How to calculate the derivative of $x^x$? I'm trying to follow an example in my textbook.
$$y=x^x$$
$$\ln (y)=\ln (x) \cdot x$$
We want to calculate the derivative with respect to x
The book makes quite a leap here and states that:
$$\frac{y'}{y}=\frac{1}{x}\cdot x+1\cdot \ln(x)$$
Since $y=x^x$ this means that:
$$y'=x^x(1+\ln(x))$$
Is this correct?
If I start from the beginning then:
$$y=x^x$$
$$\ln (y)=\ln (x) \cdot x$$
Only if we want to take the derivative this expression isn't useful, we'll have to use the full expression:
$$e^{\ln (y)}=e^{\ln (x) \cdot x}$$
Now, if we take the derivative of this with respect to x we find that:
$$\frac {1}{y}\cdot e^{\ln (y)}=e^{\ln (x) \cdot x}\cdot (1+ln(x))$$
The left hand expression would simplify to $\frac {y'}{y}$, since $e^{\ln(y)}=y$ and the derivative of $y=y'$
so:
$$\frac {y'}{y}=e^{\ln (x) \cdot x}\cdot (1+\ln(x))$$
Which can be written as:
$$\frac {y'}{x^x}=x^x\cdot (1+\ln(x))$$
Which simplifies to:
$$y'=x^{2x}\cdot (1+\ln(x))$$

You missed out $y'$ in this line:

$$\frac {1}{y}e^{\ln y}=e^{x\ln x}(1+\ln x)$$

as $$\left(e^{\ln y}\right)'=e^{\ln y}\cdot\frac1y\cdot y'=y'$$
