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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives
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Find the derivative of the given function $\sec x$

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Let $f(x) = \sec x$. Accordingly, from the first principle,
$ f'(x) = \lim\limits_{h \to 0}= \large\frac{f(x+h)-f(x)}{h}$
$ = \lim\limits_{h \to 0}= \large\frac{\sec(x+h)-\sec x}{h}$
$ = \lim\limits_{h \to 0}= \large\frac{1}{h}$$ \bigg[ \large\frac{1}{\cos (x+h)}$$- \large\frac{1}{\cos x}\bigg]$
$ = \lim\limits_{h \to 0}= \large\frac{1}{h}$$ \bigg[ \large\frac{\cos x - \cos(x+h)}{\cos x \cos(x+h)} \bigg]$
$ = \large\frac{1}{\cos x}$$. \lim\limits_{h \to 0}= \large\frac{1}{h}$$ \bigg[ \large\frac{-2 \sin \bigg( \Large\frac{x+x+h}{2} \bigg) \sin \bigg( \Large\frac{x-x-h}{2} \bigg)}{\cos (x+h)} \bigg]$
$ = \large\frac{1}{\cos x}$$. \lim\limits_{h \to 0}= \large\frac{1}{h}$$ \bigg[ \large\frac{-2 \sin \bigg( \Large\frac{2x+h}{2} \bigg) \sin \bigg( -\Large\frac{h}{2} \bigg)}{\cos (x+h)} \bigg]$
$ = \large\frac{1}{\cos x}$$. \lim\limits_{h \to 0}= \large\frac{ \bigg[\sin \bigg( \Large\frac{2x+h}{2} \bigg) \Large\frac{sin\bigg( \Large\frac{h}{2} \bigg)}{\bigg( \Large\frac{h}{2} \bigg)} \bigg]}{\cos (x+h)}$
$ = \large\frac{1}{\cos x}$$. \lim\limits_{\large\frac{h}{2} \to 0}= \large\frac{sin \bigg( \Large\frac{h}{2} \bigg)}{ \bigg( \Large\frac{h}{2} \bigg)}$$\lim\limits_{h \to 0}= \Large\frac{\sin \bigg( \Large\frac{2x+h}{2} \bigg)}{\cos (x+h)}$
$ = \large\frac{1}{\cos x}.1. \large\frac{\sin x}{\cos x}$
$ = \sec x \tan x$
answered Apr 10, 2014 by thanvigandhi_1

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