# Solve this

Question:

If $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}$ and $y=\sin ^{-1}(\sin x)$, find $\frac{d y}{d x}$

Solution:

For $\mathrm{x} \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$

$y=\sin ^{-1}(\sin x)$

$=\sin ^{-1}(\sin (\pi-(\pi-x))$

(to get $y$ in principal range of $\sin ^{-1} x$ )

i.e.,

$y=\pi-x$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=-1$

From the last problem we see that $\frac{d y}{d x}_{x \rightarrow \frac{\pi^{-}}{2}}=1$ and $\frac{d y}{d x}_{x \rightarrow \frac{\pi^{+}}{2}}=-1$

So, $y$ is not differentiable at $x=\frac{\pi}{2}$.

Extending this, we can say that $y$ is not differentiable at $x=(2 n+1) \frac{\pi}{2}$

So, for $\mathrm{X} \in\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\left\{\begin{array}{c}-1, \mathrm{x} \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right) \\ \text { does not exist at } \mathrm{x}=\frac{\pi}{2}, \frac{3 \pi}{2}\end{array}\right.$ (Ans)

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