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 , my instinct is to use the power rule:
But thats wrong! The correct answer involves :
But WHY doesnt the power rule work? The power rule says , so if , isnt it ?
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?
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2 comments
Sarah MitchellMay 5, 2026
omg the multivariable chain rule explanation finally makes sense TYSM
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
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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