Question

Derivative of $ta{n}^{-1}\left(\frac{X}{\sqrt{1-x^2}}\right)$ with respect to $si{n}^{-1}\left(3x-4x^3\right)$ 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 $u=ta{n}^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ and
$v=si{n}^{-1}\left(3x-4x^3\right)$
put $x=sin\theta$, then
$u=ta{n}^{-1}\left(\frac{sin\theta }{\sqrt{1-si{n}^2\theta }}\right)$
and $v=si{n}^{-1}\left(3sin\theta -4si{n}^3\theta \right)$
$\Rightarrow u=ta{n}^{-1}\left(\frac{sin\theta }{cos\theta }\right)$
and $v=si{n}^{-1}(sin3\theta )$
$\Rightarrow u=ta{n}^{-1}(tan\theta )$
and $v=si{n}^{-1}(sin3\theta )$
$\Rightarrow u=\theta$ and $v=3\theta$
$\Rightarrow u=si{n}^{-1}x$ and $v=3si{n}^{-1}x$
On differentiating both sides w.r.t. x, we get
$\frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}$ and $\frac{dv}{dx}=3\times \frac{1}{\sqrt{1-x^2}}$
$\therefore \frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{\frac{1}{\sqrt{1-x^2}}}{\frac{3}{\sqrt{1-x^2}}}$
$=\frac{1}{3}$