Q.Find $\frac{dy}{dx}$ in the following: $(\log x)^{\cos x}$

Concept: Implicit Differentiation (or logarithmic differentiation, since the variable appears in both the base and the exponent).

Let $y = (\log x)^{\cos x}$. Take the natural logarithm of both sides:

Differentiate both sides with respect to $x$:

Multiply through by $y$ and substitute back $y = (\log x)^{\cos x}$:

The problem asks for $\frac{dy}{dx}$ when $y = (\log x)^{\cos x}$. At first glance, this looks like a power function, but the exponent is not a constant — it's $\cos x$, which varies with $x$. Similarly, the base $\log x$ is not a constant either. So neither the standard power rule ($\frac{d}{dx} x^n = n x^{n-1}$) nor the exponential rule ($\frac{d}{dx} a^x = a^x \ln a$) applies directly.

The standard technique for such "variable base, variable exponent" functions is logarithmic differentiation. The idea is simple: take the natural log of both sides, simplify using log properties, and then differentiate implicitly. This converts the messy exponent into a product, which we can handle with the product rule.

Let’s work through it step by step.

1. Set up the equation and take logs. Start with $y = (\log x)^{\cos x}$. Take the natural logarithm of both sides:

Using the power property of logs, $\ln(a^b) = b \ln a$, we get:

2. Differentiate both sides with respect to $x$. On the left side, $\frac{d}{dx} (\ln y) = \frac{1}{y} \cdot \frac{dy}{dx}$ (by the chain rule, since $y$ is a function of $x$). On the right side, we have a product: $\cos x$ times $\ln(\log x)$. So we use the product rule:

3. Differentiate $\ln(\log x)$. Let $u = \log x$ (here $\log$ means base $e$, i.e., natural log, as is standard in calculus). Then $\frac{d}{dx} (\ln u) = \frac{1}{u} \cdot \frac{du}{dx} = \frac{1}{\log x} \cdot \frac{1}{x}$. So:

4. Put it together. The derivative of the right side becomes:

So we have:

5. Solve for $\frac{dy}{dx}$. Multiply both sides by $y$:

Now substitute back $y = (\log x)^{\cos x}$:

Method: Logarithmic Differentiation for a Variable Base Raised to a Variable Power

Use this method whenever $y = [f(x)]^{g(x)}$ — that is, both the base and the exponent depend on $x$ — since neither the power rule (which needs a constant exponent) nor the exponential rule (which needs a constant base) applies on its own.

Steps

Step 1: Take the natural log of both sides and bring the exponent down.

For $y=(\log x)^{\cos x}$, using $\log(a^b) = b\log a$:

This converts a "power with a variable exponent" into a product of two ordinary functions of $x$, which the Product Rule can handle.

Step 2: Differentiate the right-hand side using the Product Rule.

The second piece needs the Chain Rule twice: differentiating $\log(\log x)$ gives $\frac{1}{\log x}$ times the derivative of the inner $\log x$, which is $\frac{1}{x}$.

Step 3: Multiply through by $y$ and substitute the original function back in.

Whenever you spot both the base and the exponent changing with $x$, this "log, differentiate the product, multiply back by $y$" routine is the only reliable path — trying to force the power rule or exponential rule directly will always go wrong.

Common Mistakes

Mistake 1: Missing the double Chain Rule when differentiating $\log(\log x)$.

Why it's wrong: $\log(\log x)$ is a composite function — its derivative is $\frac{1}{\log x}$ (outer) times $\frac{1}{x}$ (inner), i.e. $\frac{1}{x\log x}$. Students often write just $\frac{1}{\log x}$ and drop the extra $\frac{1}{x}$. Correct approach: treat $\log(\log x)$ as $\log(u)$ with $u=\log x$ and apply the Chain Rule explicitly.

Mistake 2: Applying the ordinary power rule because $\cos x$ "looks like" a constant exponent.

Why it's wrong: $\cos x$ is not a fixed number — it changes with $x$ — so writing $\cos x\,(\log x)^{\cos x - 1}$ (as if differentiating $u^n$ for fixed $n$) is incorrect and completely misses the base's own dependence on $x$. Correct approach: whenever the exponent is a function of $x$, logarithmic differentiation is mandatory, not optional.

Mistake 3: Dropping the negative sign from $\frac{d}{dx}\cos x = -\sin x$ inside the product rule term.

Why it's wrong: this sign carries through to the final answer's first term; losing it flips that term's sign in the result. Correct approach: write $(-\sin x)\log(\log x)$ explicitly as a signed term before combining with the second part of the product rule.