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

Question

Dr. Abubacker · Accepted Answer

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

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

Explanation

Reviewed by Dr. Abubacker

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