Q.Differentiate w.r.t. $x$: $\sin\sqrt{x} + \cos^2\sqrt{x}$.

Concept: Derivative Evaluation — use the chain rule for each term separately.

Let $y = \sin\sqrt{x} + \cos^2\sqrt{x}$.

Step 1: Differentiate $\sin\sqrt{x}$.

Let $u = \sqrt{x} = x^{1/2}$, so $\frac{du}{dx} = \frac{1}{2\sqrt{x}}$.

Then $\frac{d}{dx}\sin u = \cos u \cdot \frac{du}{dx} = \cos\sqrt{x} \cdot \frac{1}{2\sqrt{x}}$.

Step 2: Differentiate $\cos^2\sqrt{x}$.

Let $v = \cos\sqrt{x}$, so $v^2$ derivative is $2v \cdot \frac{d}{dx}(\cos\sqrt{x})$.

Now $\frac{d}{dx}\cos\sqrt{x} = -\sin\sqrt{x} \cdot \frac{1}{2\sqrt{x}}$.

Thus $\frac{d}{dx}\cos^2\sqrt{x} = 2\cos\sqrt{x} \cdot \left(-\sin\sqrt{x} \cdot \frac{1}{2\sqrt{x}}\right) = -\frac{\cos\sqrt{x}\sin\sqrt{x}}{\sqrt{x}}$.

Step 3: Combine the two derivatives: …

Concept and Intuition

When you see a function like $\sin\sqrt{x}$, the key is to recognise that $\sqrt{x}$ is an inner function nested inside the outer sine function. The chain rule tells us: differentiate the outer function, then multiply by the derivative of the inner function. The same idea applies to $\cos^2\sqrt{x}$ — here the outer function is “square” and the inner is $\cos\sqrt{x}$, which itself has $\sqrt{x}$ inside. So we need two layers of chain rule.

A common mistake is to forget that $\sqrt{x}$ differentiates to $\frac{1}{2\sqrt{x}}$, not something like $\frac{1}{2}x^{-1/2}$ in a simplified form — but that’s exactly the same thing. We’ll keep it as $\frac{1}{2\sqrt{x}}$ for clarity.

Step-by-step solution

1. Differentiate $\sin\sqrt{x}$ Let $u = \sqrt{x}$. Then $\frac{d}{dx}\sin u = \cos u \cdot \frac{du}{dx}$. Since $\frac{du}{dx} = \frac{1}{2\sqrt{x}}$, we get:

2. Differentiate $\cos^2\sqrt{x}$ Write it as $(\cos\sqrt{x})^2$. Let $v = \cos\sqrt{x}$, so the term is $v^2$. Then $\frac{d}{dx}v^2 = 2v \cdot \frac{dv}{dx}$. Now $\frac{dv}{dx} = \frac{d}{dx}\cos\sqrt{x} = -\sin\sqrt{x} \cdot \frac{1}{2\sqrt{x}}$ (chain rule again). So: …

Method: Chain Rule for Sums of Composite Trigonometric Functions with a $\sqrt{x}$ Inner Function

Use this method whenever a sum contains multiple trigonometric terms, each built on the same inner function such as $\sqrt{x}$, possibly combined with an additional power (like squaring).

Steps

Step 1: Split the sum and differentiate term by term

The derivative of a sum is the sum of the derivatives, so each term — here $\sin\sqrt{x}$ and $\cos^2\sqrt{x}$ — can be handled completely independently before adding the results at the end.

Step 2: For a plain composite term, apply the chain rule once

For $\sin\sqrt{x}$, treat $\sqrt{x}$ as the inner function $u$:

Step 3: For a squared trigonometric term, apply the chain rule twice …

Common Mistakes

Mistake 1: Forgetting the $\dfrac{1}{2\sqrt{x}}$ factor from differentiating $\sqrt{x}$

Why it's wrong: $\sqrt{x}=x^{1/2}$ differentiates to $\dfrac{1}{2\sqrt{x}}$, not $1$; writing $\dfrac{d}{dx}\sin\sqrt{x}=\cos\sqrt{x}$ (stopping after the outer trig function) silently drops this inner-function factor, which is one of the single most common chain-rule slips. Correct approach: always identify the inner function first and explicitly multiply by its derivative.

Mistake 2: Treating $\cos^2\sqrt{x}$ as needing only one chain-rule step …