Question

$\frac{d}{dx}\left({\tan }^{-1}\left(\frac{\sqrt{x}-x}{1+x^{3/2}}\right)\right)$ equals $($for $x\ge 0)$

• A. $\frac{1}{2\sqrt{x}(1+x)}-\frac{1}{1+x^2}$
• B. $\frac{1}{2\sqrt{x}(1+x)}+\frac{1}{1+x^2}$
• C. $\frac{1}{1+x}-\frac{1}{1+x^2}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{2\sqrt{x}(1+x)}-\frac{1}{1+x^2}$

Let, $y={\tan }^{-1}\left(\frac{\sqrt{x}-x}{1+x^{3/2}}\right)$
$y={\tan }^{-1}\left(\sqrt{x}\right)-{\tan }^{-1}x$
$\therefore \frac{dy}{dx}=\frac{1}{(1+x)}\times \frac{1}{2\sqrt{x}}-\frac{1}{1+x^2}$