Question

If $ x={{\sin }^{-1}}(3t-4{{t}^{3}}) $ and $ y={{\cos }^{-1}}(\sqrt{1-{{t}^{2}}}), $ then $ \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

$ x={{\sin }^{-1}}(3t-4{{t}^{3}}) $ and $ y={{\cos }^{-1}}(\sqrt{1-{{t}^{2}}}) $ Put $ t=\sin \theta $ ...(i) Then, $ x={{\sin }^{-1}}(3\sin \theta -4{{\sin }^{3}}\theta ) $
$={{\sin }^{-1}}(\sin 3\theta )=3\theta =3{{\sin }^{-1}}t $ and $ y={{\cos }^{-1}}\sqrt{1-{{\sin }^{2}}\theta } $
$={{\cos }^{-1}}(\cos \theta )=\theta ={{\sin }^{-1}}t $
Now, $ \frac{dx}{dt}=\frac{3}{\sqrt{1-{{t}^{2}}}} $ $ \frac{dx}{dt}=\frac{1}{\sqrt{1-{{t}^{2}}}} $
$ \Rightarrow $ $ \frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx} $ $ \frac{dy}{dx}=\frac{1}{\sqrt{1-{{t}^{2}}}}\times \frac{\sqrt{1-{{t}^{2}}}}{3}=\frac{1}{3} $