Question:
If $\mathrm{y}=\sin ^{-1}\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right)$, write the value of $\frac{\mathrm{dy}}{\mathrm{dx}}$ for $\mathrm{x}>1$
Solution:
$y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=2 \tan ^{-1} x$
So,
$\frac{\mathrm{dy}}{\mathrm{dx}}=2 \cdot \frac{1}{1+\mathrm{x}^{2}}$
$=\frac{2}{1+x^{2}}$
So, answer is $\frac{d y}{d x}=\frac{2}{1+x^{2}}$ (Ans)
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