Question

Derivative of ${\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ with respect to ${\sin }^{-1}(3x-4x^3)$ is

• A. $\frac{1}{\sqrt{1-x^2}}$
• B. $\frac{3}{\sqrt{1-x^2}}$
• C. $3$
• D. $\frac{1}{3}$

Answer 1

The correct option is D $\frac{1}{3}$

Let $x=\sin \theta$

$\Rightarrow {\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$

$={\tan }^{-1}\left(\frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }}\right)$

$={\tan }^{-1}\left(\frac{\sin \theta }{\cos \theta }\right)=\theta =u$
$\frac{du}{d\theta }=1$

again, replacing $x=\sin \theta$

${\sin }^{-1}(3x-4x^3)$

$={\sin }^{-1}(3\sin \theta -4{\sin }^3\theta )(\because \sin 3\theta =3\sin \theta -4{\sin }^3\theta )$

$={\sin }^{-1}\left(\sin 3\theta \right)=3\theta =v$

$\frac{dv}{d\theta }=3$

$\frac{du}{dv}=\frac{1}{3}$

$\therefore$ derivative of ${\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ wrt ${\sin }^{-1}(3x-4x^3)$ is equal to $\frac{1}{3}$.