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Question: If \[y={{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)\] then \[\dfrac{dy}{dx}\] = ? A. \[\dfrac{3}{\sq...

If y=sin⁡−1(3x−4x3)y={{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right) then dydx\dfrac{dy}{dx} = ?
A. 31−x2\dfrac{3}{\sqrt{1-{{x}^{2}}}}
B. −41−x2\dfrac{-4}{\sqrt{1-{{x}^{2}}}}
C. 31+x2\dfrac{3}{\sqrt{1+{{x}^{2}}}}
D. none of these

Explanation

Solution

Hint: In the above question we will suppose the value of x is equal to sin⁡θ\sin \theta and by substituting it we get 3sin⁡θ−4sin⁡3θ3\sin \theta -4{{\sin }^{3}}\theta which is equal to sin⁡3θ\sin 3\theta . Also, we will use the property of inverse trigonometric function that sin⁡−1sin⁡=x{{\sin }^{-1}}\sin =x where −π2≤x≤π2\dfrac{-\pi }{2}\le x\le \dfrac{\pi }{2}.

Complete step-by-step answer:

We have been given y=sin⁡−1(3x−4x3)y={{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)
Let us suppose x=sin⁡θx=\sin \theta
⇒y=sin⁡−1(3sin⁡θ−4sin⁡3θ)\Rightarrow y={{\sin }^{-1}}\left( 3\sin \theta -4{{\sin }^{3}}\theta \right)
We know the formula of trigonometry, i.e. sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta
So by using this formula, we get as follows:
y=sin⁡−1sin⁡3θy={{\sin }^{-1}}\sin 3\theta
Since we know the property of inverse trigonometric function that sin⁡−1sin⁡A=A{{\sin }^{-1}}\sin A=A where −π2≤A≤π2\dfrac{-\pi }{2}\le A\le \dfrac{\pi }{2}
So by using this property, we get as follows:
⇒y=sin⁡−1sin⁡3θ=3θ\Rightarrow y={{\sin }^{-1}}\sin 3\theta =3\theta
Since x=sin⁡θx=\sin \theta
On taking sine inverse on both sides we get as follows:

& \Rightarrow {{\sin }^{-1}}x={{\sin }^{-1}}\sin \theta \\\ & \Rightarrow {{\sin }^{-1}}x=\theta \\\ \end{aligned}$$ Now substituting the value of ‘$$\theta $$’ we get as follows: $$y=3{{\sin }^{-1}}x$$ Differentiating the above function with respect to x, we get as follows: $$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( 3{{\sin }^{-1}}x \right)$$ Since, 3 is a constant, we can take it out of the differentiation. $$\Rightarrow \dfrac{dy}{dx}=3\dfrac{d}{dx}\left( {{\sin }^{-1}}x \right)$$ Since we know the derivative of $$\left( {{\sin }^{-1}}x \right)$$ is equal to $$\dfrac{1}{\sqrt{1-{{x}^{2}}}}$$ $$\Rightarrow \dfrac{dy}{dx}=\dfrac{3}{\sqrt{1-{{x}^{2}}}}$$ Therefore, the correct answer of the above question is option A. Note: We suppose that x is equal to $$\sin \theta $$ because the expression $$\left( 3x-4{{x}^{3}} \right)$$ is similar to the formula of $$\sin 3\theta $$ which helps us to simplify the given inverse trigonometric function otherwise it is very lengthy and complex to find $$\dfrac{dy}{dx}$$. Also be careful while doing calculation and take care of the sign. As there is a chance that we might take the derivative of $${{\sin }^{-1}}x$$ as $$\dfrac{1}{\sqrt{1+{{x}^{2}}}}$$ instead of $$\dfrac{1}{\sqrt{1-{{x}^{2}}}}$$ then we might end up choosing option C as the correct option.