Apply d/dx[log(f(x))] = f'(x)/f(x). Result: cot x.

Differentiate log(sin x) Using Chain Rule

Question

Find the derivative of $\log(\sin x)$ with respect to $x$ .

This is a direct application of the chain rule — a function inside another function. We treat $\log$ as the outer function and $\sin x$ as the inner function.

Solution — Step by Step

Write $y = \log(\sin x)$ and recognise it as $y = \log(u)$ where $u = \sin x$ . The moment you see a function plugged inside another function, chain rule is the tool.

The derivative of $\log(u)$ with respect to $u$ is $\dfrac{1}{u}$ . So:

\frac{dy}{du} = \frac{1}{u} = \frac{1}{\sin x}

Now differentiate $u = \sin x$ with respect to $x$ :

\frac{du}{dx} = \cos x

Chain rule says $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$ . Putting it together:

\frac{dy}{dx} = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x}

We know $\dfrac{\cos x}{\sin x} = \cot x$ . So:

\frac{d}{dx}[\log(\sin x)] = \boxed{\cot x}

Why This Works

The chain rule handles composite functions by “peeling layers.” We first ask: what happens to $\log$ as its input changes? That gives us $\frac{1}{\sin x}$ . Then we ask: how fast is that input ( $\sin x$ ) itself changing? That’s $\cos x$ .

The product $\frac{1}{\sin x} \times \cos x$ captures both effects simultaneously — the rate of the outer function scaled by the rate of the inner function.

The clean cancellation into $\cot x$ is why this appears in CBSE 12 and JEE Main so often. Examiners love results that collapse into a standard trig function. Recognising $\frac{\cos x}{\sin x}$ as $\cot x$ without hesitation is worth marks in a time-pressured exam.

Alternative Method

Use the general log-derivative formula directly:

\frac{d}{dx}[\log(f(x))] = \frac{f'(x)}{f(x)}

Here $f(x) = \sin x$ , so $f'(x) = \cos x$ . Substituting:

\frac{d}{dx}[\log(\sin x)] = \frac{\cos x}{\sin x} = \cot x

Memorise the pattern: derivative of $\log(\text{anything})$ = $\dfrac{\text{derivative of anything}}{\text{anything}}$ . This direct formula saves 30 seconds per problem in board exams — and that adds up.

This formula is just the chain rule pre-packaged. Once you’ve done enough of these, skip the intermediate steps and write the answer in one line.

Common Mistake

The most frequent error: writing $\dfrac{d}{dx}[\log(\sin x)] = \dfrac{1}{\sin x}$ and stopping there — forgetting to multiply by the derivative of $\sin x$ .

Students who write only $\frac{1}{\sin x}$ are differentiating $\log(x)$ and then substituting $\sin x$ , which is NOT the same as differentiating the composite function. The chain rule requires that second multiplication step — without $\cos x$ , the answer is wrong and will lose full marks in CBSE.

A good self-check: if your answer contains $\sin x$ in the denominator with no $\cos x$ in sight, you’ve missed the chain rule step. The final answer for this type of question should always look like a clean trig expression — $\cot x$ here, not $\frac{1}{\sin x}$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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