Question

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

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

Answer 1

Answer: (D)

$\frac{1}{3}$

Given $x={\sin }^{-1}(3t-{4t}^3)$ and $y={\cos }^{-1}\left(\sqrt{1-t^2}\right)$
On differentiating w.r.t $t$ respectively, we get
$\frac{dy}{dx}=\frac{1}{\sqrt{1-t^2}}$ and
$\frac{dy}{dt}=\frac{3}{\sqrt{1-t^2}}$
$\therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{\left(\frac{1}{\sqrt{1-t^2}}\right)}{3\left(\frac{3}{\sqrt{1-t^2}}\right)}$
$\Rightarrow \frac{dy}{dx}=\frac{1}{3}$