Question

If \[y={{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)\] then \[\dfrac{dy}{dx}\] = ?
A. \[\dfrac{3}{\sqrt{1-{{x}^{2}}}}\]
B. \[\dfrac{-4}{\sqrt{1-{{x}^{2}}}}\]
C. \[\dfrac{3}{\sqrt{1+{{x}^{2}}}}\]
D. none of these

Answer 1

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

Complete step-by-step answer:

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