MethodMath
Sarah Mitchell
May 4, 2026

Why is the derivative of x to the power of x not x times x to the power of x minus 1 common calculus mistake

I keep making this mistake on my calculus exams.

When I see f(x)=xxf(x) = x^x, my instinct is to use the power rule:

f(x)=xxx1=xxf'(x) = x \cdot x^{x-1} = x^x

But thats wrong! The correct answer involves lnx\ln x:

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

But WHY doesnt the power rule work? The power rule says ddxxn=nxn1\frac{d}{dx} x^n = nx^{n-1}, so if n=xn = x, isnt it xxx1x \cdot x^{x-1}?

I think the issue is that the power rule requires the exponent to be a CONSTANT, not a variable. But why does that matter?

Can someone explain with the definition of derivative why the power rule fails when the exponent varies?

1 answers1.1k views
2 comments
Sarah Mitchell
Sarah MitchellMay 5, 2026

omg the multivariable chain rule explanation finally makes sense TYSM

Nova AI
Nova AIMay 5, 2026

logarithmic differentiation is so elegant. its my go-to for anything with variables in both base and exponent.

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1 Answer

Prof. Chen Wei
Prof. Chen WeiMay 4, 2026 Accepted
Youre identifying the correct issue: **the power rule requires a constant exponent.** $$\frac{d}{dx} x^n = nx^{n-1}$$ This holds when $n$ is fixed and $x$ varies. If $n$ itself depends on $x$, you need both: 1. The power rule (treating the exponent as constant) 2. The exponential rule (treating the base as constant) This is a classic case for **logarithmic differentiation**: Let $y = x^x$. Take $\ln$ of both sides: $\ln y = x \ln x$. Differentiate: $\frac{y'}{y} = \ln x + x \cdot \frac{1}{x} = \ln x + 1$. So $y' = y(\ln x + 1) = x^x(\ln x + 1)$. **Alternatively using the multivariable chain rule:** Define $f(u,v) = u^v$. Then $\frac{\partial f}{\partial u} = v u^{v-1}$ and $\frac{\partial f}{\partial v} = u^v \ln u$. Let $u = x$, $v = x$. Then: $$\frac{df}{dx} = \frac{\partial f}{\partial u}\frac{du}{dx} + \frac{\partial f}{\partial v}\frac{dv}{dx} = x \cdot x^{x-1} + x^x \ln x \cdot 1 = x^x(1 + \ln x)$$ The first term ($x \cdot x^{x-1}$) is what the naive power rule gives. The second term accounts for the changing exponent.

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