Question

$\frac{d\left({\tan }^{-1}\left({\displaystyle \frac{4\sqrt{x}}{1-4x}}\right)\right)}{dx}=$

• A. $\frac{1}{\sqrt{x}\left(1+4x\right)}$
• B. $\frac{2}{\sqrt{x}\left(1+4x\right)}$
• C. $\frac{1}{\sqrt{x}\left(1+4x\right)}$$x$
• D. none of these

Answer 1

The correct option is B

$\frac{2}{\sqrt{x}\left(1+4x\right)}$

Explanation for correct option:

Step-1: Simplify the given data.

Given, $\frac{d\left({\tan }^{-1}\left({\displaystyle \frac{4\sqrt{x}}{1-4x}}\right)\right)}{dx}$

$=\frac{d\left({\tan }^{-1}\left({\displaystyle \frac{2\times 2\sqrt{x}}{1-{\left(2\sqrt{x}\right)}^2}}\right)\right)}{dx}$

Let, $$\begin{gathered} 2\sqrt{x}=\tan \left(\theta \right) \\ \theta ={\tan }^{-1}\left(2\sqrt{x}\right)...........\left(i\right) \end{gathered}$$

$=\frac{d\left({\tan }^{-1}\left({\displaystyle \frac{2\times \tan \left(\theta \right)}{1-{\left(\tan \left(x\right)\right)}^2}}\right)\right)}{dx}$

Step-2: Apply formula $\mathit{t}\mathit{a}\mathit{n}\left(2\theta \right)\mathbf{=}\left(\frac{2\times tan\left(\theta \right)}{1-{\left(tan\left(x\right)\right)}^2}\right)$

$$\begin{gathered} =\frac{d\left({\tan }^{-1}\left({\displaystyle \tan \left(2\theta \right)}\right)\right)}{dx} \\ =\frac{d\left(2\theta \right)}{dx}\left(\because {\tan }^{-1}\left(\tan \left(\theta \right)\right)=\theta \right) \\ =\frac{d\left(2{\tan }^{-1}\left(2\sqrt{x}\right)\right)}{dx}\left(fromequation\left(i\right)\right) \end{gathered}$$

Step-3: Differentiate w.r.t $x$ by chain rule.

$$\begin{gathered} =2\times \left(\frac{1}{1+{\left(2\sqrt{x}\right)}^2}\right)\times 2\times \frac{1}{2\sqrt{x}}\left[\left(\because \frac{d}{dx}\left({\tan }^{-1}\left(x\right)\right)=\frac{1}{1+x^2}\right)\right] \\ =\frac{2}{\sqrt{x}\left(1+4x\right)} \end{gathered}$$

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