Question

If $y=co{t}^{-1}\tan \left(\begin{array}{l}\frac{x}{2}\end{array}\right)$, then $\frac{dy}{dx}$ is equal to?

• A. $0$
• B. $\frac{1}{2}$
• C. $\frac{-1}{2}$
• D. $\frac{x}{2}$

Answer 1

The correct option is C

$\frac{-1}{2}$

Explanation for the correct answer:

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

Given,

$y=co{t}^{-1}\tan \left(\begin{array}{l}\frac{x}{2}\end{array}\right)$

Now differentiate the given function with respect to $x$,

Then,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}co{t}^{-1}\tan \left(\frac{x}{2}\right)\\ & =& \frac{-1}{\left(1+{\tan }^2\frac{x}{2}\right)}\times \frac{d}{dx}\tan \frac{x}{2}\left[\because \frac{d}{dx}\left(co{t}^{-1}x\right)=\frac{-1}{1+x^2}\right]\\ & =& \frac{-1}{\left(1+{\tan }^2\frac{x}{2}\right)}\times se{c}^2\frac{x}{2}\times \frac{1}{2}\left[\because \frac{d\left(\tan x\right)}{dx}=se{c}^{2}x\right]\\ & =& \frac{-1}{se{c}^2\frac{x}{2}}se{c}^2\frac{x}{2}\times \frac{1}{2}\\ & =& \frac{-1}{2}\end{array}$

Hence, the correct option is C.