Visual calculus

I've been wondering: is it possible to visualize the rules of calculus?

1. Integration by parts

To simplify things, in the above visualization $v(x_1) = 0$, so that $\int^{x'}_{x_1}\frac{dv}{dx}dx = v(x')$.  This means that $v(x_2)$ is simply the area of the cross-section facing us on the LHS, and the volume  on the LHS is $u(x_2)v(x_2)$.

The RHS shows the same volume split into two parts.  The rectangle embedded in the first is $u\frac{dv}{dx}$, and so its volume is $\int^{x_2}_{x_1}u\frac{dv}{dx}dx$.  The cross section of the 2nd part is $v(x)$, so its volume is $\int^{u_2}_{u_1}vdu$, which becomes $\int^{x_2}_{x_1}v\frac{du}{dx}dx$ when rewritten as an integral over $x$.

So the picture is visual proof that the equation in pink holds, provided that $v(x_1) = 0$.  To prove it in general we just need to check that when we replace $v$ with $v+v_1$ in the equation, both sides change by the same amount.

2. The chain rule

This picture demonstrates the chain rule for differentiation.  $z$ is a function of $y$ which is a function of $x$.  If $x$ changes by $\delta x$ then $y$ changes by, approximately, $\frac{dy}{dx}\delta x$, and so $z$ changes by, approximately, $\frac{dz}{dy}\frac{dy}{dx}\delta x$.

3. The product rule

If $u$ changes by $\delta u$ and $v$ by $\delta v$ then $uv$ changes by $v\delta u + u\delta v + \delta u\delta v$.  Dividing this by $\delta x$ and taking the limit gives the result.

Note that if you integrate both sides of the product rule you get the integration by parts rule.