{x}{1 + \tan x}$

Answer:

$Let u = x; v = 1 + \tan x Then, u' = 1; v' = \sec^{2} x Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} = \frac{(1 + \tan x) \times 1 - x \sec^{2} x}{{(1 + \tan x)}^{2}} = \frac{1 + \tan x - x \sec^{2} x}{{(1 + \tan x)}^{2}}$

Page No 30.44:

Question 7:

$\frac{1}{a x^{2} + b x + c}$

Answer:

$\frac{d}{d x} (\frac{1}{a x^{2} + b x + c}) = \frac{d}{d x} {(a x^{2} + b x + c)}^{- 1} = (- 1) {(a x^{2} + b x + c)}^{- 2} \frac{d}{d x} (a x^{2} + b x + c) (Using the chain rule) = (- 1) {(a x^{2} + b x + c)}^{- 2} (2 a x + b) = \frac{- (2 a x + b)}{{(a x^{2} + b x + c)}^{2}}$

Page No 30.44:

Question 8:

$\frac{e^{x}}{1 + x^{2}}$

Answer:

$Let u = e^{x}; v = 1 + x^{2} Then, u' = e^{x}; v' = 2 x Using the chain rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{e^{x}}{1 + x^{2}}) = \frac{(1 + x^{2}) e^{x} - e^{x} (2 x)}{{(1 + x^{2})}^{2}} = \frac{e^{x} + x^{2} e^{x} - 2 x e^{x}}{{(1 + x^{2})}^{2}} = \frac{e^{x} (1 + x^{2} - 2 x)}{{(1 + x^{2})}^{2}} = \frac{e^{x} {(1 - x)}^{2}}{{(1 + x^{2})}^{2}}$

Page No 30.44:

Question 9:

$\frac{e^{x} + \sin x}{1 + \log x}$

Answer:

$Let u = e^{x} + \sin x; v = 1 + \log x T h e n, u' = e^{x} + \cos x; v' = \frac{1}{x} Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{e^{x} + \sin x}{1 + \log x}) = \frac{(1 + \log x) (e^{x} + \cos x) - (e^{x} + \sin x) (\frac{1}{x})}{{(1 + \log x)}^{2}} = \frac{x (1 + \log x) (e^{x} + \cos x) - (e^{x} + \sin x)}{x {(1 + \log x)}^{2}}$

Page No 30.44:

Question 10:

$\frac{x \tan x}{\sec x + \tan x}$

Answer:

$Let u = x \tan x; v = \sec x + \tan x Then, u' = x \sec^{2} x + \tan x; v' = \sec x \tan x + \sec^{2} x Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{x \tan x}{\sec x + \tan x}) = \frac{(\sec x + \tan x) (x \sec^{2} x + \tan x) - x \tan x (\sec x \tan x + \sec^{2} x)}{{(\sec x + \tan x)}^{2}} = \frac{x \sec^{3} x + x \sec^{2} x \tan x + \sec x \tan x + \tan^{2} x - x \sec x \tan^{2} x - x \tan x \sec^{2} x}{{(\sec x + \tan x)}^{2}} = \frac{(\sec x + \tan x) (x \sec^{2} x + \tan x) - x \tan x \sec x (\sec x + \tan x)}{{(\sec x + \tan x)}^{2}} = \frac{x \sec^{2} x + \tan x - x \tan x \sec x}{\sec x + \tan x} = \frac{x \sec x (\sec x - \tan x) + \tan x}{\sec x + \tan x}$

Page No 30.44:

Question 11:

$\frac{x \sin x}{1 + \cos x}$

Answer:

$Let u = x \sin x; v = 1 + \cos x Then, u' = x \cos x + \sin x; v' = - \sin x Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{x \sin x}{1 + \cos x}) = \frac{(1 + \cos x) (x \cos x + \sin x) - x \sin x (- \sin x)}{{(1 + \cos x)}^{2}} = \frac{(1 + \cos x) (x \cos x + \sin x) + x \sin^{2} x}{{(1 + \cos x)}^{2}} = \frac{(1 + \cos x) (x \cos x + \sin x) + x (1 - \cos^{2} x)}{{(1 + \cos x)}^{2}} = \frac{(1 + \cos x) (x \cos x + \sin x) + x (1 + \cos x) (1 - \cos x)}{{(1 + \cos x)}^{2}} = \frac{(1 + \cos x) (x \cos x + \sin x + x - x \cos x)}{{(1 + \cos x)}^{2}} = \frac{(1 + \cos x) (x + \sin x)}{{(1 + \cos x)}^{2}}$

Page No 30.44:

Question 12:

$\frac{2^{x} \cot x}{\sqrt{x}}$

Answer:

$Let u = 2^{x} \cot x; v = \sqrt{x} Then, u' = - 2^{x} {cosec}^{2} x + 2^{x} \log 2 \cot x; v' = \frac{1}{2 \sqrt{x}} \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{2^{x} c o t x}{\sqrt{x}}) = \frac{\sqrt{x} (- 2^{x} {cosec}^{2} x + 2^{x} \log 2 \cot x) - 2^{x} \cot x (\frac{1}{2 \sqrt{x}})}{{(\sqrt{x})}^{2}} = \frac{\frac{x (- 2^{x} {cosec}^{2} x + 2^{x} \log 2 \cot x) - 2^{x - 1} \cot x}{\sqrt{x}}}{x} = \frac{2^{x} (- x {cosec}^{2} x + x \cot x \log 2 - (\frac{1}{2}) \cot x)}{x \sqrt{x}} = \frac{2^{x} (- x {cosec}^{2} x + x \cot x \log 2 - (\frac{1}{2}) \cot x)}{x^{\frac{3}{2}}}$

Page No 30.44:

Question 13:

$\frac{\sin x - x \cos x}{x \sin x + \cos x}$

Answer:

$Let u = \sin x - x \cos x; v = x \sin x + \cos x Then, u' = \cos x + x \sin x - \cos x; v' = x \cos x + \sin x - \sin x = x \sin x = x \cos x Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{\sin x - x \cos x}{x \sin x + \cos x}) = \frac{(x \sin x + \cos x) x \sin x - (\sin x - x \cos x) x \cos x}{{(x \sin x + \cos x)}^{2}} = \frac{x^{2} \sin^{2} x + x \cos x \sin x - x \cos x \sin x + x^{2} \cos^{2} x}{{(x \sin x + \cos x)}^{2}} = \frac{x^{2} (\sin^{2} x + \cos^{2} x)}{{(x \sin x + \cos x)}^{2}} = \frac{x^{2}}{{(x \sin x + \cos x)}^{2}}$

