Suggestion for answer to Question 16

[biscarri1]

First, thanks for creating this doc! I'm in the process of preparing for interviews currently and it's been extremely useful.

I wanted to suggest an alternative solution to problem 16 that I find a bit more intuitive than the one provided. It's effectively the same, but I think it follows a smoother train of thought since it beings with the reflex that many people have of immediately taking ln() when they see an expression with an exponential. I've pasted the LaTeX for the solution below.

Begin by considering,
%
\begin{align*}
    f(x) = ln(x^{x}) = xln(x).
\end{align*}
%
Then we have,
%
\begin{align*}
    f'(x) = ln(x) + 1.
\end{align*}
%
For any function, $g$, we have by the chain rule,
%
\begin{align*}
    \frac{d}{dx}g(x^{x}) = g'(x^{x})\frac{d}{dx}x^{x}.
\end{align*}
%
If we let $g = f$, and rearrange the above we get,
%
\begin{align*}
   \frac{d}{dx}x^{x} &= \frac{\frac{d}{dx}f(x^{x})}{f'(x^{x})} \\ 
   &= \frac{ln(x)+1}{\frac{1}{x^{x}}} \\
   &= x^{x}(ln(x)+1).
\end{align*}