Is this method of finding the derivative of $x^x$ acceptable in math?

Question

I wanted to find the derivative of the $x^x$, and I came up with the following, and I want to know if it's acceptable to do so:

Let's put $x^x = y$

$x^x = y $

$\ln(x^x) = \ln(y)$

$x\ln(x) = \ln(y)$

$d[x\ln(x)]/dx = [\ln(y)']/dx$

$\ln(x) + 1 = y'/y$

$y(\ln(x) + 1) = y'$

$x^x(\ln(x) + 1) = y'$

When I graph it, it comes as the correct derivative. Is it acceptable or not to do this?

It's fine. A more straightforward way is to write $x^x$ as $e^{x \ln x}$. — Ethan Bolker, Oct 27 '21 at 20:10
$\ln$ is differentiable, so that's fine, and then you're using the method of implicit differentiation. — march, Oct 27 '21 at 20:11
Per the comment of @EthanBolker, $\displaystyle \frac{d}{dx} e^{f(x)} = f'(x) e^{f(x)}.$ — user2661923, Oct 27 '21 at 20:20
@EthanBolker Right, it's easier that way I guess. Thanks for showing me! — Rayan, Oct 27 '21 at 20:51

score 1 · Answer 1 · answered Oct 27 '21 at 21:26

Your approach is reasonable, though some of the notation in the intermediate steps is suspect. For instance $[\ln(y)']/dx$ is a strange mix of Leibniz and Newton notation.

The key point is that the equation $$x \ln x = \ln y(x)$$ is true for all $x$, and so you can differentiate both sides with respect to $x$. (In general, you can differentiate both sides when they are both equal as functions of $x$. You cannot differentiate if they are equal only for one particular value of $x$).

$$\frac{d}{dx}\left(x \ln x\right) = \frac{d}{dx}\left(\ln y(x)\right),$$ where I've written $y(x)$ instead of $y$ on the RHS to remind myself that $y$ is not an independent variable: it's a function, dependent on $x$. Carrying out the derivatives then gets you to the last part of your proof, $$\ln x + 1 = \frac{1}{y(x)} \frac{dy}{dx}.$$

score 0 · Answer 2 · answered Oct 27 '21 at 21:25

Yes, it is valid; this is called logarthmic differentiation.

This works if the function is differentiable. In your case, $x^x$ is always positive, so it's pretty straighforward. But, because we're taking the logarithm, this won't work for a function with zeroes, or if the function is negative (with real numbers at least).

Usually this would be done by writing $x^x=e^{x\ln x}$, since this is more straightforward to differentiate, but what you did is still valid.

Is this method of finding the derivative of $x^x$ acceptable in math?

2 Answers2