Solve this
Question:

If $y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$, find $\frac{d y}{d x}$

Solution:

$-1<\frac{1-x^{2}}{1+x^{2}} \leq 1$ holds for all $x \in \mathbb{R}$.

So, $y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\frac{\pi}{2}$, for all $x \in \mathbb{R}$

$\left(\because \sin ^{-1} \mathrm{~m}+\cos ^{-1} \mathrm{~m}=\frac{\pi}{2}, \mathrm{~m} \in[-1,1]\right)$

Hence, $\frac{d y}{d x}=0$, for all $x \in \mathbb{R}$.

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