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Q.
The derivative of $sin^{-1}\left(\frac{2x}{1+x^{2}}\right)$ with respect to $cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$ is
Continuity and Differentiability
Solution:
Let $p=sin^{-1}\left[\frac{2x}{1+x^{2}}\right]=2\,tan^{-1}\,x$
and $q=cos^{-1}\left[\frac{1-x^{2}}{1+x^{2}}\right]=2\,tan^{-1}\,x$
$\Rightarrow \frac{dp}{dx}=\frac{2}{1+x^{2}}$,
$\frac{dq}{dx}=\frac{2}{1+x^{2}}$
$\therefore \frac{dp}{dq}=\frac{\frac{dp}{dx}}{\frac{dq}{dx}}$
$=\frac{2}{\frac{\left(1+x^{2}\right)}{\frac{2}{\left(1+x^{2}\right)}}}=1$