2x at x= 2 - Vidyadip

Question

$\sin2\text{x}$ at $\text{x}=\frac{\pi}{2}$

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Answer

We have, $\text{f}\text{(x)}=\sin2\text{x}$Therefore,
$\text{f}'\text{(a)}=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{a}+\text{h})-\text{f}\text{(a)}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}\Big(\frac{\pi}{2}+\text{h}\Big)-\text{f}\Big(\frac{\pi}{2}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\Big(\frac{\pi}{2}+\text{h}\Big)-\sin2\Big(\frac{\pi}{2}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Big(\frac{\pi}{2}\times2+2\text{h}\Big)-\sin(\pi)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-\cos2\text{h}-0}{\text{h}}$
$=-2$
Therefore, $\text{f}'\Big(\frac{\pi}{2}\Big)=-2$

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