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Question:
If $y=\sin ^{-1} x+\cos ^{-1} x$, find $\frac{d y}{d x}$.
Solution:
We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$
So, here $y=\sin ^{-1} x+\cos ^{-1} x$
$=\frac{\pi}{2}$ which is a constant.
Also, $\sin ^{-1} x$ and $\cos ^{-1} x$ exist only when $-1 \leq x \leq 1$
So, $\frac{d y}{d x}=0$ when $x \in[-1,1]$ and does not exist for all other values of $x$.
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