Derivative of x^x
The derivative of $x^x$ (for $x>0$) is $\dfrac{d}{dx}x^x = x^x(\ln x + 1)$. Neither the power rule nor the exponential rule applies here, because the variable is in the base and in the exponent.
See the derivative of x^x step by step in the calculator →
📈 Graph of x^x and its derivative
🔍 Why is the derivative of x^x equal to $x^x(\ln x + 1)$?
Logarithmic differentiation: take the natural logarithm of $y = x^x$ and differentiate implicitly:
With the exponential function: $x^x = e^{x\ln x}$. The chain rule gives $e^{x\ln x}\cdot(x\ln x)' = x^x(\ln x + 1)$.
🔗 Derivative of x^x with an inner function
If the argument is a function $u(x)$, the chain rule adds the inner derivative:
✏️ Worked examples
| Function | Derivative | Note | |
|---|---|---|---|
| $x^{2x}$ | $2x^{2x}(\ln x + 1)$ | $\ln y = 2x\ln x$. | steps → |
| $x^{\sin x}$ | $x^{\sin x}\left(\cos x\ln x + \frac{\sin x}{x}\right)$ | General formula with $u = x$, $v = \sin x$. | steps → |
| $x^2$ | $2x$ | Constant exponent: the power rule is enough. | steps → |
| $2^x$ | $2^x\ln 2$ | Constant base: the exponential rule is enough. | steps → |
⚠️ Common mistakes
Applying the power rule: $x\cdot x^{x-1} = x^x$ is wrong – it misses the $\ln x$ term.
Applying the exponential rule: $x^x\ln x$ is wrong – it misses the $+1$.
❓ Frequently asked questions
Why is the derivative of x^x not x·x^(x−1)?
The power rule needs a constant exponent. Here the exponent changes too, which adds the extra term $x^x\ln x$.
Where does x^x have its minimum?
At $x = \frac1e$, where $\ln x + 1 = 0$. The minimum value is $e^{-1/e}\approx 0.692$.
What is the derivative of x^(1/x)?
$x^{1/x}\cdot\frac{1-\ln x}{x^2}$, by logarithmic differentiation of $\ln y = \frac{\ln x}{x}$.