Page No 30.44:

Question 14:

$\frac{x^{2} - x + 1}{x^{2} + x + 1}$

Answer:

$Let u = x^{2} - x + 1; v = x^{2} + x + 1 Then, u' = 2 x - 1; v' = 2 x + 1 By quotient rule, \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{x^{2} - x + 1}{x^{2} + x + 1}) = \frac{(x^{2} + x + 1) (2 x - 1) - (x^{2} - x + 1) (2 x + 1)}{{(x^{2} + x + 1)}^{2}} = \frac{2 x^{3} + 2 x^{2} + 2 x - x^{2} - x - 1 - 2 x^{3} + 2 x^{2} - 2 x - x^{2} + x - 1}{{(x^{2} + x + 1)}^{2}} = \frac{2 x^{2} - 2}{{(x^{2} + x + 1)}^{2}} = \frac{2 (x^{2} - 1)}{{(x^{2} + x + 1)}^{2}}$

Page No 30.44:

Question 15:

$\frac{\sqrt{a} + \sqrt{x}}{\sqrt{a} - \sqrt{x}}$

Answer:

$Let u = \sqrt{a} + \sqrt{x}; v = \sqrt{a} - \sqrt{x} Then, u' = \frac{1}{2 \sqrt{x}}; v' = \frac{- 1}{2 \sqrt{x}} Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{\sqrt{a} + \sqrt{x}}{\sqrt{a} - \sqrt{x}}) = \frac{(\sqrt{a} - \sqrt{x}) \frac{1}{2 \sqrt{x}} - (\sqrt{a} + \sqrt{x}) (\frac{- 1}{2 \sqrt{x}})}{{(\sqrt{a} - \sqrt{x})}^{2}} = \frac{\sqrt{a} - \sqrt{x} + \sqrt{a} + \sqrt{x}}{2 \sqrt{x} {(\sqrt{a} - \sqrt{x})}^{2}} = \frac{2 \sqrt{a}}{2 \sqrt{x} {(\sqrt{a} - \sqrt{x})}^{2}} = \frac{\sqrt{a}}{\sqrt{x} {(\sqrt{a} - \sqrt{x})}^{2}}$

Page No 30.44:

Question 16:

$\frac{a + \sin x}{1 + a \sin x}$

Answer:

Let us use the quotient rule here.
We have:
u = a + sin x and v =1 + a sin x
u' = cos x and v'=a cos x

$Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{a + \sin x}{1 + a \sin x}) = \frac{(1 + a \sin x) (\cos x) - (a + \sin x) (a \cos x)}{(1 + a \sin x)^{2}} = \frac{\cos x + a \sin x \cos x - a^{2} \cos x - a \sin x \cos x}{(1 + a \sin x)^{2}} = \frac{\cos x - a^{2} \cos x}{(1 + a \sin x)^{2}} = \frac{(1 - a^{2}) \cos x}{(1 + a \sin x)^{2}}$

Page No 30.44:

Question 17:

$\frac{10^{x}}{\sin x}$

Answer:

$Let u = 10^{x}; v = \sin x Then, u' = 10^{x} \log 10; v' = \cos x Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{10^{x}}{\sin x}) = \frac{\sin x 10^{x} \log 10 - 10^{x} \cos x}{\sin^{2} x} = \frac{\sin x 10^{x} \log 10}{\sin^{2} x} - \frac{10^{x} \cos x}{\sin^{2} x} = 10^{x} \log 10 \cos e c x - 10^{x} cosec x \cot x = 10^{x} cosec x (\log 10 - \cot x)$

Page No 30.44:

Question 18:

$\frac{1 + 3^{x}}{1 - 3^{x}}$

Answer:

$Let u = 1 + 3^{x}; v = 1 - 3^{x} Then, u' = 3^{x} \log 3; v' = - 3^{x} \log 3 Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{1 + 3^{x}}{1 - 3^{x}}) = \frac{(1 - 3^{x}) 3^{x} \log 3 - (1 + 3^{x}) (- 3^{x} \log 3)}{{(1 - 3^{x})}^{2}} = \frac{3^{x} \log 3 - 3^{2 x} \log 3 + 3^{x} \log 3 + 3^{2 x} \log 3}{{(1 - 3^{x})}^{2}} = \frac{2 . 3^{x} \log 3}{{(1 - 3^{x})}^{2}}$

Page No 30.44:

Question 19:

$\frac{3^{x}}{x + \tan x}$

Answer:

$Let u = 3^{x}; v = x + \tan x Then, u' = 3^{x} \log 3; v' = 1 + \sec^{2} x By quotient rule, we have: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{3^{x}}{x + \tan x}) = \frac{(x + \tan x) 3^{x} \log 3 - 3^{x} (1 + \sec^{2} x)}{{(x + \tan x)}^{2}} = \frac{3^{x} [(x + \tan x) \log 3 - (1 + \sec^{2} x)]}{{(x + \tan x)}^{2}}$

Page No 30.44:

Question 20:

$\frac{1 + \log x}{1 - \log x}$

Answer:

$Let u = 1 + \log x; v = 1 - \log x Then, u' = \frac{1}{x}; v' = \frac{- 1}{x} Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{1 + \log x}{1 - \log x}) = \frac{(1 - \log x) (\frac{1}{x}) - (1 + \log x) (\frac{- 1}{x})}{{(1 - \log x)}^{2}} = \frac{1 - \log x + 1 + \log x}{x {(1 - \log x)}^{2}} = \frac{2}{x {(1 - \log x)}^{2}}$

Page No 30.44:

Question 21:

$\frac{4 x + 5 \sin x}{3 x + 7 \cos x}$

Answer:

