Find derivative of x^x using logarithmic differentiation

Question

Find $\frac{d}{dx}(x^x)$ .

Solution — Step by Step

$x^x$ is a variable raised to a variable power. Neither the power rule ( $\frac{d}{dx}x^n = nx^{n-1}$ ) nor the exponential rule ( $\frac{d}{dx}a^x = a^x \ln a$ ) applies here — those rules require either a constant exponent or a constant base. For $x^x$ , both base and exponent vary.

Logarithmic differentiation converts the problem into something manageable.

Let $y = x^x$ . Take $\ln$ of both sides:

\ln y = \ln(x^x)

Using the logarithm power rule $\ln(a^b) = b\ln a$ :

\ln y = x \ln x

Differentiate implicitly. Left side uses the chain rule:

\frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}(x \ln x)

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

So:

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

Multiply both sides by $y$ :

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

Substitute back $y = x^x$ :

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

Why This Works

$x^x$ can be rewritten as $e^{x\ln x}$ (using $a = e^{\ln a}$ ). Then:

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

This is the same answer — logarithmic differentiation is just an organised way of applying this $e^{\ln(\cdot)}$ trick. Taking $\ln$ first makes the algebra cleaner.

Logarithmic differentiation is used whenever you have:

Variable to the power of variable: $x^x$ , $x^{\sin x}$ , $(\sin x)^x$
Long products/quotients: easier to differentiate $\ln(f \cdot g)$ than $f \cdot g$ directly

The result $x^x(\ln x + 1)$ should pass a sanity check. At $x = 1$ : derivative $= 1^1(\ln 1 + 1) = 1(0 + 1) = 1$ . You can verify numerically: $1.001^{1.001} \approx 1.001001...$ , giving slope $\approx 1$ . ✓

Alternative Method — Exponential Form

Write $x^x = e^{x\ln x}$ directly.

Let $u = x\ln x$ , so $y = e^u$ .

$\frac{dy}{du} = e^u$ and $\frac{du}{dx} = \ln x + 1$ (product rule).

By chain rule: $\frac{dy}{dx} = e^u \cdot \frac{du}{dx} = e^{x\ln x}(\ln x + 1) = x^x(\ln x + 1)$

Both methods are equivalent — use whichever feels more natural.

Common Mistake

The most common error is treating $x^x$ as if the exponent is a constant: writing $\frac{d}{dx}(x^x) = x \cdot x^{x-1} = x^x$ . This incorrectly uses the power rule with the exponent “treated as constant.” The correct derivative $x^x(\ln x + 1)$ equals $x^x$ only when $\ln x + 1 = 1$ , i.e., $\ln x = 0$ , i.e., $x = 1$ . At any other $x$ , the power rule approach gives the wrong answer.

Question

Solution — Step by Step

Why This Works

Alternative Method — Exponential Form

Common Mistake

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