Derivative of xˣ: Logarithmic Differentiation

Find the derivative of xˣ. This problem is famous for one reason: every rule you know fails on it. The power rule needs a constant exponent, but here the exponent x varies. The exponential rule d/dx[aˣ] = aˣ ln a needs a constant base, but here the base x varies. When both the base and the exponent move, students typically guess one of two wrong answers: x·xˣ⁻¹ (the power rule applied illegally) or xˣ ln x (the exponential rule applied illegally).

The correct move is to take the natural logarithm of both sides first. That drags the variable exponent down into a product, and products are something the standard rules can handle. One note on domain: the derivation uses ln x, so it holds for x > 0.

Worked solution

Set y = xˣ. Take the natural log of both sides: ln y = ln(xˣ) = x ln x. (Logarithm power rule: the exponent becomes a multiplier.)
Differentiate both sides with respect to x. The left side needs the chain rule: d/dx[ln y] = (1/y)·y′ = y′/y.
The right side needs the product rule: d/dx[x ln x] = 1·ln x + x·(1/x) = ln x + 1.
Equate the two sides: y′/y = ln x + 1.
Multiply through by y: y′ = y(ln x + 1) = xˣ(ln x + 1). This is the answer.

Why this method works

The logarithm converts exponentiation into multiplication: ln(xˣ) = x ln x. Products are exactly what the product rule and chain rule are built for. Implicit differentiation then lets us recover y′ without ever needing a rule for “a variable base raised to a variable power”, because no such elementary rule exists. Take the log, differentiate normally, solve for y′.

Common mistakes

Applying the power rule: d/dx[xˣ] = x·xˣ⁻¹ = xˣ. Wrong. The power rule d/dx[xⁿ] = nxⁿ⁻¹ requires n to be a constant.
Applying the exponential rule: d/dx[xˣ] = xˣ ln x. Wrong. That rule requires a constant base a.
Forgetting to multiply by y at the end, leaving the answer as ln x + 1. That is y′/y, not y′.
Ignoring the domain. The derivation uses ln x, so the result is valid for x > 0.

Check (never skip)

Verify with a second, independent method. Rewrite xˣ as e^(x ln x), since e^(ln z) = z. Then d/dx[e^(x ln x)] = e^(x ln x)·d/dx[x ln x] by the chain rule = e^(x ln x)(ln x + 1) = xˣ(ln x + 1). ✓ Two independent routes, same answer. As a numeric sanity check, at x = 1: y′ = 1¹(ln 1 + 1) = 1·(0 + 1) = 1, which is exactly the slope you get from a quick finite-difference check.

Why the two wrong answers fail

The two most common wrong answers each break one condition of a real rule. x·xˣ⁻¹ comes from the power rule d/dx[xⁿ] = nxⁿ⁻¹, whose proof fixes n as a constant before differentiating; here the “n” is x itself, moving as you differentiate. xˣ ln x comes from d/dx[aˣ] = aˣ ln a, whose proof fixes the base a; here the base x moves too. Each wrong answer applies half the truth: the power rule handles a varying base, the exponential rule handles a varying exponent, but neither handles both varying at once. That gap is precisely what logarithmic differentiation fills, by converting the problem into a product before any rule is applied.

A quick numeric test kills both wrong answers at once. At x = 2, the correct formula gives 2²(ln 2 + 1) ≈ 6.77. The power-rule guess gives 2·2¹ = 4. The exponential-rule guess gives 4·ln 2 ≈ 2.77. A rough slope check, (2.01^2.01 − 1.99^1.99)/0.02, lands near 6.78, matching only the logarithmic-differentiation answer.

FAQ

Q: What is logarithmic differentiation? A: The technique of taking ln of both sides before differentiating. Reach for it whenever the variable appears in both the base and the exponent, or when facing a messy product or quotient that a logarithm would flatten.

Q: Why exactly does the power rule fail here? A: The power rule is proved for constant exponents. With exponent x, the “n” in nxⁿ⁻¹ is itself changing as x changes, so the proof collapses. Rules have conditions; this problem violates the power rule’s condition.

Q: What is the derivative of xˣ at x = e? A: Plug in: eᵉ(ln e + 1) = eᵉ(1 + 1) = 2eᵉ.

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