$Let u = 4 x + 5 \sin x; v = 3 x + 7 \cos x Then, u' = 4 + 5 \cos x; v' = 3 - 7 \sin x Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{4 x + 5 \sin x}{3 x + 7 \cos x}) = \frac{(3 x + 7 \cos x) (4 + 5 \cos x) - (4 x + 5 \sin x) (3 - 7 \sin x)}{{(3 x + 7 \cos x)}^{2}} = \frac{12 x + 15 x \cos x + 28 \cos x + 35 \cos^{2} x - 12 x + 28 x \sin x - 15 \sin x + 35 \sin^{2} x}{{(3 x + 7 \cos x)}^{2}} = \frac{15 x \cos x + 28 x \sin x + 28 \cos x 15 \sin x + 35 (\sin^{2} x + \cos^{2} x)}{{(3 x + 7 \cos x)}^{2}} = \frac{15 x \cos x + 28 x \sin x + 28 \cos x 15 \sin x + 35}{{(3 x + 7 \cos x)}^{2}}$

Page No 30.44:

Question 22:

$\frac{x}{1 + \tan x}$

Answer:

$Let u = x; v = 1 + \tan x Then, u' = 1; v' = \sec^{2} x Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{x}{1 + \tan x}) = \frac{(1 + \tan x) 1 - x \sec^{2} x}{{(1 + \tan x)}^{2}} = \frac{1 + \tan x - x \sec^{2} x}{{(1 + \tan x)}^{2}}$

Page No 30.44:

Question 23:

$\frac{a + b \sin x}{c + d \cos x}$

Answer:

$Let u = a + b \sin x; v = c + d \cos x Then, u' = b \cos x; v' = - d \sin x Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{a + b \sin x}{c + d \cos x}) = \frac{(c + d \cos x) b \cos x - (a + b \sin x) (- d \sin x)}{{(c + d \cos x)}^{2}} = \frac{b c \cos x + b d \cos^{2} x + a d \sin x + b d \sin^{2} x}{{(c + d \cos x)}^{2}} = \frac{b c \cos x + a d \sin x + b d (\sin^{2} x + \cos^{2} x)}{{(c + d \cos x)}^{2}} = \frac{b c \cos x + a d \sin x + b d}{{(c + d \cos x)}^{2}}$

Page No 30.44:

Question 24:

$\frac{p x^{2} + q x + r}{a x + b}$

Answer:

$Let u = p x^{2} + q x + r; v = a x + b Then, u' = 2 p x + q; v' = a Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{p x^{2} + q x + r}{a x + b}) = \frac{(a x + b) (2 p x + q) - (p x^{2} + q x + r) a}{{(a x + b)}^{2}} = \frac{2 a p x^{2} + a q x + 2 b p x + b q - a p x^{2} - a q x - a r}{{(a x + b)}^{2}} = \frac{a p x^{2} + 2 b p x + b q - a r}{{(a x + b)}^{2}}$

Page No 30.44:

Question 25:

$\frac{\sec x - 1}{\sec x + 1}$

Answer:

$Let u = \sec x - 1; v = \sec x + 1 Then, u' = \sec x t an x; v' = \sec x \tan x Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{s e c x - 1}{s e c x + 1}) = \frac{(\sec x + 1) \sec x \tan x - (\sec x - 1) \sec x \tan x}{{(s e c x + 1)}^{2}} = \frac{\sec^{2} x \tan x + \sec x \tan x - \sec^{2} x \tan x + \sec x \tan x}{{(\sec x + 1)}^{2}} = \frac{2 \sec x \tan x}{{(\sec x + 1)}^{2}}$

Page No 30.44:

Question 26:

$\frac{x^{5} - \cos x}{\sin x}$

Answer:

$Let u = x^{5} - \cos x; v = \sin x Then, u' = 5 x^{4} + \sin x; v' = \cos x Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{x^{5} - \cos x}{\sin x}) = \frac{\sin x (5 x^{4} + \sin x) - (x^{5} - \cos x) \cos x}{\sin^{2} x} = \frac{- x^{5} \cos x + 5 x^{4} \sin x + (\sin^{2} x + \cos^{2} x)}{\sin^{2} x} = \frac{- x^{5} \cos x + 5 x^{4} \sin x + 1}{\sin^{2} x}$

Page No 30.44:

Question 27:

$\frac{x + \cos x}{\tan x}$

Answer:

$Let u = x + \cos x; v = \tan x Then, u' = 1 - \sin x; v' = \sec^{2} x Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{x + \cos x}{\tan x}) = \frac{\tan x (1 - \sin x) - (x + \cos x) \sec^{2} x}{\tan^{2} x}$

Page No 30.44:

Question 28:

$\frac{x}{\sin^{n} x}$

Answer:

$Let u = x; v = \sin^{n} x Then, u' = 1; v' = n \sin^{n - 1} x . \cos x Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{x}{\sin^{n} x}) = \frac{\sin^{n} x . 1 - x n \sin^{n - 1} x . \cos x}{{(\sin^{n} x)}^{2}} = \frac{\sin^{n - 1} x (\sin x - n x . \cos x)}{\sin^{2 n} x} = \frac{\sin x - n x . \cos x}{\sin^{2 n - (n - 1)} x} = \frac{\sin x - n x \cos x}{\sin^{n + 1} x}$

Page No 30.44:

Question 29:

$\frac{a x + b}{p x^{2} + q x + r}$

Answer:

$Let u = a x + b; v = p x^{2} + q x + r Then, u' = a; v' = 2 p x + q Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{a x + b}{p x^{2} + q x + r}) = \frac{(p x^{2} + q x + r) a - (a x + b) (2 p x + q)}{{(p x^{2} + q x + r)}^{2}} = \frac{a p x^{2} + a q x + a r - 2 a p x^{2} - 2 b p x - a q x - b q}{{(p x^{2} + q x + r)}^{2}} = \frac{- a p x^{2} - 2 b p x + a r - b q}{{(p x^{2} + q x + r)}^{2}}$

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Question 30:

$\frac{1}{a x^{2} + b x + c}$

Answer:

$Let u = 1; v = a x^{2} + b x + c Then, u' = 0; v' = 2 a x + b Using the quotient rule: \frac{d}{d x} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} \frac{d}{d x} (\frac{1}{a x^{2} + b x + c}) = \frac{(a x^{2} + b x + c) 0 - 1 (2 a x + b)}{{(a x^{2} + b x + c)}^{2}} = \frac{- (2 a x + b)}{{(a x^{2} + b x + c)}^{2}}$

Page No 30.46:

Question 1:

Mark the correct alternative in each of the following:

Let f(x) = x − [x], x ∈ R, then $f' (\frac{1}{2})$ is

(a) $\frac{3}{2}$ (b) 1 (c) 0 (d) −1

Answer:

Given: f(x) = x − [x], x ∈ R

Now,

For 0 ≤ x < 1, [x] = 0.

