This is a visualization of the product rule in calculus, which says that, for differentiable functions [math]f[/math] and [math]g[/math],[br][br][math]\left(fg\right)'=f'g+g'f[/math][br]Or, equvalently,[br][math]\frac{df\left(t\right)g\left(t\right)}{dt}=\frac{df\left(t\right)}{dt}g\left(t\right)+\frac{dg\left(t\right)}{dt}f\left(t\right)[/math][br][br][br]The basic idea of this visualization is that any product [math]ab[/math] can be visualized as the area of an [math]a\times b[/math] rectangle. In this case, the rectangles will be [math]f\left(t\right)\times g\left(t\right)[/math] rectangles for different values of [math]t[/math].[br][br]For small [math]h[/math], [math]\left(fg\right)'[/math] can be approximated as [math]\frac{f\left(t+h\right)g\left(t+h\right)-f\left(t\right)g\left(t\right)}{h}[/math]. We can visualize the numerator as a change in rectangle area from an [math]f\left(t\right)\times g\left(t\right)[/math] rectangle to an [math]f\left(t+h\right)\times g\left(t+h\right)[/math] rectangle. That change can be visualized as the L-shaped area made of three rectangles colored red, brown, and blue in the activity above. If we define [math]\Delta f=f\left(t+h\right)-f\left(t\right)[/math] and [math]\Delta g=f\left(t+h\right)-f\left(t\right)[/math], then it's clear that the area of the red rectangle is [math]f\left(t\right)\Delta g[/math], the area of the blue rectangle is [math]g\left(t\right)\Delta f[/math], and the area of the brown rectangle is [math]\Delta f\Delta g[/math]. [br][br]Now drag the [math]h[/math] slider to the left, making is smaller. What you can see is that, as [math]h[/math] gets smaller, the brown rectangle becomes less and less significant in the L-shaped area representing the difference in rectangle areas. Ignoring that brown area, we can see that[br][br][math]\frac{f\left(t+h\right)g\left(t+h\right)-f\left(t\right)g\left(t\right)}{h}\approx\frac{f\left(t\right)\Delta g+g\left(t\right)\Delta f}{h}=f\left(t\right)\frac{\Delta g}{h}+g\left(t\right)\frac{\Delta f}{h}[/math].[br][br]In the limiting case as [math]h\longrightarrow0[/math],[br][br] [math]\frac{d\left(f\left(t\right)g\left(t\right)\right)}{dt}=f\left(t\right)\frac{dg\left(t\right)}{dt}+g\left(t\right)\frac{df\left(t\right)}{dt}[/math][br][br]One slightly confusing thing about the picture is that the graph shows a parametric curve where [math]t[/math] varies invisibly while the curve represents values of [math]\left(f\left(t\right),g\left(t\right)\right)[/math]. Don't let that confusion get in the way of the visualization.