Question

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

• A. $\frac{3}{\sqrt{1-x^2}}$
• B. $\frac{-1}{\sqrt{1-x^2}}$
• C. $1$
• D. None of These

Answer 1

The correct option is A

$\frac{3}{\sqrt{1-x^2}}$

Explanation for the correct option:

Differentiating the given expression:

Let, $\mathrm{y}={\sin }^{-1}(3\mathrm{x}-4{\mathrm{x}}^3)$

Put, $x=\sin \theta$

$\theta ={\sin }^{-1}x$

Therefore,

$\begin{array}{rcl}& & {\sin }^{-1}(3\sin \theta -4{\sin }^3\theta )\\ & =& {\sin }^{-1}\left(\sin 3\theta \right)\\ & =& 3\theta \\ y& =& 3{\sin }^{-1}x\end{array}$

Now,

$\Rightarrow \frac{dy}{dx}=\frac{3}{\sqrt{1-x^2}}$

Hence, correct option is option (A).