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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives
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Find the derivative of the given function from first principle : $ \cos \bigg( x - \large\frac{\pi}{8} \bigg)$

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Let $ f(x) = \cos \bigg( x - \large\frac{\pi}{8} \bigg) $. Accordingly, $ f(x+h) = \cos \bigg( x + h - \large\frac{\pi}{8} \bigg)$
By first principle
$ f'(x) = \lim\limits_{ h \to 0} \large\frac{f(x+h)-f(x)}{h}$
$ = \lim\limits_{ h \to 0} \large\frac{1}{h}$$ \bigg[ \cos \bigg(x+h - \large\frac{\pi}{8} \bigg)$$ - \cos \bigg( x - \large\frac{\pi}{8} \bigg) \bigg]$
$ = \lim\limits_{ h \to 0} \large\frac{1}{h}$$ \bigg[-2\sin \large\frac{\bigg(x+h-\Large\frac{\pi}{8}+x-\Large\frac{\pi}{8} \bigg)}{2}$$\sin \large\frac{\bigg(x+h-\Large\frac{\pi}{8}-x+\Large\frac{\pi}{8} \bigg)}{2} \bigg]$
$ = \lim\limits_{ h \to 0} \large\frac{1}{h}$$ \bigg[-2\sin\bigg( \large\frac{2x+h-\Large\frac{\pi}{4}}{2}\bigg)$$\sin \large\frac{h}{2} \bigg]$
$ = \lim\limits_{ h \to 0} \bigg[-\sin \bigg(\large\frac{2x+h-\Large\frac{\pi}{4} }{2}\bigg)$$\large\frac{\sin \bigg(\large\frac{h}{2}\bigg)}{\bigg(\Large\frac{h}{2}\bigg) }\bigg]$
$ = \lim\limits_{ h \to 0} \bigg[-\sin\bigg( \large\frac{2x+h-\Large\frac{\pi}{4} }{2}\bigg)\bigg]$$\lim\limits_{ \large\frac{h}{2} \to 0}\large\frac{\sin \bigg(\large\frac{h}{2}\bigg)}{\bigg(\Large\frac{h}{2}\bigg) }\qquad \bigg[ As \: h \to 0 \Rightarrow \large\frac{h}{2} \to 0 \bigg]$
$ = -\sin \bigg(\large\frac{2x+0-\Large\frac{\pi}{4} }{2}\bigg)$$.1$
$ = -\sin \bigg( x - \large\frac{\pi}{8} \bigg)$
answered Apr 12, 2014 by thanvigandhi_1
 

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