Is this method of finding the derivative of $x^x$ acceptable in math? 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?
 A: 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,
$$\ln x + 1 = \frac{1}{y(x)} \frac{dy}{dx}.$$
A: 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.
