Question

If $x={\sin }^{-1}(3t-4t^3)$ and $y={\cos }^{-1}\left(\sqrt{1-t^2}\right)$, $\frac{dy}{dx}$ is equal to:

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{2}{5}$

Answer 1

The correct option is C

$\frac{1}{3}$

Explanation for the correct option

we know that, $\mathbf{3}\mathbf{}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{\theta }\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathbf{3}\mathit{\theta }\mathbf{-}\mathbf{4}{\mathit{\theta }}^{\mathbf{3}}\mathbf{)}$

So, $x=3{\sin }^{-1}t$

Now,

$\begin{array}{rcl}\frac{dx}{dt}& =& 3\frac{d}{dt}\left({\sin }^{-1}t\right)\\ & =& \frac{3}{\sqrt{1-t^2}}\end{array}$

Let $t=\sin \theta$, then

$\begin{array}{rcl}y& =& {\cos }^{-1}\left(\sqrt{1-{\sin }^2\theta }\right)\\ & =& {\cos }^{-1}\cos \theta \\ & =& \theta \\ & =& {\sin }^{-1}t\\ & \Rightarrow & \frac{dy}{dt}=\frac{1}{\sqrt{1-t^2}}\end{array}$

Now,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{dy}{dt}\times \frac{dt}{dx}\\ & =& \frac{1}{\sqrt{1-t^2}}\times \frac{\sqrt{1-t^2}}{3}\\ & =& \frac{1}{3}\end{array}$

Hence, option C is correct.