Question

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

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

Answer 1

Answer: (D)

Let $u=tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)$ and
$v=sin ^{-1}\left(3x-4x^{3}\right)$
put $x=sin \,\theta$, then
$u=tan ^{-1}\left(\frac{sin \,\theta}{\sqrt{1-sin ^{2} \theta}}\right)$
and $v=sin ^{-1}\left(3 \,sin \,\theta-4 \, sin ^{3} \,\theta\right) $
$ \Rightarrow u=tan ^{-1}\left(\frac{sin\, \theta}{cos \,\theta}\right)$
and $ v= sin ^{-1} (sin \,3 \,\theta)$
$ \Rightarrow u=tan ^{-1}(tan\, \theta)$
and $ v=sin ^{-1}(sin \,3 \,\theta)$
$ \Rightarrow u=\theta$ and $ v=3 \,\theta$
$ \Rightarrow u=sin ^{-1} \,x$ and $v=3\,sin ^{-1} \,x $
On differentiating both sides w.r.t. x, we get
$\frac{du}{dx}=\frac{1}{\sqrt{1-x^{2}}}$ and $\frac{dv}{dx}=3 \times \frac{1}{\sqrt{1-x^{2}}}$
$\therefore \frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{\frac{1}{\sqrt{1-x^{2}}}}{\frac{3}{\sqrt{1-x^{2}}}}$
$=\frac{1}{3}$