Q.Differentiate w.r.t. $x$: $\sin^m x \cdot \cos^n x$.

Concept: Chain Rule — treat each factor separately, then apply the product rule.

Let $y = \sin^m x \cdot \cos^n x$.

Take $\log y = m \log(\sin x) + n \log(\cos x)$.

Differentiate: $\frac{1}{y} \frac{dy}{dx} = m \cdot \frac{\cos x}{\sin x} + n \cdot \frac{-\sin x}{\cos x} = m \cot x - n \tan x$.

Multiply through by $y$: …

We have a product of two functions, each raised to a power. The key is to see that each factor is a function of a function: $\sin^m x$ means $(\sin x)^m$, and similarly for $\cos^n x$. So when we differentiate, we need the chain rule inside the product rule.

Let’s set:

We want $\frac{d}{dx}(u \cdot v) = u'v + uv'$.

1. Differentiate $u = \sin^m x$.

Think of it as $(\sin x)^m$. The outer function is “raise to the power $m$”, the inner function is $\sin x$. By the chain rule:

2. Differentiate $v = \cos^n x$.

Similarly, this is $(\cos x)^n$. Outer: power $n$, inner: $\cos x$. Chain rule gives:

3. Apply the product rule.

4. Simplify each term. …

Method: Logarithmic Differentiation for a Product of Power Functions

Use this whenever you must differentiate a product of two or more factors, each raised to a power — like $\sin^m x \cdot \cos^n x$. Taking logs turns the product into a sum and the powers into coefficients, avoiding a clumsy repeated product rule.

Steps

Step 1: Take the natural log of both sides

Step 2: Differentiate term by term

Each term needs the chain rule, since $\log(\sin x)$ and $\log(\cos x)$ are themselves composite functions:

Step 3: Multiply through by $y$ and substitute back

$\frac{dy}{dx} = \sin^m x \cos^n x\,(m\cot x - n\tan x)$ …

Common Mistakes

Mistake 1: Dropping the negative sign from the derivative of $\cos x$

Why it's wrong: $\dfrac{d}{dx}\cos x = -\sin x$, not $\sin x$ — if this sign is lost while differentiating $\cos^n x$, the corresponding term of the final answer comes out with the wrong sign. Correct approach: always write the derivative of $\cos x$ with its negative sign explicit before simplifying further.

Mistake 2: Forgetting to multiply by $y$ after logarithmic differentiation

Why it's wrong: differentiating $\log y$ only ever produces $\dfrac{1}{y}\dfrac{dy}{dx}$; stopping at $m\cot x - n\tan x$ and calling that the derivative skips the final multiplication step. Correct approach: multiply the simplified expression by the original $y=\sin^m x\cos^n x$ before reporting the answer. …