Differentiate the function $\tan^{-1}\left( \frac{\sin x}{1 + \cos x} \right)$ with respect to $x$.

$\frac{1}{2}$

The correct answer is Option (4) → $\frac{1}{2}$ ##

Let $f(x) = \tan^{-1}\left( \frac{\sin x}{1 + \cos x} \right)$. Observe that this function is defined for all real numbers, where $\cos x \neq -1$; i.e., at all odd multiples of $\pi$.

We may rewrite this function as

$ f(x) = \tan^{-1}\left( \frac{\sin x}{1 + \cos x} \right)$

$= \tan^{-1} \left[ \frac{2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)}{2 \cos^2 \frac{x}{2}} \right]$

$= \tan^{-1} \left[ \tan \left( \frac{x}{2} \right) \right] = \frac{x}{2}$

Observe that we could cancel $\cos \left( \frac{x}{2} \right)$ in both numerator and denominator as it is not equal to zero. Thus,

$f'(x) = \frac{1}{2}$