Question

The value of $\frac{d}{dx}\left[{\tan }^{-1}\frac{\sqrt{x}(3-x)}{(1-3x)}\right]$is

• A. $\frac{1}{2(1+x)\sqrt{x}}$
• B. $\frac{3}{(1+x)\sqrt{x}}$
• C. $\frac{2}{(1+x)\sqrt{x}}$
• D. $\frac{3}{2(1+x)\sqrt{x}}$

Answer 1

The correct option is D

$\frac{3}{2(1+x)\sqrt{x}}$

Explanation for the correct option:

Find the required derivative.

Given: $\frac{d}{dx}\left[{\tan }^{-1}\frac{\sqrt{x}(3-x)}{(1-3x)}\right]$

Let $y={\tan }^{-1}\frac{\sqrt{x}(3-x)}{(1-3x)}$

Put $\sqrt{x}=\tan \theta$, So

$$\begin{gathered} y={\tan }^{-1}\left(\frac{\tan \theta (3-{\tan }^2\theta )}{(1-3{\tan }^2\theta )}\right) \\ ={\tan }^{-1}\left(\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right) \\ ={\tan }^{-1}\left(\tan 3\theta \right)\left[\because \tan 3\theta =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right] \\ =3\theta \\ =3{\tan }^{-1}\sqrt{x}\left[\because \sqrt{x}=\tan \theta \right] \end{gathered}$$

$\therefore y=3{\tan }^{-1}\sqrt{x}$

Now, by differentiating $y$ w.r.t. $x$, we get

$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(3{\tan }^{-1}\sqrt{x}\right) \\ =3\times \frac{1}{1+{\left(\sqrt{x}\right)}^2}\times \frac{1}{2\sqrt{x}} \\ =\frac{3}{2(1+x)\sqrt{x}} \\ \therefore \frac{d}{dx}\left[{\tan }^{-1}\frac{\sqrt{x}(3-x)}{(1-3x)}\right]=\frac{3}{2(1+x)\sqrt{x}} \end{gathered}$$

Hence, option (D) is the correct answer.