question_answer

If \[y={{\sin }^{-1}}(\sin \,x),\] \[x\in \left[ \frac{\pi }{2},\,\,\frac{3\pi }{2} \right],\] then evaluate \[\frac{dy}{dx}.\]

Answer:

We have, \[y={{\sin }^{-1}}(\sin x),\] \[x\in \left[ \frac{\pi }{2},\,\,\frac{3\pi }{2} \right]\]             \[={{\sin }^{-1}}[\sin (\pi -x)]\] \[\Rightarrow \]   \[y=\pi -x,\] \[x\in \left[ \frac{\pi }{2},\,\,\frac{3\pi }{2} \right]\] \[\because \]       \[\frac{\pi }{2}\le x\le \frac{3\pi }{2}\] \[\Rightarrow \]   \[-\,\frac{3\pi }{2}\le -\,x\le -\,\frac{\pi }{2}\] \[\Rightarrow \]   \[-\,\frac{\pi }{2}\le \pi -\,x\le \frac{\pi }{2}\] and \[{{\sin }^{-1}}(sin\theta )=\theta ,\] \[\forall \,\,\theta \ \in \left[ -\,\frac{\pi }{2},\,\,\frac{\pi }{2} \right]\] \[\therefore \]      \[\frac{dy}{dx}=-\,1\]