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

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Let $f(x) = \sin x \: \cos x$. 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{\sin (x+h) \cos (x+h)-\sin x \cos x}{h}$
$ = \lim\limits_{h \to 0}\large\frac{1}{2h}$$[ 2 \sin(x+h) \cos (x+h)-2\sin x\cos x]$
$ = \lim\limits_{h \to 0}\large\frac{1}{2h}$$[\sin 2(x+h)- \sin 2x]$
$ = \lim\limits_{h \to 0}\large\frac{1}{2h}$$ \bigg[2 \cos \large\frac{2x+2h+2x}{2}$$.\sin\large\frac{2x+2h-2x}{2} \bigg]$
$ = \lim\limits_{h \to 0}\large\frac{1}{h}$$\bigg[ \cos \large\frac{4x+2h}{2}$$\sin\large\frac{2h}{2}\bigg]$
$ = \lim\limits_{h \to 0}\large\frac{1}{h}$$[\cos (2x+h) \sin h]$
$ = \lim\limits_{h \to 0} \cos (2x+h). \lim\limits_{h \to 0}\large\frac{\sin h}{h}$
$ = \cos(2x+0).1$
$ =\cos 2x$
answered Apr 9, 2014 by thanvigandhi_1

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