Question

What is the derivative of ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$ with respect to ${\sin }^{-1}\left(3x-4x^3\right)$?

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

The correct option is A

$\frac{2}{3}$

Explanation of correct answer :

Finding derivative of ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$ with respect to ${\sin }^{-1}\left(3x-4x^3\right)$ :

Let $u$ as a function of $x$ as $u\left(x\right)=$${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$

Put, $x=\sin \varnothing$

$\Rightarrow \varnothing ={\sin }^{-1}\left(x\right)$

Now solving $u$.

$$\begin{gathered} \Rightarrow u={\sin }^{-1}\left(2\sin \varnothing \sqrt{1-{\sin }^2\varnothing }\right) \\ ={\sin }^{-1}\left(2\sin \varnothing \sqrt{{\cos }^2\varnothing }\right) \\ ={\sin }^{-1}\left(2\sin \varnothing \cos \varnothing \right) \\ ={\sin }^{-1}(\sin 2\varnothing )(\text{as we know},\sin 2\varnothing =2\sin \varnothing \cos \varnothing ) \\ =2\varnothing \\ =2{\sin }^{-1}x \end{gathered}$$

Let $v$ as a function of $x$ as $v\left(x\right)=$${\sin }^{-1}\left(3x-4x^3\right)$

Put, $x=\sin \varnothing$

$\Rightarrow \varnothing ={\sin }^{-1}\left(x\right)$

Now solving $v$

$$\begin{gathered} \Rightarrow v={\sin }^{-1}\left(3\sin \varnothing -4{\sin }^3\varnothing \right) \\ ={\sin }^{-1}\left(\sin 3\varnothing \right){\mathbf{\text{(as we know, sin 3∅ =3 sin∅ - 4 sin}}}^{\mathbf{3}}\mathbf{\varnothing }\mathbf{)} \\ =3\varnothing \\ =3{\sin }^{-1}x \end{gathered}$$

Now finding, $\frac{du}{dx}=\frac{d\left(2{\sin }^{-1}x\right)}{dx}$

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

Now finding, $\frac{dv}{dx}=\frac{d\left(3{\sin }^{-1}x\right)}{dx}$

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

We need, $\frac{du}{dv}=\frac{\left({\displaystyle \frac{du}{dx}}\right)}{\left({\displaystyle \frac{dv}{dx}}\right)}$

$$\begin{gathered} \Rightarrow \frac{du}{dv}=\frac{\left({\displaystyle \frac{2}{\sqrt{1-x^2}}}\right)}{\left({\displaystyle \frac{3}{\sqrt{1-x^2}}}\right)} \\ =\frac{2}{3} \end{gathered}$$

Hence, the derivative of ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$ with respect to ${\sin }^{-1}\left(3x-4x^3\right)$ is $\frac{2}{3}$.

Hence, the correct answer is option(A).