Question

If $y={\tan }^{-1}\left(\frac{\sqrt{a}-\sqrt{x}}{1+\sqrt{ax}}\right)$, then $\frac{dy}{dx}=$

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

Answer 1

The correct option is C

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

Explanation for the correct option.

Find the value of $\frac{dy}{dx}$.

Using the identity ${\tan }^{-1}\left(\frac{x-y}{1+xy}\right)={\tan }^{-1}x-{\tan }^{-1}y$, the equation $y={\tan }^{-1}\left(\frac{\sqrt{a}-\sqrt{x}}{1+\sqrt{ax}}\right)$ can be written as:

$$\begin{gathered} y={\tan }^{-1}\left(\frac{\sqrt{a}-\sqrt{x}}{1+\sqrt{ax}}\right) \\ \Rightarrow y={\tan }^{-1}\left(\sqrt{a}\right)-{\tan }^{-1}\left(\sqrt{x}\right) \end{gathered}$$

Now differentiate it with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}\left({\tan }^{-1}\left(\sqrt{a}\right)-{\tan }^{-1}\left(\sqrt{x}\right)\right)\\ & =& 0-\frac{1}{1+{\left(\sqrt{x}\right)}^2}\frac{1}{2\sqrt{x}}\\ & =& -\frac{1}{2\sqrt{x}(1+x)}\end{array}$

Hence, the correct option is C.