1. ## Derivative of x^x

I read the theory over and over again, but i don't get it.
You want the derivative of x^x. The theory says

"If , then . The chain rule is then used, to differentiate ln(y) with respect to x (y'/y). The product rule is then used on the right. Then solve for y'"

So you'd get
ln (y)'--> (1/y) =ln(x)+1<-- xln(x)'
Thus y= 1/ln(x)+1, which is wrong. The derivative of x^x should be:x^x(ln(x)+1)
1) Where is the mistake?
2)as the graph of y=f(x) is the same as the graph of ln(y)=ln(f(x)), why can you not simply derive the "right side", why do you have to derive both sides? In "normal" deriviation, you also don't derive the y!

2. Hello, Schdero!

I can't connect to that link.

Differentiate: . $y \:=\:x^x$

Take logs: . $\ln y \:=\:\ln(x^x) \quad\Rightarrow\quad \ln y \:=\:x\ln x$

Differentiate implicitly: . $\frac{1}{y}\!\cdot\!\frac{dy}{dx} \:=\:x\!\cdot\!\frac{1}{x} + \ln x$

We have: . $\frac{1}{y}\!\cdot\!\frac{dy}{dx} \;=\;1 + \ln x \quad\Rightarrow\quad \frac{dy}{dx} \;=\;y\left(1 + \ln x\right)$

Therefore: . $\frac{dy}{dx}\;=\;x^x\left(1 + \ln x\right)$

3. $
\frac{1}{y}\!\cdot\!\frac{dy}{dx}
$

why is it necessary, do derive ln(y) at all? $\frac{dy}{dx}
$
means the right side derived, but why do you also have to derive the ln(y)?

4. Originally Posted by Schdero
$
\frac{1}{y}\!\cdot\!\frac{dy}{dx}
$

why is it necessary, do derive ln(y) at all? $\frac{dy}{dx}
$
means the right side derived, but why do you also have to derive the ln(y)?

I'm not exactly sure what your problem is. There seem to be two issues involved. One is how to determine the derivative $y'(x)$ of $y(x) :=x^x$ with respect to x. You can do this without reference to the "rule of logarithmic differentiation", using chain rule and product rule only, like this:

$y'(x)=\left(x^x\right)'=\left(\mathrm{e}^{x\cdot \ln x}\right)'=\mathrm{e}^{x\cdot\ln x}\cdot (1\cdot \ln x+x\cdot \tfrac{1}{x})=x^x\cdot (\ln x+1)$

The other issue is the "rule of logarithmic differentiation" which says that $y'(x)=y(x) \cdot\left(\ln y(x)\right)'$.
This follows from differentiation of $\ln y(x)$ with respect to x by the chain rule: $\left(\ln y(x)\right)' =\frac{1}{y(x)}\cdot y'(x)$. Solving for $y'(x)$ gives the above rule.

And, yes, differentiation of $y(x)=x^x$ can be done by referring to this rule of logarithmic differentation (if you happen to know it):
$y'(x)=y(x)\cdot\left(\ln y(x)\right)' =x^x\cdot \left(\ln x^x\right)' = x^x\cdot \left(x\cdot \ln x\right)'=x^x\cdot (\ln x+1)$

5. I didn't see, respectively understand, the thing with the chain rule, but it's clear now. thanks for both your patience and explenations ;-)
Schdero