Question

Derivative of ${\sin }^{-1}\left(3x-4x^3\right)$ with respect to ${\sin }^{-1}\left(x\right)$ is

• A. $3,\left|x\right|<1$
• B. $3,\left|x\right|\le \frac{1}{2}\mathrm{and}-3,\frac{1}{2}<\left|\mathrm{x}\right|<1$
• C. $-3,\left|x\right|<1$
• D. $3,\left|x\right|\le \frac{1}{2}\mathrm{and}3,\frac{1}{2}\le \left|\mathrm{x}\right|\le 1$

Answer 1

The correct option is B

$3,\left|x\right|\le \frac{1}{2}\mathrm{and}-3,\frac{1}{2}<\left|\mathrm{x}\right|<1$

Explanation for correct option:

Step-1: Differentiate ${\sin }^{-1}\left(3x-4x^3\right)$ with respect to $x$.

Let, $u={\sin }^{-1}\left(3x-4x^3\right)$

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

$$\begin{gathered} \Rightarrow u={\sin }^{-1}\left(3\sin \left(\theta \right)-4{\sin }^3\left(\theta \right)\right) \\ \Rightarrow u={\sin }^{-1}\left(\sin \left(3\theta \right)\right)\left(\because 3\sin \left(\theta \right)-4{\sin }^3\left(\theta \right)=\sin \left(3\theta \right)\right) \\ \Rightarrow u=3\theta \\ \Rightarrow u=3{\sin }^{-1}\left(x\right) \end{gathered}$$

Differentiate with respect to $x$.

$\Rightarrow \frac{du}{dx}=\frac{3}{\sqrt{1-x^2}}........\left(i\right)$

Step-2: Differentiate ${\sin }^{-1}\left(x\right)$ with respect to $x$ :

Let, $v={\sin }^{-1}\left(x\right)$

Differentiate with respect to $x$

$\Rightarrow \frac{dv}{dx}=\frac{1}{\sqrt{1-x^2}}........\left(ii\right)$

Step-3: Divide the equation $\left(i\right)$ by $\left(ii\right)$

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}=\frac{{\displaystyle \frac{3}{\sqrt{1-x^2}}}}{{\displaystyle \frac{1}{\sqrt{1-x^2}}}} \\ \Rightarrow \frac{du}{dv}=3,\left|x\right|\le \frac{1}{2}\mathrm{and}-3,\frac{1}{2}<\left|\mathrm{x}\right|<1 \end{gathered}$$

Hence, correct answer is option $\left(B\right)$