Q.Differentiate w.r.t. $x$: $\tan^{-1}(\sec x + \tan x),\ -\dfrac{\pi}{2} < x < \dfrac{\pi}{2}$.

Concept: Chain Rule – differentiate the outer inverse tangent function, then multiply by the derivative of the inner function.

Step 1: Let $y = \tan^{-1}(\sec x + \tan x)$. Then

$\frac{dy}{dx} = \frac{1}{1 + (\sec x + \tan x)^2} \cdot \frac{d}{dx}(\sec x + \tan x)$.

Step 2: Compute the inner derivative:

$\frac{d}{dx}(\sec x + \tan x) = \sec x \tan x + \sec^2 x = \sec x (\tan x + \sec x)$.

Step 3: Simplify the denominator:

$1 + (\sec x + \tan x)^2 = 1 + \sec^2 x + \tan^2 x + 2\sec x \tan x$.

Using $\sec^2 x = 1 + \tan^2 x$, this becomes …

The key insight here is that the function inside the inverse tangent can be rewritten in a much simpler form. When you see $\sec x + \tan x$, think about the half-angle formulas — specifically, $\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)$ expands to exactly this sum. This is not a coincidence; it's a standard trick for such problems.

Once you recognise that $\tan^{-1}(\sec x + \tan x) = \frac{\pi}{4} + \frac{x}{2}$ (for $-\frac{\pi}{2} < x < \frac{\pi}{2}$), differentiation becomes trivial. The chain rule is still at work, but the inner function's derivative is just $\frac{1}{2}$, and the derivative of $\tan^{-1}$ of its argument is $1$ because the argument is already an angle whose tangent matches.

Let's walk through it step by step.

1. Rewrite the inner expression using a trigonometric identity. Recall that $\sec x = \frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$. So:

Now use the identities $\sin x = \frac{2\tan(x/2)}{1+\tan^2(x/2)}$ and $\cos x = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$. Substituting:

Simplify numerator: $1 + \frac{2t}{1+t^2} = \frac{1+t^2+2t}{1+t^2} = \frac{(1+t)^2}{1+t^2}$.

Denominator: $\frac{1-t^2}{1+t^2}$.

So the whole fraction becomes:

Hence $\sec x + \tan x = \frac{1 + \tan(x/2)}{1 - \tan(x/2)}$.

2. Recognise the tangent addition formula. Notice that $\frac{1 + \tan\theta}{1 - \tan\theta} = \tan\left(\frac{\pi}{4} + \theta\right)$ because:

Here $\theta = \frac{x}{2}$, so:

3. Apply the inverse tangent. Since $-\frac{\pi}{2} < x < \frac{\pi}{2}$, we have $\frac{\pi}{4} + \frac{x}{2}$ lying between $0$ and $\frac{\pi}{2}$ (specifically, from $0$ to $\frac{\pi}{2}$ exclusive). In this range, $\tan^{-1}(\tan\theta) = \theta$. Therefore: …

Method: Recognise $\sec x + \tan x$ as a Tangent of a Shifted Angle

Whenever you see $\sec x + \tan x$ (or a similar combination) inside an inverse trig function, check whether it matches a known tangent half-angle/addition pattern before differentiating directly — recognising the identity turns a messy chain-rule computation into differentiating a simple linear function.

Steps

Step 1: Rewrite in terms of $t=\tan(x/2)$

Using $\sin x = \dfrac{2t}{1+t^2}$ and $\cos x = \dfrac{1-t^2}{1+t^2}$ (the standard tangent half-angle substitutions), $\sec x+\tan x$ simplifies to $\dfrac{1+t}{1-t}$.

Step 2: Match against the tangent addition formula

With $\theta = x/2$, this is exactly the form found in Step 1, so $\sec x+\tan x = \tan\left(\dfrac\pi4+\dfrac x2\right)$.

Step 3: Apply $\tan^{-1}(\tan\theta)=\theta$, checking the domain …

Common Mistakes

Mistake 1: Applying $\tan^{-1}(\tan\theta)=\theta$ without checking the domain

Why it's wrong: this simplification is only valid when $\theta$ lies in the principal branch $(-\pi/2,\pi/2)$; the given restriction on $x$ exists precisely to guarantee this, and ignoring it risks an incorrect simplified form. Correct approach: always verify the stated domain places the simplified angle inside the principal branch before applying the identity.

Mistake 2: Going straight to direct chain-rule differentiation and making a sign or algebra slip

Why it's wrong: expanding $1+(\sec x+\tan x)^2$ and simplifying it down to $2\sec x(\sec x+\tan x)$ involves several trig-identity substitutions where a sign error easily creeps in, producing an answer that looks plausible but is wrong. Correct approach: prefer spotting the $\tan\left(\frac\pi4+\frac x2\right)$ identity first; if using the direct route, verify each algebraic step against a known identity rather than expanding freehand. …