To start, we need some way to understand the function $$A(x) = f(x) \cdot g(x).$$ 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 $$\frac {dA}{dx} = f(x) g'(x) + g(x) f'(x) + f'(x) g'(x) dx$$ and letting $dx$ go to zero we see $$A'(x) = f(x) g'(x) + g(x) f'(x).$$

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 $$\frac {df}{dx} = f'(g(x))g'(x).$$