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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives
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Find the derivative of the given function $ 5\sin x-6 \cos x+7$

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Let $f(x) = 5\sin x-6 \cos x+7$. 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{1}{h}$$ [5 \sin (x+h)-6 \cos(x+h)+7-5\sin x+6 \cos x-7]$
$ = \lim\limits_{h \to 0} \large\frac{1}{h}$$ [5 \{ \sin (x+h)-\sin x\}-6\{ \cos (x+h)-\cos x\}]$
$ = 5 \lim\limits_{h \to 0} \large\frac{1}{h}$$ [\sin (x+h)-\sin x] -6 \lim\limits_{h \to 0} \large\frac{1}{h}$$ [\cos (x+h)-\cos x]$
$ = 5 \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ 2\cos \bigg( \large\frac{x+h+x}{2}\bigg)$$ \sin \bigg( \large\frac{x+h-x}{2} \bigg) \bigg]$$- 6 \lim\limits_{h \to 0} \large\frac{\cos x \cos h- \sin x \sin h- \cos x}{h}$
$ = 5 \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ 2\cos \bigg( \large\frac{2x+h}{2}\bigg)$$ \sin \large\frac{h}{2} \bigg]$$- 6 \lim\limits_{h \to 0} \bigg[\large\frac{-\cos x (1-\cos h)- \sin x \sin h}{h}\bigg]$
$ 5 \lim\limits_{h \to 0} \bigg[ \cos \bigg(\Large\frac{2x+h}{2} \bigg) \Large\frac{sin \large\frac{h}{2}}{\large\frac{h}{2}} \bigg]$$- 6 \lim\limits_{h \to 0} \bigg[\large\frac{-\cos x (1-\cos h)}{h}-\large\frac{ \sin x \sin h}{h}\bigg]$
$ 5 \bigg[ \lim\limits_{h \to 0}\cos \bigg(\large\frac{2x+h}{2} \bigg)\bigg] \bigg[\lim\limits_{\Large\frac{h}{2} \to 0}\Large\frac{sin \large\frac{h}{2}}{\large\frac{h}{2}} \bigg]$$- 6 \bigg[(-\cos x) \bigg(\lim\limits_{h \to 0}\large\frac{ 1-\cos h}{h}\bigg)$$-\sin x \lim\limits_{h \to 0}\bigg(\large\frac{\sin h}{h}\bigg)\bigg]$
$ = 5 \cos x.1-6[ ( - \cos x).(0)- \sin x.1]$
$ = 5 \cos x+ 6 \sin x$
answered Apr 10, 2014 by thanvigandhi_1
 

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