Q. The value of $\frac{d}{dx}\left\lbrack \tan^{- 1}\left( \frac{\sqrt{x}(3 - x)}{1 - 3x} \right) \right\rbrack\ is\$
A) $\frac{1}{2(1 + x)\sqrt{x}}$
B) $\frac{3}{(1 + x)\sqrt{x}}$
C) $\frac{2}{(1 + x)\sqrt{x}}$
D) $\frac{3}{2(1 + x)\sqrt{x}}$
Explanation
(d)
Let $y = \tan^{- 1}\left\{ \frac{3\sqrt{x} - x^{3/2}}{1 - 3x} \right\}$
Again let $\sqrt{x} = \tan t$
$\therefore\ \ y = \tan^{- 1}{\left\{ \frac{3\tan{t - \tan^{3}t}}{1 - 3\tan^{2}t} \right\} = \tan^{- 1}{(\tan{3t)}}}$
$\Rightarrow \ \ y = 3\tan^{- 1}\sqrt{x}$
$\Rightarrow \ \ \frac{dy}{dx} = \frac{3}{1 + x}.\frac{1}{2\sqrt{x}} = \frac{3}{2(1 + x)\sqrt{x}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.