∴ f(x) = x − 0 = x, ∀ x ∈ [0, 1)

Differentiating both sides with respect to x, we get

f '(x) = 1, ∀ x ∈ [0, 1)

$∴ f' (\frac{1}{2}) = 1$

Hence, the correct answer is option (b).

Page No 30.47:

Question 2:

Mark the correct alternative in each of the following:

If $f (x) = \frac{x - 4}{2 \sqrt{x}}$ , then f '(1) is

(a) $\frac{5}{4}$ (b) $\frac{4}{5}$ (c) 1 (d) 0

Answer:

$f (x) = \frac{x - 4}{2 \sqrt{x}} = \frac{1}{2} \sqrt{x} - \frac{2}{\sqrt{x}} = \frac{1}{2} x^{\frac{1}{2}} - 2 x^{- \frac{1}{2}}$

Differentiating both sides with respect to x, we get

$f' (x) = \frac{1}{2} \times \frac{1}{2} x^{\frac{1}{2} - 1} - 2 \times (- \frac{1}{2}) x^{- \frac{1}{2} - 1} [f (x) = x^{n} \Rightarrow f' (x) = n x^{n - 1}] \Rightarrow f' (x) = \frac{1}{4} x^{- \frac{1}{2}} + x^{- \frac{3}{2}} ∴ f' (1) = \frac{1}{4} \times 1 + 1 = \frac{5}{4}$

Hence, the correct answer is option (a).

Page No 30.47:

Question 3:

Mark the correct alternative in each of the following:

If $y = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . .$ , then $\frac{d y}{d x} =$

(a) y + 1 (b) y − 1 (c) y (d) y²

Answer:

$y = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . .$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = \frac{d}{d x} (1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . .) = \frac{d}{d x} (1) + \frac{d}{d x} (\frac{x}{1!}) + \frac{d}{d x} (\frac{x^{2}}{2!}) + \frac{d}{d x} (\frac{x^{3}}{3!}) + \frac{d}{d x} (\frac{x^{4}}{4!}) + . . . = \frac{d}{d x} (1) + \frac{1}{1!} \frac{d}{d x} (x) + \frac{1}{2!} \frac{d}{d x} (x^{2}) + \frac{1}{3!} \frac{d}{d x} (x^{3}) + \frac{1}{4!} \frac{d}{d x} (x^{4}) + . . . = 0 + \frac{1}{1!} \times 1 + \frac{1}{2!} \times 2 x + \frac{1}{3!} \times 3 x^{2} + \frac{1}{4!} \times 4 x^{3} + . . . (y = x^{n} \Rightarrow \frac{d y}{d x} = n x^{n - 1})$
$= 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . [\frac{n}{n!} = \frac{1}{(n - 1)!}] = y$

$∴ \frac{d y}{d x} = y$

Hence, the correct answer is option (c).

Page No 30.47:

Question 4:

Mark the correct alternative in each of the following:

If $f (x) = 1 - x + x^{2} - x^{3} + . . . - x^{99} + x^{100}$ , then $f' (1)$ equals

(a) 150 (b) −50 (c) −150 (d) 50

Answer:

$f (x) = 1 - x + x^{2} - x^{3} + . . . - x^{99} + x^{100}$

Differentiating both sides with respect to x, we get

$f' (x) = \frac{d}{d x} (1 - x + x^{2} - x^{3} + . . . - x^{99} + x^{100}) = \frac{d}{d x} (1) - \frac{d}{d x} (x) + \frac{d}{d x} (x^{2}) - \frac{d}{d x} (x^{3}) + . . . - \frac{d}{d x} (x^{99}) + \frac{d}{d x} (x^{100}) = 0 - 1 + 2 x - 3 x^{2} + . . . - 99 x^{98} + 100 x^{99} = - 1 + 2 x - 3 x^{2} + . . . - 99 x^{98} + 100 x^{99}$

Putting x = 1, we get

$f' (1) = - 1 + 2 - 3 + . . . - 99 + 100 = (- 1 + 2) + (- 3 + 4) + (- 5 + 6) + . . . + (- 99 + 100) = 1 + 1 + 1 + . . . + 1 (50 terms) = 50$

Hence, the correct answer is option (d).

Page No 30.47:

Question 5:

Mark the correct alternative in each of the following:

If $y = \frac{1 + \frac{1}{x^{2}}}{1 - \frac{1}{x^{2}}}$ , then $\frac{d y}{d x} =$

(a) $- \frac{4 x}{{(x^{2} - 1)}^{2}}$ (b) $- \frac{4 x}{x^{2} - 1}$ (c) $\frac{1 - x^{2}}{4 x}$ (d) $\frac{4 x}{x^{2} - 1}$

Answer:

$y = \frac{1 + \frac{1}{x^{2}}}{1 - \frac{1}{x^{2}}} = \frac{x^{2} + 1}{x^{2} - 1}$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = \frac{(x^{2} - 1) \times \frac{d}{d x} (x^{2} + 1) - (x^{2} + 1) \times \frac{d}{d x} (x^{2} - 1)}{{(x^{2} - 1)}^{2}} (Quotient rule) = \frac{(x^{2} - 1) \times (2 x + 0) - (x^{2} + 1) \times (2 x - 0)}{{(x^{2} - 1)}^{2}} = \frac{2 x^{3} - 2 x - 2 x^{3} - 2 x}{{(x^{2} - 1)}^{2}} = \frac{- 4 x}{{(x^{2} - 1)}^{2}}$

Hence, the correct answer is option (a).

