To start, we need some way to understand the function One interpretation of multiplication is it is the area of a \(f(x) \times g(x)\) rectangle: To understand the derivative of the product, we must understand how the area, \(A\), changes as \(x\) changes. If we change the inputs of \(f\) and \(g\) by \(dx\), then the size of the rectangle changes: However, we know from our previous work that \begin{align*} f(x+dx) &\approx f(x) + df,\\ g(x+dx) &\approx g(x) + dg,\\ \end{align*}

so now our picture becomes:

Note, if we think of \(A(x) = f(x)\cdot g(x)\), then we can also label our picture as follows: Finally, from the pictures above and recalling that \begin{align*} df &= f'(x) dx\\ dg &= g'(x) dx, \end{align*}

we see that \begin{align*} dA &= f(x) dg + g(x) df + df dg\\ &=f(x) g'(x) dx + g(x) f'(x) dx + f'(x) g'(x) \left (\d x\right )^2. \end{align*}

Dividing both sides by \(dx\) we see

and letting \(dx\) go to zero we see

We’ll try to understand this geometrically. In what follows, the functions \(f\) and \(g\) look like lines; however, the young mathematician should realize that we are not looking at true lines, instead we are looking at \(f\) and \(g\) sufficiently “zoomed-in” so that they appear to be lines. First consider a graph of \(g\) with respect to \(x\): Now consider a graph of \(f\) with respect to \(g\): If we combine these graphs, by laying the graph of \(g\) on its side, we obtain: Ah! From this we see that \begin{align*} df &= f'(g) dg\\ &= f'(g(x)) g'(x) dx, \end{align*}

so