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Integration as Antiderivative

Differentiate...

Question

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)

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Solution

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

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

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

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

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

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

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

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

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

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

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

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