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Mathematics Question on Differentiability

If x=sin−1(3t−4t3)x=sin^{-1}\left(3t-4t^{3}\right) and y=cos−1(1−t2)y=cos^{-1}\left(\sqrt{1-t^{2}}\right), 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}\left(3 t-4 t^{3}\right)
and y=cos⁡−1(1−t2)y=\cos ^{-1}\left(\sqrt{1-t^{2}}\right)
Put t=sin⁡θ      …(i)t=\sin \theta \,\,\,\,\,\,\dots(i)
Then, x=sin⁡−1(3sin⁡θ−4sin⁡3θ)x =\sin ^{-1} \left(3 \sin \theta-4 \sin ^{3} \theta\right)
=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{d x}{d t} =\frac{3}{\sqrt{1-t^{2}}}
dydt=11−t2\frac{d y}{d t}=\frac{1}{\sqrt{1-t^{2}}}
⇒ dydx=dydt⋅dtdx\Rightarrow \,\frac{d y}{d x} =\frac{d y}{d t} \cdot \frac{d t}{d x}
dydx=11−t2×1−t23\frac{d y}{d x} =\frac{1}{\sqrt{1-t^{2}}} \times \frac{\sqrt{1-t^{2}}}{3}
=13=\frac{1}{3}