Question

Differential coefficient of ${\sin }^{-1}\left(\frac{1-x}{1+x}\right)$ w.r.t. $\sqrt{x}$ is

• A. $\frac{1}{2\sqrt{x}}$
• B. $\frac{\sqrt{x}}{\sqrt{\left(1-x\right)}}$
• C. $1$
• D. $-\frac{2}{\left(1+x\right)}$

Answer 1

The correct option is D

$-\frac{2}{\left(1+x\right)}$

Explanation for the correct option:

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

Let, the given, $u={\sin }^{-1}\left(\frac{1-x}{1+x}\right)$

$\Rightarrow \frac{du}{dx}=\frac{1}{\sqrt{1-{\left({\displaystyle \frac{1-x}{1+x}}\right)}^2}}\times \frac{\left(1+x\right)\times \left(-1\right)-\left(1-x\right)\left(1\right)}{{\left(1+x\right)}^2}$

$$\begin{gathered} \Rightarrow \frac{du}{dx}=\frac{1}{\sqrt{{\displaystyle \frac{{\left(1+x\right)}^2-{\left(1-x\right)}^2}{{\left(1+x\right)}^2}}}}\times \frac{-1-x-1+x}{{\left(1+x\right)}^2} \\ \Rightarrow \frac{du}{dx}=\frac{\left(1+x\right)}{\sqrt{1+x^2+2x-\left(1+x^2-2x\right)}}\times \frac{-2}{{\left(1+x\right)}^2} \\ \Rightarrow \frac{du}{dx}=-\frac{1}{\left(1+x\right)\sqrt{x}}...........\left(i\right) \end{gathered}$$

Step-2: Differentiate $\sqrt{x}$ with respect to $x$.

Let the given, $v=\sqrt{x}$

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

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

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

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