Page No 30.47:

Question 6:

Mark the correct alternative in each of the following:

If $y = \sqrt{x} + \frac{1}{\sqrt{x}}$ , then $\frac{d y}{d x}$ at x = 1 is

(a) 1 (b) $\frac{1}{2}$ (c) $\frac{1}{\sqrt{2}}$ (d) 0

Answer:

$y = \sqrt{x} + \frac{1}{\sqrt{x}} = x^{\frac{1}{2}} + x^{- \frac{1}{2}}$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = \frac{d}{d x} (x^{\frac{1}{2}} + x^{- \frac{1}{2}}) = \frac{d}{d x} (x^{\frac{1}{2}}) + \frac{d}{d x} (x^{- \frac{1}{2}}) = \frac{1}{2} x^{\frac{1}{2} - 1} + (- \frac{1}{2}) x^{- \frac{1}{2} - 1} (y = x^{n} \Rightarrow \frac{d y}{d x} = n x^{n - 1}) = \frac{1}{2} x^{- \frac{1}{2}} - \frac{1}{2} x^{- \frac{3}{2}}$

Putting x = 1, we get

${(\frac{d y}{d x})}_{x = 1} = \frac{1}{2} \times 1 - \frac{1}{2} \times 1 = 0$

Thus, $\frac{d y}{d x}$ at x = 1 is 0.

Hence, the correct answer is option (d).

Page No 30.47:

Question 7:

Mark the correct alternative in each of the following:

If $f (x) = x^{100} + x^{99} + . . . + x + 1$ , then $f' (1)$ is equal to

(a) 5050 (b) 5049 (c) 5051 (d) 50051

Answer:

$f (x) = x^{100} + x^{99} + . . . + x + 1$

Differentiating both sides with respect to x, we get

$f' (x) = \frac{d}{d x} (x^{100} + x^{99} + . . . + x + 1) = \frac{d}{d x} (x^{100}) + \frac{d}{d x} (x^{99}) + . . . + \frac{d}{d x} (x^{2}) + \frac{d}{d x} (x) + \frac{d}{d x} (1) = 100 x^{99} + 99 x^{98} + . . . + 2 x + 1 + 0 (y = x^{n} \Rightarrow \frac{d y}{d x} = n x^{n - 1}) = 100 x^{99} + 99 x^{98} + . . . + 2 x + 1$

Putting x = 1, we get

$f' (1) = 100 + 99 + 98 + . . . + 2 + 1 = \frac{100 (100 + 1)}{2} (S_{n} = \frac{n (n + 1)}{2}) = 50 \times 101 = 5050$

Hence, the correct answer is option (a).

Page No 30.47:

Question 8:

Mark the correct alternative in each of the following:

If $f (x) = 1 + x + \frac{x^{2}}{2} + . . . + \frac{x^{100}}{100}$ , then $f' (1)$ is equal to

(a) $\frac{1}{100}$ (b) 100 (c) 50 (d) 0

Answer:

$f (x) = 1 + x + \frac{x^{2}}{2} + . . . + \frac{x^{100}}{100}$

Differentiating both sides with respect to x, we get

$f' (x) = \frac{d}{d x} (1 + x + \frac{x^{2}}{2} + . . . + \frac{x^{100}}{100}) = \frac{d}{d x} (1) + \frac{d}{d x} (x) + \frac{d}{d x} (\frac{x^{2}}{2}) + . . . + \frac{d}{d x} (\frac{x^{100}}{100}) = \frac{d}{d x} (1) + \frac{d}{d x} (x) + \frac{1}{2} \frac{d}{d x} (x^{2}) + . . . + \frac{1}{100} \frac{d}{d x} (x^{100}) = 0 + 1 + \frac{1}{2} \times 2 x + . . . + \frac{1}{100} \times 100 x^{99} (y = x^{n} \Rightarrow \frac{d y}{d x} = n x^{n - 1}) = 1 + x + x^{2} + . . . + x^{99}$

Putting x = 1, we get

$f' (1) = 1 + 1 + 1 + . . . + 1 (100 terms) = 100$

Hence, the correct answer is option (b).

Page No 30.47:

Question 9:

Mark the correct alternative in each of the following:

If $y = \frac{\sin x + \cos x}{\sin x - \cos x}$ , then $\frac{d y}{d x}$ at x = 0 is

(a) −2 (b) 0 (c) $\frac{1}{2}$ (d) does not exist

Answer:

$y = \frac{\sin x + \cos x}{\sin x - \cos x}$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = \frac{(\sin x - \cos x) \times \frac{d}{d x} (\sin x + \cos x) - (\sin x + \cos x) \times \frac{d}{d x} (\sin x - \cos x)}{{(\sin x - \cos x)}^{2}} (Quotient rule) = \frac{(\sin x - \cos x) \times [\frac{d}{d x} (\sin x) + \frac{d}{d x} (\cos x)] - (\sin x + \cos x) \times [\frac{d}{d x} (\sin x) - \frac{d}{d x} (\cos x)]}{{(\sin x - \cos x)}^{2}} = \frac{(\sin x - \cos x) (\cos x - \sin x) - (\sin x + \cos x) (\cos x + \sin x)}{{(\sin x - \cos x)}^{2}} = \frac{- (\cos^{2} x + \sin^{2} x - 2 \cos x \sin x) - (\sin^{2} x + \cos^{2} x + 2 \sin x \cos x)}{{(\sin x - \cos x)}^{2}}$
$= \frac{- 1 + 2 \cos x \sin x - 1 - 2 \sin x \cos x}{{(\sin x - \cos x)}^{2}} = \frac{- 2}{{(\sin x - \cos x)}^{2}}$

Putting x = 0, we get

${(\frac{d y}{d x})}_{x = 0} = \frac{- 2}{{(\sin 0 - \cos 0)}^{2}} = \frac{- 2}{{(0 - 1)}^{2}} = - 2$

Thus, $\frac{d y}{d x}$ at x = 0 is −2.

Hence, the correct answer is option (a).

