Q.Differentiate w.r.t. $x$: $\tan^{-1}\left(\sqrt{\dfrac{1 - \cos x}{1 + \cos x}}\right),\ -\dfrac{\pi}{4} < x < \dfrac{\pi}{4}$.

Concept: Chain rule after simplifying the argument.

With the half-angle identities $1-\cos x=2\sin^2\frac{x}{2}$ and $1+\cos x=2\cos^2\frac{x}{2}$,

The absolute value matters: the square root is never negative, but $\tan\frac{x}{2}$ is negative when $x<0$. On $-\frac{\pi}{4}<x<\frac{\pi}{4}$, $\frac{x}{2}\in(-\frac{\pi}{8},\frac{\pi}{8})$, where $\tan\frac{x}{2}$ has the same sign as $x$. Hence

$y=\tan^{-1}\left|\tan\tfrac{x}{2}\right|=\frac{|x|}{2}.$ …

Set up

Let

Direct differentiation would be ugly; simplifying the inside first makes it easy.

Simplify the argument

Using $1-\cos x=2\sin^2\frac{x}{2}$ and $1+\cos x=2\cos^2\frac{x}{2}$,

The absolute value is essential: a principal square root cannot be negative, but $\tan\frac{x}{2}$ is negative for $x<0$.

Resolve the sign

On $-\frac{\pi}{4}<x<\frac{\pi}{4}$ we have $\frac{x}{2}\in(-\frac{\pi}{8},\frac{\pi}{8})$, where $\tan\frac{x}{2}$ has the same sign as $x$: …

Method: Half-Angle Substitution With Explicit Sign (Absolute Value) Handling

Use this method whenever a square root of a ratio of $(1\pm\cos x)$ appears inside an inverse trig function — the half-angle identities collapse it to a single tangent, but you must track the sign carefully because a square root is never negative while $\tan(x/2)$ can be.

Steps

Step 1: Apply the half-angle identities

so the ratio under the root becomes $\tan^2\dfrac x2$.

Step 2: Take the square root carefully — introduce the absolute value

A principal square root is never negative, but $\tan(x/2)$ is negative whenever $x<0$, so this absolute value is not optional.

Step 3: Resolve the sign on the given interval

Determine the sign of $\tan(x/2)$ throughout the given domain. Here $x/2$ stays in a small interval around $0$ where $\tan(x/2)$ has the same sign as $x$, so the problem splits into two cases: $x>0$ and $x<0$. …

Common Mistakes

Mistake 1: Writing $\sqrt{\tan^2(x/2)} = \tan(x/2)$ without the absolute value

Why it's wrong: a square root is defined to be non-negative, but $\tan(x/2)$ is negative for $x<0$ — skipping the absolute value silently produces the wrong sign (and hence the wrong derivative) on half the domain. Correct approach: always write $\sqrt{u^2}=|u|$, then resolve the sign using the specific interval given.

Mistake 2: Treating the function as differentiable everywhere on the given interval

Why it's wrong: $y=|x|/2$ has a corner at $x=0$ — the left- and right-hand derivatives disagree ($-\tfrac12$ vs $\tfrac12$), so the derivative genuinely does not exist there, even though $x=0$ lies inside the stated domain. Correct approach: report the derivative piecewise and explicitly note the point where it fails to exist, rather than quoting a single formula for the whole interval. …