Q.Differentiate w.r.t. $x$: $\log\left(x + \sqrt{x^2 + a}\right)$.

The key idea is to differentiate a composite logarithmic function using the chain rule.

Let $y = \log\left(x + \sqrt{x^2 + a}\right)$. Differentiate step by step:

1. Derivative of $\log u$ is $\frac{1}{u} \cdot \frac{du}{dx}$, where $u = x + \sqrt{x^2 + a}$.
2. For $\frac{du}{dx}$, differentiate $x$ (gives $1$) and $\sqrt{x^2 + a}$ (gives $\frac{1}{2\sqrt{x^2 + a}} \cdot 2x = \frac{x}{\sqrt{x^2 + a}}$). …

Why This Works

The function $\log\left(x + \sqrt{x^2 + a}\right)$ is a classic form — it’s actually the inverse hyperbolic sine function $\sinh^{-1}(x/\sqrt{a})$ up to a constant, but we don’t need that fact. The key insight is that the expression inside the log simplifies beautifully when differentiated: the derivative of $x + \sqrt{x^2 + a}$ turns out to be $\frac{\sqrt{x^2 + a} + x}{\sqrt{x^2 + a}}$, which is exactly the original inside divided by $\sqrt{x^2 + a}$. That cancellation is what gives the clean result.

Let’s work through it step by step.

1. Set up the chain rule. Let $y = \log u$, where $u = x + \sqrt{x^2 + a}$. Then

2. Differentiate $u$. $u = x + (x^2 + a)^{1/2}$. The derivative of $x$ is $1$. For the square root term, use the chain rule:

So

3. Combine into a single fraction. Write $1$ as $\frac{\sqrt{x^2 + a}}{\sqrt{x^2 + a}}$:

4. Apply the chain rule.

5. Cancel the common factor. The $x + \sqrt{x^2 + a}$ cancels completely, leaving …

Method: Chain Rule for $\log\big(x+\sqrt{x^2+a}\big)$-Type Expressions

This method applies to logarithms of a sum of a linear term and a square root of a quadratic — a recurring NCERT pattern where the derivative simplifies dramatically through cancellation.

Steps

Step 1: Identify the inner function

Let $u = x+\sqrt{x^2+a}$, so the given expression is $\log u$, and the chain rule gives $\dfrac{dy}{dx}=\dfrac{1}{u}\cdot\dfrac{du}{dx}$.

Step 2: Differentiate the inner function term by term

Differentiate $x$ (giving $1$) and $\sqrt{x^2+a}=(x^2+a)^{1/2}$ using the chain rule (giving $\dfrac{x}{\sqrt{x^2+a}}$):

Step 3: Combine the inner derivative into a single fraction

Write the $1$ with the common denominator $\sqrt{x^2+a}$ so the numerator matches the structure of $u$ itself:

$\frac{du}{dx} = \frac{\sqrt{x^2+a}+x}{\sqrt{x^2+a}}$ …

Common Mistakes

Mistake 1: Forgetting the chain rule when differentiating $\sqrt{x^2+a}$

Why it's wrong: writing $\dfrac{d}{dx}\sqrt{x^2+a} = \dfrac{1}{2\sqrt{x^2+a}}$ (missing the extra factor of $2x$ from differentiating $x^2+a$ on the inside) gives a derivative that is off by a factor of $x$. Correct approach: always apply the chain rule fully — differentiate the outer square root, then multiply by the derivative of $x^2+a$, which is $2x$.

Mistake 2: Not combining terms into a single fraction before simplifying …