Page No 30.47:

Question 10:

Mark the correct alternative in each of the following:

If $y = \frac{\sin (x + 9)}{\cos x}$ , then $\frac{d y}{d x}$ at x = 0 is

(a) cos 9 (b) sin 9 (c) 0 (d) 1

Answer:

$y = \frac{\sin (x + 9)}{\cos x}$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = \frac{\cos x \times \frac{d}{d x} \sin (x + 9) - \sin (x + 9) \times \frac{d}{d x} \cos x}{\cos^{2} x} (Quotient rule) = \frac{\cos x \times \cos (x + 9) - \sin (x + 9) \times (- \sin x)}{\cos^{2} x} = \frac{\cos (x + 9) \cos x + \sin (x + 9) \sin x}{\cos^{2} x} = \frac{\cos (x + 9 - x)}{\cos^{2} x} = \frac{\cos 9}{\cos^{2} x}$

Putting x = 0, we get

${(\frac{d y}{d x})}_{x = 0} = \frac{\cos 9}{\cos^{2} 0} = \cos 9$ (cos 0 = 1)

Thus, $\frac{d y}{d x}$ at x = 0 is cos 9.

Hence, the correct answer is option (a).

Page No 30.47:

Question 11:

Mark the correct alternative in each of the following:

If $f (x) = \frac{x^{n} - a^{n}}{x - a}$ , then $f' (a)$ is

(a) 1 (b) 0 (c) $\frac{1}{2}$ (d) does not exist

Answer:

Given: $f (x) = \frac{x^{n} - a^{n}}{x - a}$

Now, f(x) is not defined at x = a. Therefore, f(x) is not differentiable at x = a.

So, $f' (a)$ does not exist.

Hence, the correct answer is option (d).

Page No 30.47:

Question 12:

Mark the correct alternative in each of the following:

If f(x) = x sinx, then $f' (\frac{π}{2}) =$

(a) 0 (b) 1 (c) −1 (d) $\frac{1}{2}$

Answer:

f(x) = x sinx

Differentiating both sides with respect to x, we get

$f' (x) = x \times \frac{d}{d x} (\sin x) + \sin x \times \frac{d}{d x} (x) (Product rule) = x \times \cos x + \sin x \times 1 = x \cos x + \sin x$

Putting $x = \frac{π}{2}$ , we get

$f' (\frac{π}{2}) = \frac{π}{2} \times \cos (\frac{π}{2}) + \sin (\frac{π}{2}) = \frac{π}{2} \times 0 + 1 = 1$

Hence, the correct answer is option (b).

Page No 30.48:

Question 1:

If y = 1 + $\frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . . . ., then \frac{d y}{d x} =______________________________.$

Answer:

If y = 1 + $\frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . . . .$
$\frac{d y}{d x} = \frac{d}{d x} (1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . . . .) = \frac{d}{d x} (1) + \frac{d}{d x} (\frac{x}{1!}) + \frac{d}{d x} (\frac{x^{2}}{2!}) + \frac{d}{d x} (\frac{x^{3}}{3!}) + . . . . = 0 + \frac{1}{1!} \frac{d}{d x} (x) + \frac{1}{2!} \frac{d}{d x} (x^{2}) + \frac{1}{3!} \frac{d}{d x} (x^{3}) + . . . . . = 0 + \frac{1}{1!} \times 1 + \frac{1}{2!} \times 2 x + \frac{1}{3!} \times 3 x^{2} + . . . . . . = 1 + \frac{x}{1!} + \frac{1}{3 \times 2!} 3 x^{2} + \frac{1}{4 \times 3!} 4 x^{3} + . . . . . i . e \frac{d y}{d x} = 1 + \frac{x}{1!} + \frac{1}{2!} x^{2} + \frac{1}{3!} x^{3} + . . . . . .$

Page No 30.48:

Question 2:

$If y = 1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \frac{x^{4}}{4!} - \frac{x^{5}}{5!} + . . . ., then \frac{dy}{dx} =________________________.$

Answer:

If $y = 1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \frac{x^{4}}{4!} - \frac{x^{5}}{5!} + . . . . . . .$
Then $\frac{d y}{d x} = \frac{d}{d x} [1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \frac{x^{4}}{4!} - . . . . . . .]$
$= \frac{d}{d x} (1) - \frac{d}{d x} (\frac{x}{1!}) + \frac{d}{d x} (\frac{x^{2}}{2!}) - \frac{d}{d x} (\frac{x^{3}}{3!}) + . . . . . . . . . . . = 0 - \frac{1}{1!} \frac{d}{d x} (x) + \frac{1}{2!} \frac{d}{d x} (x^{2}) - \frac{1}{3!} \frac{d}{d x} (x^{3}) + . . . . . . . . . . = - \frac{1}{1!} \times 1 + \frac{1}{2!} \times 2 x - \frac{1}{3!} \times 3 x^{2} + \frac{1}{4!} \times 4 x^{3} - . . . . . . . . = - \frac{1}{1!} + \frac{x}{1!} - \frac{1}{2!} x^{2} + \frac{1}{3!} x^{3} - . . . . . . . . . . . i . e \frac{d y}{d x} = - (1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \frac{x^{4}}{4!} - . . . . . . . .)$

Page No 30.48:

Question 3:

Let f(x) = x – [x], x ∈ R. Then f ' $(\frac{1}{3})$ =__________________.

Answer:

Given, f(x) = x – [x]
For $\frac{1}{3}$ we know $[\frac{1}{3}]$ = 0
i.e f(x) = x – [x] = {x} = x since x = $\frac{1}{3}$
∴ f'(x) = 1 at x = $\frac{1}{3}$
i.e f' $(\frac{1}{3})$ = 1

Page No 30.48:

Question 4:

If f(x) = x + $|x|$ , then f'(1) = _________________________.

Answer:

for f(x) = x + $|x|$
for x = 1, $|x|$ = x
i.e f(x) = x + x
i.e f(x) = 2x
i.e f'(x) = 2 at x = 1
i.e f'(1) = 2

Page No 30.48:

Question 5:

If f(x) = x + $|x|$ , then f'(–1) = _________________________.

Answer:

For f(x) = x + $|x|$ given
at x = –1, $|x|$ = – x
⇒ f(x) = x – x = 0
i.e f'(x) = 0 at x = –1
i.e f'(–1) = 0

Page No 30.48:

Question 6:

If f(x) = x $|x|$ , then f'(–2) = _________________________.

