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Standard Formulae - 1

Differentiate...

Question

Differentiate the following functions from first principles:

(i) $e^{\sqrt{\cot x}}$
(ii) x²e^x
(iii) log cosec x
(iv) sin⁻¹ (2x + 3)

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Solution

$(i) Let f (x) = e^{\sqrt{\cot x}} \Rightarrow f (x + h) = e^{\sqrt{\cot (x + h)}} ∴ \frac{d}{d x} \{f (x)\} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{e^{\sqrt{\cot (x + h)}} - e^{\sqrt{\cot x}}}{h} = \lim_{h \to 0} \frac{e^{\sqrt{\cot x}} (e^{\sqrt{\cot (x + h)} - \sqrt{\cot x}} - 1)}{h} = e^{\sqrt{\cot x}} \lim_{h \to 0} (\frac{e^{\sqrt{\cot (x + h)} - \sqrt{\cot x}} - 1}{\sqrt{\cot (x + h)} - \sqrt{\cot x}}) \times (\frac{\sqrt{\cot (x + h)} - \sqrt{\cot x}}{h}) = e^{\sqrt{\cot x}} \lim_{h \to 0} \frac{(\sqrt{\cot (x + h)} - \sqrt{\cot x})}{h} \times \frac{\sqrt{\cot (x + h)} + \sqrt{\cot x}}{\sqrt{\cot (x + h)} + \sqrt{\cot x}} [∵ \lim_{x \to 0} \frac{e^{x} - 1}{x} = 1 and rationalizing the numerator] = e^{\sqrt{\cot x}} \lim_{h \to 0} \frac{\cot (x + h) - \cot x}{h (\sqrt{\cot (x + h)} + \sqrt{\cot x})} = e^{\sqrt{\cot x}} \lim_{h \to 0} \frac{\frac{\cot (x + h) \cot x + 1}{\cot (x - x - h)}}{h (\sqrt{\cot (x + h)} + \sqrt{\cot x})} [∵ \cot (A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A}] = e^{\sqrt{\cot x}} \lim_{h \to 0} \frac{\cot (x + h) \cot x + 1}{\cot (- h) \times h (\sqrt{\cot (x + h)} + \sqrt{\cot x})} = - e^{\sqrt{\cot x}} \lim_{h \to 0} \frac{\cot (x + h) \cot x + 1}{(\frac{h}{\tan h}) (\sqrt{\cot (x + h)} + \sqrt{\cot x})} = \frac{e^{\sqrt{\cot x}} \times (\cot^{2} x + 1)}{2 \sqrt{\cot x}} [∵ \lim_{x \to 0} \frac{\tan x}{x} = 1] = - \frac{e^{\sqrt{\cot x}} \times {cosec}^{2} x}{2 \sqrt{\cot x}} [∵ (1 + \cot^{2} x) = {cosec}^{2} x] ∴ \frac{d}{d x} (e^{\sqrt{c o t x}}) = - \frac{e^{\sqrt{\cot x}} \times {cosec}^{2} x}{2 \sqrt{\cot x}}$

$(i i) Let f (x) = x^{2} e^{x} \Rightarrow f (x + h) = {(x + h)}^{2} e^{(x + h)} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{{(x + h)}^{2} e^{(x + h)} - x^{2} e^{x}}{h} = \lim_{h \to 0} (\frac{x^{2} e^{x + h} - x^{2} e^{x}}{h} + \frac{2 x h e^{(x + h)}}{h} + \frac{h^{2} e^{(x + h)}}{h}) = \lim_{h \to 0} (\frac{x^{2} e^{x} (e^{(x + h) - x} - 1)}{h} + 2 x e^{(x + h)} + h e^{(x + h)}) = \lim_{h \to 0} (x^{2} e^{x} \frac{(e^{h} - 1)}{h} + 2 x e^{(x + h)} + h e (x + h)) = x^{2} e^{x} + 2 x e^{x} + 0 x e^{x} [∵ \lim_{x \to 0} \frac{e^{x} - 1}{x} = 1] ∴ \frac{d}{d x} (x^{2} e^{x}) = e^{x} (x^{2} + 2 x)$

