Q.Differentiate w.r.t. $x$: $(\sin x)^{\cos x}$.

We treat this as an exponential function of the form $y = (\sin x)^{\cos x}$. The key idea is implicit differentiation (or logarithmic differentiation), because the variable appears in both the base and the exponent.

Step 1: Take the natural logarithm of both sides:

Step 2: Differentiate implicitly with respect to $x$:

Step 3: Simplify the right-hand side:

$\frac{1}{y} \frac{dy}{dx} = -\sin x \log(\sin x) + \frac{\cos^2 x}{\sin x}.$ …

When you see a function where both the base and the exponent are functions of $x$ — like $(\sin x)^{\cos x}$ — the standard power rule or exponential rule alone won't work. The power rule $\frac{d}{dx} x^n = n x^{n-1}$ assumes a constant exponent. The exponential rule $\frac{d}{dx} a^x = a^x \log a$ assumes a constant base. Here, both are moving.

The trick is to take the natural logarithm first. This brings the exponent down as a product, turning the problem into something we can differentiate using the product rule and chain rule. Then we solve for $\frac{dy}{dx}$ by multiplying through by the original function.

Let’s do it step by step.

1. Set up the function and take $\log$ of both sides

   Let $y = (\sin x)^{\cos x}$.

   Taking the natural logarithm:

Using the logarithm power rule: $\log(a^b) = b \log a$, we get:

2. Differentiate implicitly with respect to $x$

   On the left side, $\frac{d}{dx} (\log y) = \frac{1}{y} \cdot \frac{dy}{dx}$ (chain rule).

   On the right side, we have a product: $\cos x$ times $\log(\sin x)$. Use the product rule:

The derivative of $\log(\sin x)$ is $\frac{1}{\sin x} \cdot \cos x = \cot x$, but I’ll keep it as $\frac{\cos x}{\sin x}$ for clarity.

So the right side becomes:

Simplify the second term: $\cos x \cdot \frac{\cos x}{\sin x} = \frac{\cos^2 x}{\sin x}$.

Therefore:

3. Solve for $\frac{dy}{dx}$

   Multiply both sides by $y$: …

Method: Logarithmic Differentiation for Variable Base, Variable Exponent

Use this method whenever a function has the form $[f(x)]^{g(x)}$, where both the base and the exponent depend on $x$. Neither the power rule (which needs a constant exponent) nor the exponential rule (which needs a constant base) applies directly — logarithms convert the exponent into a coefficient so ordinary differentiation rules work again.

Steps

Step 1: Take the natural log of both sides

For $y = [f(x)]^{g(x)}$, write

using the rule $\log(a^b) = b\log a$. This is always the first move — it turns a variable exponent into an ordinary product.

Step 2: Differentiate both sides with respect to $x$ …

Common Mistakes

Mistake 1: Applying the power rule with the exponent treated as constant

Why it's wrong: writing $\dfrac{d}{dx}(\sin x)^{\cos x} = \cos x\,(\sin x)^{\cos x - 1}\cos x$ silently assumes $\cos x$ doesn't change — but it does, so this drops an entire term of the derivative. Correct approach: whenever the exponent is itself a function of $x$, logarithmic differentiation is mandatory, not optional.

Mistake 2: Applying the exponential rule with the base treated as constant

Why it's wrong: writing $\dfrac{d}{dx}(\sin x)^{\cos x} = (\sin x)^{\cos x}\log(\sin x)\cdot(-\sin x)$ assumes $\sin x$ is a fixed base like $e$ — but the base is also varying with $x$, so this drops the contribution from differentiating the base. Correct approach: take logs first so both varying pieces are accounted for together. …