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

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Q. The derivative of $sin^{-1}\left(\frac{2x}{1+x^{2}}\right)$ with respect to $tan^{-1}\left(\frac{2x}{1-x^{2}}\right)$ is

Continuity and Differentiability

Solution:

Let $u=sin^{-1}\left(\frac{2x}{1+x^{2}}\right)=2\,tan^{-1}\,x$
and $v=tan^{-1}\left(\frac{2x}{1-x^{2}}\right)=2\,tan^{-1}\,x$
$\Rightarrow u = v$
$\Rightarrow \frac{du}{dx} = \frac{dv}{dx}$
$\therefore \frac{du}{dv} = \frac{\left(du/dx\right)}{\left(dvdx/\right)} = 1$