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Q.
The derivative of $\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$ with respect to $\tan ^{-1}\left(\frac{2 x \sqrt{1- x ^{2}}}{1-2 x ^{2}}\right)$ at $x =\frac{1}{2}$ is :
JEE MainJEE Main 2020Continuity and Differentiability
Solution:
Let $f =\tan ^{-1}\left(\frac{\sqrt{1+ x ^{2}}-1}{ x }\right)$
Put $x=\tan \theta $
$\Rightarrow \theta=\tan ^{-1} x$
$f =\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)$
$f =\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)=\frac{\theta}{2}$
$f =\frac{\tan ^{-1} x }{2} $
$\Rightarrow \frac{ df }{ dx }=\frac{1}{2\left(1+ x ^{2}\right)} \dots$(i)
Let $g=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)$
Put $x=\sin \theta $
$\Rightarrow \theta=\sin ^{-1} x$
$g =\tan ^{-1}\left(\frac{2 \sin \theta \cos \theta}{1-2 \sin ^{2} \theta}\right)$
$g=\tan ^{-1}(\tan 2 \theta)=2 \theta$
$g=2 \sin ^{-1} x$
$\frac{ dg }{ dx }=\frac{2}{\sqrt{1- x ^{2}}} \dots$(ii)
$\frac{d f}{d g}=\frac{1}{2\left(1+x^{2}\right)} \frac{\sqrt{1-x^{2}}}{2}$
at $x=\frac{1}{2}\left(\frac{d f}{d g}\right)_{x=\frac{1}{2}}=\frac{\sqrt{3}}{10}$