Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do you differentiate $$\large{f(x) = x^x}$$

The working I got was $$\ln f(x) = x \ln x$$

which I am pretty fine...but I do not know how it advances to

$$\frac{f'(x)}{f(x)} = x\begin{pmatrix} \frac 1 x\end{pmatrix} + \ln x$$

although the final answer can be, by multiplying $f(x)$ on both sides of the equation,

$${f'(x)} = x^x\begin{bmatrix}x\begin{pmatrix} \frac 1 x\end{pmatrix} + \ln x\end{bmatrix}$$


Indeed, $$\ln f(x) = x \ln x$$ Differentiate both sides of the equation w.r.t $x$ $$\frac{f'(x)}{f(x)} = x\begin{pmatrix} \frac 1 x\end{pmatrix} + \ln x = 1 + \ln x$$ Bring the $f(x)$ over and you'll finally get

$$f'(x) = x^x\begin {pmatrix} 1 + \ln x\end{pmatrix}$$

share|cite|improve this question
Yes............but why won't you better write $\,x\frac{1}{x}=1\,$ ...? – DonAntonio Feb 3 '13 at 12:55
It was directly lifted off from the textbook. I know what you meant, but perhaps some help to get there before I change it back to $1 + \ln x$ – bryansis2010 Feb 3 '13 at 12:57
Some help...where? What you did is correct! – DonAntonio Feb 3 '13 at 12:58
If you have any function to a variable exponent in a calculus problem, say $f(x)^{g(x)}$ I find it almost always pays to replace it by $e^{f(x)g(x)}$. Example: ${d\over dx}2^x={d\over dx}e^{x\ln2}=(\ln 2)e^x$ and $\int 2^x dx=\int e^{x\ \ln 2}dx=e^{x \ln 2}+C$ by an elementary substitution which is much better than memorizing where that $\ln 2$ goes and then putting it in the wrong place. – Barbara Osofsky Feb 3 '13 at 14:22
up vote 1 down vote accepted

By the chain rule

$$\frac{d}{dx} \ln{f(x)} = \frac{1}{f(x)} f'(x)$$

share|cite|improve this answer

Another way (You already used the chain rule and $[\ln f(x)]'$: $$x^x=e^{x\ln x}$$

share|cite|improve this answer

The following came from a mathoverflow answer (I cannot recall the question):

A student was asked to differentiate $y=x^x$. Not remembering how to do logarithmic differentiation the student reasoned as follows: "well, I do not know how to differntiate $x^x$, but I do know how $x^r$. This is $rx^{r-1}$. But this is not right since the exponent is $x$. But I also know the derivative of $r^x$. This is $ln(r)r^x$. This is not right either for the same reason. So I'll split the difference and add them together and get $rx^{r-1}+ln(r)r^x$. But that $r$ is actually $x$ so the answer must be $x^x+ln(x)x^x$

The funny thing is that the result is right, but the reasoning is incomplete. What you need is multivariable chain rule for a composite of the form, $$\mathbb{R}\to\mathbb{R}^2\to\mathbb{R}$$, where the first map is the diagonal, and the second will be called $f$. The answer that you get in the case is $$\frac{df(x,x)}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}_{x=y}$$
Where $\frac{\partial f}{\partial y}_{x=y}$ is the composite $\frac{\partial f}{\partial y}(\iota)$, where $\iota(y)=x$.

share|cite|improve this answer
You would not necessarily see this type of solution in a first calculus course. – Baby Dragon Feb 3 '13 at 14:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.