Question

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 $x=0$, is

• A. $\frac{1}{8}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. 1

Answer 1

Answer: (B)

Let $y=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$
Putting $x=\tan \theta$, we get
$y =\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\tan \frac{\theta}{2}\right) $
$=\frac{1}{2} \tan ^{-1} x$
On differentiating both sides w.r.t. $x$, we get
$\frac{d y}{d x}=\frac{1}{2\left(1+x^{2}\right)}$
and $z=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)$
Putting $x=\sin \theta$, we get
$z =\tan ^{-1}\left(\frac{2 \sin \theta \cos \theta}{\cos 2 \theta}\right)=\tan ^{-1}(\tan 2 \theta) $
$\Rightarrow z =2 \theta=2 \sin ^{-1} x$
On differentiating both sides w.r.t. $x$, we get
$\frac{d z}{d x}=\frac{2}{\sqrt{1-x^{2}}}$
Thus, $\frac{d y}{d z}=\frac{d y / d x}{d z / d x}=\frac{1}{4\left(1+x^{2}\right)} \sqrt{1-x^{2}}$
$\therefore \left[\frac{d y}{d z}\right]_{x=0}=\frac{1}{4}$