We have, (x)=2 '(a)= _ h 0 (a+h)-f(a) h = _ h 0 f ( 2 +h )-f ( 2 ) h = _ h 0 f ( 2 +h )-2 2 h = _ h 0 -2sinh-0 h =-2 [ _ 0 =1 ] ' ( 2 )=-2

2 cosx at x = 2

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2 cosx at x = $\frac{\pi}{2}$

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Answer

$$We have,
$\because\text{f(x)}=2\cos\text{x}$
$\therefore\text{f}'\text{(a)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{(a+h)}-\text{f(a)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}\Big(\frac{\pi}{2}+\text{h}\Big)-\text{f}\Big(\frac{\pi}{2}\Big)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}\Big(\frac{\pi}{2}+\text{h}\Big)-2\cos\frac{\pi}{2}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-2\text{sinh}-0}{\text{h}}$
$=-2\ \Big[\because\lim_\limits{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$\therefore\text{f}'\Big(\frac{\pi}{2}\Big)=-2$

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