Question

$\frac{d\left({\sin }^{-1}\left(3x-4x^3\right)\right)}{dx}=$

• A. $\frac{3}{\sqrt{\left(1-x^2\right)}}$
• B. $-\frac{1}{\sqrt{\left(1-x^2\right)}}$
• C. $1$
• D. none of these

Answer 1

The correct option is A

$\frac{3}{\sqrt{\left(1-x^2\right)}}$

Explanation for correct option:

Step-1: Simplify the given data.

Given, $\frac{d\left({\sin }^{-1}\left(3x-4x^3\right)\right)}{dx}$

Put, $x=\sin \left(\theta \right)\Rightarrow \theta ={\sin }^{-1}\left(x\right)$

$\Rightarrow \frac{d\left({\sin }^{-1}\left(3x-4x^3\right)\right)}{dx}=\frac{d\left({\sin }^{-1}\left(3\sin \left(\theta \right)-4{\left(\sin \left(\theta \right)\right)}^3\right)\right)}{dx}$

Apply formula,

$3\sin \left(\theta \right)-4{\sin }^3\left(\theta \right)=\sin \left(3\theta \right)$

$$\begin{gathered} =\frac{d\left({\sin }^{-1}\left(\sin \left(3\theta \right)\right)\right)}{dx} \\ =\frac{d\left(3{\sin }^{-1}\left(x\right)\right)}{dx} \end{gathered}$$

Step-2: Differentiate w.r.t $x$

$$\begin{gathered} \Rightarrow \frac{d\left(3{\sin }^{-1}\left(x\right)\right)}{dx}=3\times \frac{1}{\sqrt{1-x^2}}\left(\because {\sin }^{-1}\left(x\right)=\frac{1}{\sqrt{1-x^2}}\right) \\ \Rightarrow \frac{d\left(3{\sin }^{-1}\left(x\right)\right)}{dx}=\frac{3}{\sqrt{1-x^2}} \end{gathered}$$

Therefore, correct answer is option $\left(A\right)$