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  4. The derivative of sin -1 (2 x/1+x2) with respect to cos -1 (1-x2/1+x2) is

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Q. The derivative of $\sin ^{-1} \frac{2 x}{1+x^{2}}$ with respect to $\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}$ is

BITSATBITSAT 2007

Solution:

Let $P =\sin ^{-1} \frac{2 x }{1+ x ^{2}}=2 \tan ^{-1} x$
And $q =\cos ^{-1} \frac{1- x ^{2}}{1+ x ^{2}}=2 \tan ^{-1} x$
$\therefore \frac{ dp }{ dx }=\frac{2}{1+ x ^{2}}=$ and $\frac{ dq }{ dx }=\frac{2}{1+ x ^{2}}$
$\Rightarrow \frac{ dp }{ dq }=\frac{\frac{ dp }{ dx }}{\frac{2}{ dq }}=\frac{\frac{2}{1+ x ^{2}}}{\frac{2}{1+ x ^{2}}}=1$