Answer:

For f(x) = x $|x|$ given
at x = –2
$|x|$ = – x
∴ f(x) = x (– x)
i.e f(x) = – x²
∴ f'(x) = –2x at x = –2
i.e f'(x) = (–2) (–2)
i.e f'(x) = 4

Page No 30.48:

Question 7:

$\frac{d}{d x} {(\sqrt{x} + \frac{1}{\sqrt{x}})}^{2} =______________.$

Answer:

$\frac{d}{d x} {(\sqrt{x} + \frac{1}{\sqrt{x}})}^{2}$
$= \frac{d}{d x} (x + \frac{1}{x} + 2 x \frac{1}{x}) = \frac{d}{d x} (x + \frac{1}{x} + 2) = \frac{d}{d x} (x) + \frac{d}{d x} (x^{- 1}) + \frac{d}{d x} (2) = 1 - 1 x^{- 1 - 1} + 0 i . e \frac{d}{d x} {(\sqrt{x} + \frac{1}{\sqrt{x}})}^{2} = 1 - \frac{1}{x^{2}}$

Page No 30.48:

Question 8:

$\frac{d}{d x} (\log |x|) =_______________, x \neq 0 .$

Answer:

$\frac{d}{d x} (\log |x|)$
case (i) if x > 0
i.e $|x|$ = x
⇒ $\frac{d}{d x} (\log x) = \frac{1}{x}$
case (ii) if x < 0
i.e $|x|$ = – x
$\Rightarrow \frac{d}{d x} (\log (- x)) = \frac{1}{(- x)} \times \frac{d}{d x} (- x) = \frac{- 1}{x} \times (- 1) = \frac{1}{x}$

Page No 30.48:

Question 9:

If f(x) = mx + c, f(0) = f'(0) = 1, then f(2) = ___________________.

Answer:

Given f(x) = mx + c
Since f(0) = 1
⇒ f(0) = m(0) + c
i.e 1 = m(0) + c
⇒ c = 1
i.e f(x) = mx + 1
∴ f'(x) = (mx + 1)'
   = (mx)' +(1)'
           = m(x)' +(1)'
           = m × 1 + 0
i.e f'(x) = m
also, given f'(0) = 1
⇒ f'(0) = m
i.e 1 = m
∴ f(x) = x + 1
i.e f(2) = 2 + 1 = 3

Page No 30.48:

Question 10:

$\frac{d}{d x} (\frac{1}{\log_{x} e}) =________________.$

Answer:

$\frac{d}{d x} (\frac{1}{\log_{x} e})$
$\begin{array}{rcl} Since \log_{x} e & = & \frac{\log_{e} e}{\log_{e} x} = \frac{1}{\log_{e} x} \\ = & \frac{d}{d x} (\frac{1}{(\frac{1}{\log_{e} x})}) \\ = & \frac{d}{d x} (\log_{e} x) \\ i . e \frac{d}{d x} (\frac{1}{\log_{x} e}) & = & \frac{1}{x} \end{array}$

Page No 30.48:

Question 11:

$\frac{d}{d x} (\frac{1}{\log_{10} x}) =______________.$

Answer:

Check Answer
$\frac{d}{d x} (\frac{1}{\log_{10} x})$
$Since \log_{10} x = \frac{\log_{e} x}{\log_{e} 10} i . e \frac{d}{d x} (\frac{\log_{e} 10}{\log_{e} x})$

Page No 30.48:

Question 12:

$\frac{d}{d x} (\log_{x^{n}} x^{m}) =_____________.$

Answer:

$\frac{d}{d x} (\log_{x n} x^{m})$
$\begin{array}{rcl} Since \log_{x^{n}} x^{m} & = & \frac{\log_{e} x^{m}}{\log_{e} x^{n}} \\ = & \frac{m \log_{e} x}{n \log_{e} x} (By property \log, \log p^{q} = q \log p q l o g p) \\ i . e \log_{x^{n}} x^{m} & = & \frac{m}{n} \\ ∴ \frac{d}{d x} (\log_{x^{n}} x^{m}) & = & \frac{d}{d x} (\frac{m}{n}) \\ ∴ \frac{d}{d x} (\log_{x^{n}} x^{m}) & = & 0 \end{array}$

Page No 30.48:

Question 1:

Write the value of $\lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ .

Answer:

$Using the definition of derivative, we have: \lim_{x \to c} \frac{f (x) - f (x)}{x - c} = f' (c)$

Page No 30.48:

Question 2:

Write the value of $\lim_{x \to a} \frac{x f (a) - a f (x)}{x - a}$ .

Answer:

$\lim_{x \to a} \frac{x f (a) - a f (x)}{x - a} = \lim_{x \to a} \frac{x f (a) - a f (x) - x f (x) + x f (x)}{x - a} = \lim_{x \to a} \frac{x f (a) - x f (x) + x f (x) - a f (x)}{x - a} = \lim_{x \to a} \frac{- x (f (x) - f (a)) + (x - a) f (x)}{x - a} = \lim_{x \to a} - x \lim_{x \to a} \frac{f (x) - f (a)}{x - a} + \lim_{x \to a} \frac{(x - a) f (x)}{x - a} = - a f' (a) + f (a)$

Page No 30.48:

Question 3:

If x < 2, then write the value of $\frac{d}{d x} (\sqrt{x^{2} - 4 x + 4)}$ .

Answer:

$Given: x<2 ∴ 2-x>0 \frac{d}{d x} (\sqrt{x^{2} - 4 x + 4}) = \frac{d}{d x} (\sqrt{{(2 - x)}^{2}}) = \frac{d}{d x} (2 - x) (∵ 2-x>0) = 0 - 1 = - 1$

Page No 30.48:

Question 4:

If $\frac{π}{2}$ < x < π, then find $\frac{d}{d x} (\sqrt{\frac{1 + \cos 2 x}{2}})$ .

Answer:

$\sqrt{\frac{1 + \cos 2 x}{2}} = \sqrt{\frac{2 \cos^{2} x}{2}} = \sqrt{\cos^{2} x} = - \cos x (∵ \frac{π}{2} <x<π) \frac{d}{d x} (\sqrt{\frac{1 + \cos 2 x}{2}}) = \frac{d}{d x} (- \cos x) = - (- \sin x) = \sin x$

Page No 30.49:

Question 5:

Write the value of $\frac{d}{d x} (x |x|)$ .

