calculus - 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.

Answer

First let's try to understand why the derivative of the function $f$ given by $f(x) = x^2$ is equal to $2x$ and not to $x$. (The product rule and the power rule are both generalizations of this.)

Imagine that you have a square whose sides have length $x$. Now imagine what happens to its area if we increase the length of each side by a small amount $\Delta x$. We can do this by adding three regions to the picture: two thin rectangles measuring $x$ by $\Delta x$ (say one on the right of the square and another on the top) and one small square measuring $\Delta x$ by $\Delta x$ (say added in the top right corner.) So the change in the area $x^2$ is equal to $2x \cdot \Delta x + (\Delta x)^2$. If we divide this by $\Delta x$ and take the limit as $\Delta x$ approaches zero, we get $2x$.

So geometrically what is happening is that the small square in the corner is too small to matter, but you have to count both rectangles. If you only count one of them, you will get the answer $x$; however, this only tells you what happens when you lengthen, say, the horizontal sides and not the vertical sides of your square to get a rectangle. This is a different problem than the one under consideration, which asks (after we put it in geometrical terms) how the area varies as we lengthen all the sides.