Take log both sides: log y = x log x, then differentiate.

Logarithmic Differentiation — Find dy/dx if y = xˣ

Question

Find $\dfrac{dy}{dx}$ if $y = x^x$ .

Solution — Step by Step

We cannot differentiate $x^x$ directly — the base and exponent are both variables. Taking $\log$ converts this into something we can handle.

\log y = \log(x^x) = x \log x

Left side uses the chain rule: $\dfrac{d}{dx}(\log y) = \dfrac{1}{y} \cdot \dfrac{dy}{dx}$ .

Right side uses the product rule: $\dfrac{d}{dx}(x \log x) = 1 \cdot \log x + x \cdot \dfrac{1}{x} = \log x + 1$ .

So we get:

\frac{1}{y} \cdot \frac{dy}{dx} = \log x + 1

Multiply both sides by $y$ :

\frac{dy}{dx} = y(\log x + 1)

We always substitute back — leaving $y$ in the answer is incomplete.

\boxed{\frac{dy}{dx} = x^x(\log x + 1)}

Why This Works

The function $y = x^x$ is neither a pure power function (like $x^n$ , where only the base varies) nor a pure exponential (like $a^x$ , where only the exponent varies). Standard differentiation rules don’t apply directly because both the base and exponent contain $x$ .

Logarithmic differentiation is the standard tool for such forms. When we write $\log y = x \log x$ , the right side is now a simple product of two functions — perfectly handled by the product rule.

This technique extends naturally to products of many functions and to forms like $y = f(x)^{g(x)}$ in general. Master this once and you handle a whole family of JEE-favourite questions.

Alternative Method

We can rewrite $x^x$ using the exponential identity $a^b = e^{b \ln a}$ :

y = x^x = e^{x \ln x}

Now differentiate using the chain rule:

\frac{dy}{dx} = e^{x \ln x} \cdot \frac{d}{dx}(x \ln x)

Since $\dfrac{d}{dx}(x \ln x) = \ln x + 1$ :

\frac{dy}{dx} = e^{x \ln x}(\ln x + 1) = x^x(\ln x + 1)

Same answer, slightly different route. The logarithmic differentiation method (Step-by-Step above) is faster in exams — fewer chances of sign errors.

In JEE Main 2023, this exact function appeared as part of a two-step problem. Students who knew the $x^x$ derivative directly saved 30–40 seconds per question across the paper. Memorise the final form: $\dfrac{d}{dx}(x^x) = x^x(1 + \ln x)$ .

Common Mistake

The most frequent error: treating $x^x$ like $x^n$ and writing $\dfrac{dy}{dx} = x \cdot x^{x-1} = x^x$ .

This is wrong. The power rule $\dfrac{d}{dx}(x^n) = nx^{n-1}$ only applies when the exponent is a constant. Here, the exponent is $x$ — a variable. Using the power rule here is like applying a formula outside its validity condition, which NCERT explicitly warns against in Chapter 5.

Always ask yourself: “Is the exponent a constant or a variable?” If variable — reach for logarithmic differentiation.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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