Answer:

$Case 1: x > 0 | x | = x Thus, we have: \frac{d}{d x} (x | x |) = \frac{d}{d x} (x . x) = \frac{d}{d x} (x^{2}) = 2 x (1) Case 2: x < 0 | x | = - x Thus, we have: \frac{d}{d x} (x | x |) = \frac{d}{d x} (x . (- x)) = \frac{d}{d x} (- x^{2}) = - 2 x (2) From (1) and (2), we have: \frac{d}{d x} (x | x |) = \{\begin{cases} 2 x, if x > 0 \\ - 2 x, if x < 0 \end{cases}$

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Question 6:

Write the value of $\frac{d}{d x} \{(x + |x|) |x|\}$ .

Answer:

$Case 1: x > 0 |x| = x (x + |x|) |x| = (x + x) x = 2 x^{2} \frac{d}{d x} [(x + |x|) |x|] = \frac{d}{d x} (2 x^{2}) = 4 x (1) Case 2: x < 0 |x| = - x (x + |x|) |x| = (x - x) x = 0 \frac{d}{d x} [(x + |x|) |x|] = \frac{d}{d x} (0) = 0 (2) From (1) and (2), we have: \frac{d}{d x} [(x + |x|) |x|] = \{\begin{cases} 4 x, if x > 0 \\ 0, if x < 0 \end{cases}$

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Question 7:

If f (x) = |x| + |x−1|, write the value of $\frac{d}{d x} (f (x))$ .

Answer:

$f (x) = |x| + |x - 1| Case 1: x<0 (∴ x-1<-1<0) |x| = - x; |x - 1| = - (x - 1) = - x + 1 f (x) = - x + (- x + 1) = - 2 x f' (x) = - 2 Case 2: 0< x <1 (∴ x>0 and x-1<0) |x| = x; |x - 1| = - (x - 1) = 1 - x f (x) = x + 1 - x = 1 f' (x) = 0 Case 3: x>1 ∴ x>1>0 ⇒ x>0) |x| = x; |x - 1| = x - 1 f (x) = x + x - 1 = 2 x - 1 f' (x) = 2 From case 1, case 2 and case 3, we have: f' (x) = \{\begin{cases} \begin{matrix} - 2, when x < 0 \\ 0, when 0 < x < 1 \\ 2, when x > 1 \end{matrix} \end{cases}$

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Question 8:

Write the value of the derivative of f (x) = |x − 1| + |x − 3| at x = 2.

Answer:

$Let x = 2 We know: 2>1 and 2<3 ∴ x>1 and x<3 |x - 1| = x - 1 and |x - 3| = - (x - 3) = - x + 3 f (x) = |x - 1| + |x - 3| = x - 1 - x + 3 = 2 f' (x) = 0$

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Question 9:

If f (x) = $\frac{x^{2}}{|x|}, write \frac{d}{d x} (f (x))$ .

Answer:

$Case 1: x>0 |x| = x f (x) = \frac{x^{2}}{|x|} = \frac{x^{2}}{x} = x f' (x) = 1 Case 2: x<0 |x| = - x f (x) = \frac{x^{2}}{|x|} = \frac{x^{2}}{- x} = - x f' (x) = - 1 From case 1 and case 2, we have: f' (x) = \{\begin{cases} 1, if x > 0 \\ - 1, if x < 0 \end{cases}$

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Question 10:

Write the value of $\frac{d}{d x} (\log |x|)$ .

Answer:

Case 1: x>0 |x| = x . . . (1) \frac{d}{d x} (\log |x|) = \log x = \frac{1}{x} = \frac{1}{|x|} (from (1)) Case 2: x <0 |x| = - x . . . (2) \frac{d}{d x} (\log |x|) = \log (- x) = \frac{1}{- x} = \frac{1}{|x|} (from (2)) From case (1) and case(2), \frac{d}{d x} (\log |x|) = \frac{1}{|x|}

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Question 11:

If f (1) = 1, f' (1) = 2, then write the value of $\lim_{x \to 1} \frac{\sqrt{f (x)} - 1}{\sqrt{x} - 1}$ .

Answer:

$\lim_{x \to 1} \frac{\sqrt{f (x)} - 1}{\sqrt{x} - 1} = \lim_{x \to 1} \frac{\sqrt{f (x)} - 1}{\sqrt{x} - 1} \times \frac{\sqrt{f (x)} + 1}{\sqrt{f (x)} + 1} \times \frac{\sqrt{x} + 1}{\sqrt{x} + 1} = \lim_{x \to 1} \frac{(f (x) - 1) (\sqrt{x} + 1)}{(x - 1) (\sqrt{f (x)} + 1)} = \lim_{x \to 1} \frac{f (x) - 1}{x - 1} \times \lim_{x \to 1} \frac{(\sqrt{x} + 1)}{(\sqrt{f (x)} + 1)} = f' (1) \times \frac{1 + 1}{\sqrt{f (1)} + 1} = 2 \times \frac{2}{1 + 1} = 2$

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Question 12:

Write the derivative of f (x) = 3 |2 + x| at x = −3.

Answer:

$Let x = -3 We know: -3<-2 Thus, we have: x<-2 It gives x+2<0. ∴ |2 + x| = |x + 2| = - (x + 2) = - x - 2 f (x) = 3 |2 + x| = 3 (- x - 2) = - 3 x - 6 f' (x) = - 3 \frac{d}{d x} (x) - \frac{d}{d x} (6) = - 3$

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Question 13:

If |x| < 1 and y = 1 + x + x² + x³ + ..., then write the value of $\frac{d y}{d x}$ .

Answer:

The given series is a geometric series where a = 1 and r = x.
$f (x) = 1 + x + x^{2} + x^{3} + . . . = \frac{1}{1 - x} (Sum of the infinite series of a geometric series is \frac{a}{1 - r} .) f' (x) = \frac{- 1}{(1 - x)^{2}} \frac{d}{d x} (1 - x) = \frac{- 1}{(1 - x)^{2}} (- 1) = \frac{1}{(1 - x)^{2}}$

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Question 14:

If f (x) = $\log_{x_{2}}$ x³, write the value of f' (x).

Answer:

$f (x) = \log_{x^{2}} x^{3} = \frac{\log x^{3}}{\log x^{2}} (Change of base property) = \frac{3 \log x}{2 \log x} = \frac{3}{2} f' (x) = 0 (Since \frac{3}{2} is a constant)$

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