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

Concept: Implicit Differentiation — when the function is a sum of variable-exponent terms, take logs of each term separately and differentiate.

Let $y = (\log x)^x + x^{\log x}$. Write $u = (\log x)^x$ and $v = x^{\log x}$, so $y = u + v$.

For $u = (\log x)^x$, take $\log u = x \log(\log x)$. Differentiate:

Thus $\frac{du}{dx} = (\log x)^x \left( \log(\log x) + \frac{1}{\log x} \right)$.

For $v = x^{\log x}$, take $\log v = (\log x)(\log x)$. Differentiate:

$\frac{1}{v} \frac{dv}{dx} = \frac{1}{x} \cdot \log x + \log x \cdot \frac{1}{x} = \frac{\log x + \log x}{x}$ …

We have $y = (\log x)^x + x^{\log x}$. The variable $x$ appears in both the base and the exponent of each term — that’s the classic signal that direct differentiation rules (like the power rule or exponential rule) won’t work alone. Instead, we use logarithmic differentiation: take the natural log of each term, differentiate implicitly, then solve for the derivative.

Let’s break it into two parts:

Let $u = (\log x)^x$ and $v = x^{\log x}$, so $y = u + v$, and $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$.

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Step 1: Differentiate $u = (\log x)^x$

Take natural log on both sides:

Now differentiate both sides with respect to $x$. On the left, by the chain rule: $\frac{d}{dx} \log u = \frac{1}{u} \cdot \frac{du}{dx}$. On the right, use the product rule:

The last term simplifies: $x \cdot \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{\log x}$.

So:

Multiply through by $u$:

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Step 2: Differentiate $v = x^{\log x}$

Again, take natural log:

Differentiate both sides:

Use the chain rule: derivative of $(\log x)^2$ is $2(\log x) \cdot \frac{1}{x}$.

So:

Multiply by $v$:

$\frac{dv}{dx} = x^{\log x} \cdot \frac{2\log x}{x}$ …

Method: Logarithmic Differentiation for Two Terms Where $\log x$ Plays Different Roles

This is a variant of term-by-term logarithmic differentiation, used here because both terms — $(\log x)^x$ and $x^{\log x}$ — involve $\log x$ in a different role (once as the base, once as the exponent), which makes it easy to conflate the two.

Steps

Step 1: Split into two separate terms and log-differentiate the first, $u=(\log x)^x$.

Differentiate with the Product Rule; the derivative of $\log(\log x)$ needs the Chain Rule twice — once for the outer log, once for the inner $\log x$:

Step 2: Log-differentiate the second term, $v=x^{\log x}$.

This is now a single composite function of $x$ (not a product), so differentiate with the Chain Rule directly:

$\frac{1}{v}\frac{dv}{dx} = 2\log x\cdot\frac{1}{x} = \frac{2\log x}{x}$ …

Common Mistakes

Mistake 1: Treating $\log v = (\log x)^2$ as a product needing the Product Rule.

Why it's wrong: $(\log x)^2$ is a single function squared, not two different functions multiplied together — it should be differentiated with the Chain Rule (power rule on the outside, $\log x$ on the inside), giving $2\log x\cdot\frac1x$, not via the Product Rule. Correct approach: recognise $(\log x)\cdot(\log x)$ as $(\log x)^2$ and apply the Chain Rule to that squared form directly.

Mistake 2: Dropping the extra $\frac1x$ factor when differentiating $\log(\log x)$ for the first term.

Why it's wrong: $\log(\log x)$ is a composite function whose derivative is $\frac{1}{\log x}\cdot\frac1x$; writing only $\frac{1}{\log x}$ loses the inner Chain Rule factor. Correct approach: always identify the innermost function ($\log x$) and multiply by its derivative ($\frac1x$) even when it looks like a "small" extra step. …