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

Concept: Chain Rule — we differentiate the first function with respect to the second by writing it as a derivative ratio.

Let $u = \tan^{-1}\left(\dfrac{\sqrt{1 + x^2} - 1}{x}\right)$ and $v = \tan^{-1} x$. We need $\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx}$.

Step 1: Simplify $u$ using the identity $\sqrt{1+x^2} - 1 = \frac{x^2}{\sqrt{1+x^2}+1}$. Then

Let $x = \tan\theta$, so $\sqrt{1+x^2} = \sec\theta$. Then

$\frac{x}{\sqrt{1+x^2}+1} = \frac{\tan\theta}{\sec\theta+1} = \frac{\sin\theta}{1+\cos\theta} = \tan\frac{\theta}{2}.$ …

The core idea here is the Chain Rule for parametric differentiation. When we need the derivative of one function $u$ with respect to another function $v$ (both of $x$), we compute:

This is just the chain rule in disguise — we differentiate both with respect to $x$ and then divide. The trick is often to simplify $u$ first, so the differentiation becomes clean.

Let’s set:

We want $\frac{du}{dv}$.

1. Simplify $u$ using a trigonometric substitution. The expression inside the arctan looks like it comes from a tangent half-angle or a double-angle identity. Let $x = \tan \theta$, so $\theta = \tan^{-1} x$. Then $\sqrt{1 + x^2} = \sqrt{1 + \tan^2 \theta} = \sec \theta$ (taking the positive root since $x$ can be any real, but we’ll handle sign later). So:

2. Rewrite in terms of sine and cosine. $\sec \theta = \frac{1}{\cos \theta}$, $\tan \theta = \frac{\sin \theta}{\cos \theta}$. Then:

3. Use the half-angle identity. Recall: $1 - \cos \theta = 2 \sin^2(\theta/2)$ and $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$. So:

Therefore:

4. Handle the principal value carefully. …

Method: Differentiating One Function with Respect to Another

This method applies whenever a question asks for $\dfrac{du}{dv}$ — the derivative of one function $u(x)$ with respect to a different function $v(x)$ of the same variable $x$ — rather than $\dfrac{du}{dx}$ directly.

Steps

Step 1: Recognise the two functions and set up the ratio

If you're asked to differentiate $u$ with respect to $v$, both being functions of $x$, the chain rule tells you that

This works because $\dfrac{du}{dx} = \dfrac{du}{dv}\cdot\dfrac{dv}{dx}$ (chain rule), so dividing both sides by $\dfrac{dv}{dx}$ isolates $\dfrac{du}{dv}$.

Step 2: Simplify $u$ first, if possible, before differentiating …

Common Mistakes

Mistake 1: Attacking the messy expression directly instead of simplifying first

Students often try to differentiate $\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ head-on with the quotient rule buried inside the arctan derivative formula. Why it's wrong: this produces an extremely messy expression that is very easy to botch algebraically. Correct approach: substitute $x=\tan\theta$ first, use $\sqrt{1+x^2}=\sec\theta$, and simplify to $\tan(\theta/2)$ — the derivative becomes trivial only after this simplification.

Mistake 2: Forgetting the "derivative w.r.t. another function" formula

When asked to differentiate $u$ with respect to $v$ (not $x$), students sometimes just compute $\frac{du}{dx}$ and stop, or divide $u$ by $v$ directly. Why it's wrong: differentiating "w.r.t. $v$" means $\frac{du}{dv}=\frac{du/dx}{dv/dx}$, both derivatives must be taken with respect to $x$ first, then divided. Correct approach: compute $\frac{du}{dx}$ and $\frac{dv}{dx}$ separately, then form the ratio. …