Find the derivative of $y=x^x$. Find the derivative of $y=x^x$.
My Attempt:
$$y=x^x$$
Taking $\textrm {ln}$ on both sides, we get:
$$\textrm {ln} y= \textrm {ln} x^x$$
$$\textrm {ln} y = x \textrm {ln} x$$
How do I procees further?
 A: Here is an alternative way to do it, without the use of implicit differentiation:

One can start by writing:
$$y=x^x=(e^{\ln{x}})^x=e^{x\ln{x}}$$
Hence, it now becomes extremely obvious to apply the chain rule to obtain:
$$\frac{dy}{dx}=\frac{d}{dx}(e^{x\ln{x}})=\color{green}{\frac{d}{dx}(x\ln{x})}\cdot e^{x\ln{x}}$$
Now all you need to do is apply the product rule on the green one.
A: your start is very good: $$\frac{1}{y}y'=\ln(x)+x\frac{1}{x}$$
A: Differentiate with respect to $x$ and on right hand side use product rule. 
$\frac 1y \frac{dy}{dx} = \frac{d}{dx} x \cdot \ln x + x \cdot \frac{d}{dx} \ln x$
$\frac 1y \frac{dy}{dx} = 1 \cdot \ln x + x \cdot \frac 1x$
$\frac 1y \frac{dy}{dx} = \ln x + 1$
$\frac{dy}{dx} = y(\ln x + 1)$
$\frac{dy}{dx} = x^x(\ln x + 1)$
A: Despite being an answer
Instead of "How do I proceed further?" one may ask as well "How can I preprocess?".
Here's a proposal: Searching for $\,$ x^x $\,$ in $\,$ https://approach0.xyz/search/ $\,$ yields many$^\text{many}$ hits. Ordered by age – or ripeness? – you may choose & click


*

*Finding the derivative of $x^x$

*Derivative of $x^x$ at $x=1$ from first principles

*Taylor expansion of $x^x-1$ around 1

*explaining the derivative of $x^x$

*Derivative of $x^x$ and the chain rule

*... and so on
Worthwhile & close to it is
Is $x^x$ an exponential function?
(just having received an upvote).
Still link no. 7
How effectively to search through MSE resources
before posting this one.
A: Remark that $y = x^x=e^{x \log x}$ and then proceed with the product rule $\frac{dy}{dx} = e^{x \log x} \frac{d}{dx} (x \log x) = x^x (\log x + 1)$.
