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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 : $ \sin (x+1)$

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Let $f(x) = \sin (x+1)$. Accordingly, $ f(x+h)= \sin (x+h+1)$
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}$$ [\sin (x+h+1)- \sin (x+1)]$
$ = \lim\limits_{h \to 0} \large\frac{1}{h}$$ \bigg[ 2 \cos \bigg( \large\frac{x+h+1+x+1}{2} \bigg) $$\sin \bigg( \large\frac{x+h+1-x-1}{2} \bigg) \bigg]$
$ = \lim\limits_{h \to 0} \large\frac{1}{h}$$ \bigg[ 2 \cos \bigg( \large\frac{2x+h+2}{2} \bigg) $$\sin \bigg( \large\frac{h}{2} \bigg) \bigg]$
$ = \lim\limits_{h \to 0} \bigg[ \cos \bigg( \large\frac{2x+h+2}{2} \bigg) $$. \large\frac{sin\bigg( \large\frac{h}{2}\bigg)}{\bigg(\large\frac{h}{2}\bigg)} \bigg]$
$ = \lim\limits_{h \to 0} \cos \bigg( \large\frac{2x+h+2}{2} \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 \qquad \bigg[ As \: h \to 0 \Rightarrow \large\frac{h}{2}$$ \to 0 \bigg]$
$ = \cos \bigg( \large\frac{2x+0+2}{2} \bigg) $$.1 \qquad \qquad \bigg[ \lim\limits_{x \to 0} \large\frac{ \sin x}{x}$$=1 \bigg]$
$ = \cos (x+1)$
answered Apr 11, 2014 by thanvigandhi_1
 

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