Q.Differentiate w.r.t. $x$: $\tan^{-1}\left(\dfrac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}}\right),\ -1 < x < 1,\ x \ne 0$.

Concept: Chain Rule — first simplify the expression inside $\tan^{-1}$ with a trigonometric substitution, then differentiate.

Step 1 — Substitute. Let $x^2 = \cos 2\theta$. Since $0 < x^2 < 1$ for $-1<x<1,\ x\ne 0$, we have $2\theta \in (0,\pi/2)$, so $\theta \in (0,\pi/4)$ and $\cos\theta,\sin\theta > 0$. Then

Step 2 — Reduce the fraction.

$\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}} = \frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta} = \frac{1+\tan\theta}{1-\tan\theta} = \tan\left(\frac{\pi}{4}+\theta\right).$ …

Attacking this directly with the quotient rule and two nested square roots would be brutal. The smart move is to simplify the argument of $\tan^{-1}$ before differentiating. Whenever you see $\sqrt{1+x^2}$ and $\sqrt{1-x^2}$ appearing together, reach for a substitution that turns both radicals into clean trig functions.

1. Pick the right substitution

Let $x^2 = \cos 2\theta$. Because $-1 < x < 1$ with $x \ne 0$, we have $0 < x^2 < 1$, so $\cos 2\theta \in (0,1)$, giving $2\theta \in (0,\pi/2)$ and therefore

Using the half-angle identities $1+\cos 2\theta = 2\cos^2\theta$ and $1-\cos 2\theta = 2\sin^2\theta$:

The positivity of $\cos\theta$ and $\sin\theta$ on $(0,\pi/4)$ lets us drop the absolute values safely.

2. Simplify the fraction

Substitute and cancel the common $\sqrt{2}$:

Divide top and bottom by $\cos\theta$ (nonzero here):

The tangent addition formula $\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$ with $\alpha=\tfrac{\pi}{4}$ (so $\tan\alpha=1$) and $\beta=\theta$ gives exactly

3. Remove the inverse tangent — carefully

So the function is

The identity $\tan^{-1}(\tan u)=u$ is only valid when $u \in (-\tfrac{\pi}{2},\tfrac{\pi}{2})$, so we must check the range. …

Method: Trigonometric Substitution Inside Inverse Trig Functions

Use this whenever the argument of $\sin^{-1}$, $\cos^{-1}$, or $\tan^{-1}$ contains $\sqrt{1+x^2}$ and/or $\sqrt{1-x^2}$ together — a direct quotient/chain-rule attack on such an expression is usually a mess, so simplify the argument first.

Steps

Step 1: Spot the pattern and choose a substitution

Whenever you see $\sqrt{1+x^2}$ paired with $\sqrt{1-x^2}$, the half-angle identities $1+\cos 2\theta = 2\cos^2\theta$ and $1-\cos 2\theta = 2\sin^2\theta$ suggest setting

Always note the domain of $x$ first and pin down the resulting range of $\theta$ — this determines the signs of $\sin\theta,\cos\theta$ and which branch of the inverse function is valid later.

Step 2: Rewrite both radicals as trig functions

Substitute these into the original quotient; the $\sqrt{2}$ factors cancel, and dividing numerator and denominator by $\cos\theta$ turns the expression into a ratio of $1\pm\tan\theta$.

Step 3: Collapse to a single angle using a standard identity …

Common Mistakes

Mistake 1: Attacking the fraction directly with the quotient rule on two nested square roots

Why it's wrong: the raw quotient rule on $\dfrac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}$ produces a huge, error-prone expression full of nested radicals. Correct approach: recognize the $\sqrt{1+x^2},\sqrt{1-x^2}$ pair as a signal to substitute $x^2=\cos2\theta$, which turns both radicals into $\sqrt2\cos\theta,\sqrt2\sin\theta$ and collapses the whole fraction to $\tan\left(\tfrac{\pi}{4}+\theta\right)$ before any differentiation happens.

Mistake 2: Removing $\tan^{-1}(\tan u)$ without checking the range of $u$ …