$(i i i) Let f (x) = \log cosec x \Rightarrow f (x + h) = \log cosec (x + h) ∴ \frac{d}{d x} \{f (x)\} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\log cosec (x + h) - \log cosec x}{h} = \lim_{h \to 0} \frac{\log \{\frac{cosec (x + h)}{cosec x}\}}{h} = \lim_{h \to 0} \frac{\log \{1 + (\frac{\sin x}{\sin (x + h)} - 1)\}}{h} = \lim_{h \to 0} \{\frac{\log \{1 + (\frac{\sin x - \sin (x + h)}{\sin (x + h)})\}}{\{\frac{\sin x - \sin (x + h)}{\sin (x + h)}\}}\} \frac{\{\frac{\sin x - \sin (x + h)}{\sin (x + h)}\}}{h} = \lim_{h \to 0} \frac{2 \cos (\frac{x + x + h}{2}) \sin (\frac{x - x - h}{2})}{\sin (x + h) h} [∵ \lim_{x \to 0} \frac{\log (1 + x)}{x} = 1 and \sin A - \sin B = 2 \cos (\frac{A + B}{2}) \sin (\frac{A - B}{2})] = \lim_{h \to 0} \frac{2 \cos (\frac{2 x + h}{2})}{\sin (x + h) (- 2)} \{\frac{\sin (- \frac{h}{2})}{- \frac{h}{2}}\} [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] = - \cot x ∴ \frac{d}{d x} (\log cosec x) = - \cot x$

$(i v) Let f (x) = \sin^{- 1} (2 x + 3) \Rightarrow f (x + h) = \sin^{- 1} (2 (x + h) + 3) \Rightarrow f (x + h) = \sin^{- 1} (2 x + 2 h + 3) ∴ \frac{d}{d x} \{f (x)\} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\sin^{- 1} (2 x + 2 h + 3) - \sin^{- 1} (2 x + 3)}{h} = \lim_{h \to 0} \frac{\sin^{- 1} [(2 x + 2 h + 3) \sqrt{1 - {(2 x + 3)}^{2}} - (2 x + 3) \sqrt{1 - {(2 x + 2 h + 3)}^{2}}]}{h} [∵ \sin^{- 1} x - \sin^{- 1} y = \sin^{- 1} [x \sqrt{1 - y^{2}} - y \sqrt{1 - x^{2}}]] = \lim_{h \to 0} \frac{\sin^{- 1} z}{z} \times \frac{z}{h} where, z = (2 x + 2 h + 3) \sqrt{1 - {(2 x + 3)}^{2}} - (2 x + 3) \sqrt{1 - {(2 x + 2 h + 3)}^{2}} and \lim_{h \to 0} \frac{\sin^{- 1} h}{h} = 1 = \lim_{h \to 0} \frac{z}{h} = \lim_{h \to 0} \frac{(2 x + 2 h + 3) \sqrt{1 - {(2 x + 3)}^{2}} - (2 x + 3) \sqrt{1 - {(2 x + 2 h + 3)}^{2}}}{h} = \lim_{h \to 0} \frac{{(2 x + 2 h + 3)}^{2} \{1 - {(2 x + 3)}^{2}\} - {(2 x + 3)}^{2} \{1 - {(2 x + 2 h + 3)}^{2}\}}{h \{(2 x + 2 h + 3) \sqrt{1 - {(2 x + 3)}^{2}} + (2 x + 3) \sqrt{1 - {(2 x + 2 h + 3)}^{2}}\}} [Rationalizing numerator] = \lim_{h \to 0} \frac{[{(2 x + 3)}^{2} + 4 h^{2} + 4 h (2 x + 3)] \{1 - {(2 x + 3)}^{2}\} - {(2 x + 3)}^{2} [1 - {(2 x + 3)}^{2} - 4 h^{2} - 4 h (2 x + 3)]}{h \{(2 x + 2 h + 3) \sqrt{1 - {(2 x + 3)}^{2}} + (2 x + 3) \sqrt{1 - {(2 x + 2 h + 3)}^{2}}\}} = \lim_{h \to 0} \frac{[{(2 x + 3)}^{2} + 4 h^{2} + 4 h (2 x + 3) - {(2 x + 3)}^{4} - 4 h^{2} {(2 x + 3)}^{2} - 4 h {(2 x + 3)}^{3} - {(2 x + 3)}^{2} + {(2 x + 3)}^{4} + 4 h^{2} {(2 x + 3)}^{2} + 4 h {(2 x + 3)}^{3}]}{h \{(2 x + 2 h + 3) \sqrt{1 - {(2 x + 3)}^{2}} + (2 x + 3) \sqrt{1 - {(2 x + 2 h + 3)}^{2}}\}} = \lim_{h \to 0} \frac{4 h [h + (2 x + 3)]}{h \{(2 x + 2 h + 3) \sqrt{1 - {(2 x + 3)}^{2}} + (2 x + 3) \sqrt{1 - {(2 x + 2 h + 3)}^{2}}\}} = \frac{4 (2 x + 3)}{(2 x + 3) \sqrt{1 - {(2 x + 3)}^{2}} + (2 x + 3) \sqrt{1 - {(2 x + 3)}^{2}}} = \frac{4 (2 x + 3)}{2 (2 x + 3) \sqrt{1 - {(2 x + 3)}^{2}}} = \frac{2}{\sqrt{1 - {(2 x + 3)}^{2}}} ∴ \frac{d}{d x} \{\sin^{- 1} (2 x + 3)\} = \frac{2}{\sqrt{1 - {(2 x + 3)}^{2}}}$

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