Derivative of xˣ: Formula, Proof and Examples

The derivative of x^x is x^x(\ln x + 1). It is the classic example of a function that needs logarithmic differentiation.

Why the usual rules fail

Applying the power rule gives x\cdot x^{x-1} = x^x, and applying the exponential rule gives x^x\ln x. Both are wrong. Interestingly, the correct answer is their sum.

Method 1: logarithmic differentiation

Let y = x^x. Take the natural log of both sides to bring the exponent down:

Differentiate both sides. The left side needs implicit differentiation, and the right side needs the product rule (it is the derivative of x ln x):

Multiply by y = x^x:

Method 2: rewrite with base e

Since x = e^{\ln x}, we have x^x = e^{x\ln x}. Now use the chain rule with outer function e^u:

Both methods are really the same idea. Use whichever feels more natural.

Graph of xˣ and its derivative

Since x^x > 0, the sign of the derivative depends only on \ln x + 1. It is negative for x < \frac{1}{e} and positive after, so the minimum is at x = \frac{1}{e}. Its value is \left(e^{-1}\right)^{1/e} = e^{-1/e} \approx 0.692.

Worked examples

Example 1: slope at x = 2. \frac{d}{dx}x^x\big|_{x = 2} = 2^2(\ln 2 + 1) = 4(1.693) \approx 6.77.

Example 2: x to the sin x. Let y = x^{\sin x}. Then \ln y = \sin x\ln x, and

Example 3: x to the 1/x. Let y = x^{1/x}. Then \ln y = \frac{\ln x}{x}, and the quotient rule gives

This is zero at x = e, so x^{1/x} is largest at x = e, with value e^{1/e} \approx 1.445.

Example 4: x to the x squared. For y = x^{x^2}, \ln y = x^2\ln x, so y' = x^{x^2}(2x\ln x + x).

Common mistakes

1. Power rule: x\cdot x^{x-1} = x^x. Wrong, because the exponent is not constant.
2. Exponential rule: x^x\ln x. Wrong, because the base is not constant.
3. Forgetting to multiply by y at the end of logarithmic differentiation, leaving just \ln x + 1.

For more on this method, see implicit differentiation problems and the derivative of 2ˣ and aˣ.