Question

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

• A. ${\cos }^{-1}\left(\frac{x}{a}\right)$
• B. $-{\cos }^{-1}\left(\frac{x}{a}\right)$
• C. $\frac{1}{2}{\cos }^{-1}\left(\frac{x}{a}\right)$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option.

Step 1. Substitute the value of $x$.

Let $x=a\cos 2\theta$, then $\theta =\frac{1}{2}{\cos }^{-1}\left(\frac{x}{a}\right)$.

Now in the equation $y={\tan }^{-1}\left(\sqrt{\frac{a-x}{a+x}}\right)$ substitute $a\cos 2\theta$ for $x$.

$\begin{array}{rcl}y& =& {\tan }^{-1}\left(\sqrt{\frac{a-a\cos 2\theta }{a+a\cos 2\theta }}\right)\\ & =& {\tan }^{-1}\left(\sqrt{\frac{1-\cos 2\theta }{1+\cos 2\theta }}\right)\\ & =& {\tan }^{-1}\left(\sqrt{\frac{2{\sin }^2\theta }{2{\cos }^2\theta }}\right)\left[\because \cos 2A=2{\cos }^{2}A-1=1-2{\sin }^{2}A\right]\\ & =& {\tan }^{-1}\left(\sqrt{{\tan }^2\theta }\right)\\ & =& {\tan }^{-1}\left(\tan \theta \right)\\ & =& \theta \end{array}$

But $\theta =\frac{1}{2}{\cos }^{-1}\left(\frac{x}{a}\right)$, so $y=\frac{1}{2}{\cos }^{-1}\left(\frac{x}{a}\right)$.

Step 2. Find the value of $\frac{dy}{dx}$.

Differentiate $y=\frac{1}{2}{\cos }^{-1}\left(\frac{x}{a}\right)$ with respect to $x$.

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

Hence, the correct option is D.