TITLE: Why does the power rule work?
QUESTION [11 upvotes]: 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.

REPLY [3 votes]: 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.