The Product Rule
In calculus textbooks, the product rule for differentiation is often presented as a theorem and then proven using the definition of the derivative. Other authors choose to simply write out the steps of the proof and present the product rule as a useful identity without motivating the mathematical steps (see MathWorld for an example). For those of us who prefer to think visually, both of these approaches are somewhat unsatisfactory. On this page, I present the product rule in a more visual / geometric way. The explanation presented below may not be considered "rigorous" enough for a mathematician, however it's a suitable explanation for a physicist.

Suppose the product we are differentiating is the area of a rectangle,
. If the width and length of the rectangle are
and
, then

,
, and
. Further suppose that, in a time interval
, the area of the rectangle changes by
.


, we see that
,
, and
also approach zero. In fact, if
,
, and
are continuous functions of
:



with respect to
is
![\displaystyle\frac{{\rm d}}{{\rm d}t}\left[x(t)y(t)\right]=\frac{{\rm d}A}{{\rm d}t}=\lim_{\Delta t\rightarrow0}\left[ y\frac{\Delta x}{\Delta t}+x\frac{\Delta y}{\Delta t} +\frac{\Delta x\Delta y}{\Delta t}\right]](http://www.idius.net/wp-content/plugins/latex/cache/tex_edbac001d8f66a3714b0a1801844582c.gif)
![\displaystyle\frac{{\rm d}}{{\rm d}t}\left[x(t)y(t)\right]=y\frac{{\rm d}x}{{\rm d}t}+x\frac{{\rm d} y}{{\rm d}t}](http://www.idius.net/wp-content/plugins/latex/cache/tex_b91faf584f3e4f427b71215fb5f18ebd.gif)
