Direct answerCalculus IIntermediate7 min read

What is the derivative of x to the x?

The base and exponent both vary, so neither the ordinary power rule nor the simple exponential rule works alone.

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

Direct answer

The result first, followed by the reasoning and a clean verification.

Reviewed July 11, 2026

AnswerStart here

\boxed{\frac{d}{dx}x^x=x^x(\ln x+1)}\qquad(x>0)

Take logarithms first so the variable exponent becomes a product.

Why the usual rules do not fit

The power rule assumes a constant exponent. The derivative of a to the x assumes a constant base. In x to the x, both move, so applying either rule by itself drops part of the change.

Logarithmic differentiation turns the exponent into multiplication, where the product rule can see both contributions.

The logarithmic step

Set y equal to x to the x and take natural logs: ln y = x ln x. Differentiate implicitly, remembering that y is a function of x.

Multiplying by y at the end restores the original exponential expression.

\frac{y'}y=\ln x+1

Domain note

The real-valued logarithmic derivation is cleanest for x greater than zero. Extensions to selected negative rational inputs require more careful domain analysis and do not define one smooth real function on all negative x.

Worked exampleDifferentiate and evaluate

Find the slope of y = xˣ at x = e.

y'=x^x(\ln x+1)

Use logarithmic differentiation.

y'(e)=e^e(\ln e+1)

Substitute x = e.

\ln e=1

Simplify the logarithm.

Result

\boxed{2e^e}

Watch for

Common mistakes

Writing x·x^(x−1) and treating the exponent as constant.
Forgetting the chain-rule factor y′/y after differentiating ln y.
Omitting the positive-domain assumption.

Keep

Three takeaways

Variable exponents suggest logarithms.
Differentiate x ln x with the product rule.
Substitute y = xˣ back only at the end.