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Mathematics Question on Derivatives

If x=sin⁡−1(3t−4t3)x={{\sin }^{-1}}(3t-4{{t}^{3}}) and y=cos⁡−1(1−t2),y={{\cos }^{-1}}(\sqrt{1-{{t}^{2}}}), then dydx\frac{dy}{dx} is equal to

A

12\frac{1}{2}

B

23\frac{2}{3}

C

13\frac{1}{3}

D

25\frac{2}{5}

Answer

13\frac{1}{3}

Explanation

Solution

x=sin⁡−1(3t−4t3)x={{\sin }^{-1}}(3t-4{{t}^{3}}) and y=cos⁡−1(1−t2)y={{\cos }^{-1}}(\sqrt{1-{{t}^{2}}}) Put t=sin⁡θt=\sin \theta ...(i) Then, x=sin⁡−1(3sin⁡θ−4sin⁡3θ)x={{\sin }^{-1}}(3\sin \theta -4{{\sin }^{3}}\theta )
=sin⁡−1(sin⁡3θ)=3θ=3sin⁡−1t={{\sin }^{-1}}(\sin 3\theta )=3\theta =3{{\sin }^{-1}}t and y=cos⁡−11−sin⁡2θy={{\cos }^{-1}}\sqrt{1-{{\sin }^{2}}\theta }
=cos⁡−1(cos⁡θ)=θ=sin⁡−1t={{\cos }^{-1}}(\cos \theta )=\theta ={{\sin }^{-1}}t
Now, dxdt=31−t2\frac{dx}{dt}=\frac{3}{\sqrt{1-{{t}^{2}}}} dxdt=11−t2\frac{dx}{dt}=\frac{1}{\sqrt{1-{{t}^{2}}}}
⇒\Rightarrow dydx=dydt.dtdx\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx} dydx=11−t2×1−t23=13\frac{dy}{dx}=\frac{1}{\sqrt{1-{{t}^{2}}}}\times \frac{\sqrt{1-{{t}^{2}}}}{3}=\frac{1}{3}