Question

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

• A. $1/3$
• B. $2/5$
• C. $3/2$
• D. $2/3$

Hint:

$\sin 3\theta = 3\sin\theta - 4\sin^3\theta$, $\cos 2\theta = 2\cos^2\theta - 1$.

Answer 1

The Correct Option is D

Step 1: $\sin^{-1}(3t-4t^3) = 3\sin^{-1}t$, $\cos^{-1}(2t-1) = 2\cos^{-1}\sqrt{t}$.}
Step 2: $\frac{dx}{dt} = \frac{3}{\sqrt{1-t^2}}$, $\frac{dy}{dt} = -\frac{2}{\sqrt{t}\sqrt{1-t}}$. Then $\frac{dy}{dx} = \frac{-2/\sqrt{t(1-t)}}{3/\sqrt{1-t^2}} = \frac{2}{3}$.}
Step 3: Final Answer: $\frac{2}{3}$.}