Why does the power rule work? If $$f(x)=x^u$$ then the derivative function will always be $$f'(x)=u*x^{u-1}$$
I've been trying to figure out why that makes sense and I can't quite get there.
I know it can be proven with limits, but I'm looking for something more basic, something I can  picture in my head.
The derivative should be the slope of the function.
If $$f(x)=x^3=x^2*x$$ then the slope should be $x^2$. But it isn't. The power rule says it's $3x^2$.
I understand that it has to do with having variables where in a more simple equation there would be a constant. I'm trying to understand how that exactly translates into the power rule.

This may be too advanced for you right now,
but knowing about the derivative of the log function
can be very helpful.
The basic idea is that
$(\ln(x))' = 1/x$,
where $\ln$ is the natural log.
Applying the chain rule,
$(\ln(f(x))' = f'(x)/f(x)$.
For this case,
set $f(x) = x^n$.
Then $\ln(f(x)) = \ln(x^n) = n \ln(x)$.
Taking derivatives on both sides,
$(\ln(x^n))' = f'(x)/f(x) = f'(x)/x^n$
and $(\ln(x^n))' = (n \ln(x))' = n/x$,
so
$f'(x)/x^n = n/x$
or $f'(x) = n x^{n-1}$.
More generally,
if $f(x) = \prod a_i(x)/\prod b_i(x)$,
then
$\ln(f(x)) = \sum \ln(a_i(x))-\sum \ln(b_i(x))$
so
$\begin{align}
f'(x)/f(x) &= (\ln(f(x))'\\
 &= \sum (\ln(a_i(x))'-\sum (\ln(b_i(x))'\\
&=\sum (a_i(x))'/a_i(x)-\sum (b_i(x))'/b_i(x)\\
\end{align}
$,
so
$f'(x) = f(x)\left(\sum (a_i(x))'/a_i(x)-\sum (b_i(x))'/b_i(x)\right)
$.
Note that this technique,
called logarithmic differentiation,
 generalizes both the
product and quotient rules for derivatives.
