Let \(u = f(x), v = g(x)\text{.}\) Then, as \(x\) changes, both \(f\) and \(g\) change, and this causes \(uv = f(x)g(x)\) to change. The product rule shows how the product \(uv\) changes as \(x\) changes, in terms of \(f\) and \(g\text{.}\) Suppose that \(x\) is incremented by some \(\Delta x\text{,}\) and let \(\Delta u, \Delta v\) be the corresponding increments in \(u\) and \(v\text{.}\) Then, consider the change in the product, \(\Delta(uv)\text{,}\)
\begin{equation*}
\Delta(uv) = \underbrace{(u + \Delta u)(v + \Delta v)}_{\text{new }} - \underbrace{uv}_{\text{old }}
\end{equation*}
Then, expanding and simplifying,
\begin{align*}
\Delta(uv) \amp = uv + \Delta u v + u \Delta v + \Delta u \Delta v - uv\\
\amp = \Delta u v + u \Delta v + \Delta u \Delta v
\end{align*}
Then, dividing both sides by \(\Delta x\text{,}\)
\begin{align*}
\frac{\Delta(uv)}{\Delta x} \amp = \frac{\Delta u v}{\Delta x} + \frac{u \Delta v}{\Delta x} + \frac{\Delta u \Delta v}{\Delta x}\\
\amp = \frac{\Delta u}{\Delta x} v + u \frac{\Delta v}{\Delta x} + \Delta u \frac{\Delta v}{\Delta x}
\end{align*}
Then, as \(\Delta x \to 0\text{,}\)
\begin{align*}
\lim_{\Delta x \to 0} \frac{\Delta(uv)}{\Delta x} \amp = \lim_{\Delta x \to 0} \brac{\frac{\Delta u}{\Delta x} v + u \frac{\Delta v}{\Delta x} + \Delta u \frac{\Delta v}{\Delta x}}\\
\frac{d(uv)}{dx} \amp = \lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} \cdot v + \lim_{\Delta x \to 0} u \cdot \frac{\Delta v}{\Delta x} + \lim_{\Delta x \to 0} \Delta u \frac{\Delta v}{\Delta x}
\end{align*}
The first term approaches \(\frac{du}{dx} v\text{,}\) the second approaches \(u \frac{dv}{dx}\text{,}\) and for the third, \(\lim_{\Delta x \to 0} \frac{\Delta v}{\Delta x} = \frac{dv}{dx}\text{,}\) however \(\Delta u\) also depends on \(\Delta x\text{,}\) and since \(f\) is continuous, as the change in \(x\) goes to 0, the change in \(u\) goes to 0 also. Thus,
\begin{align*}
\frac{d(uv)}{dx} \amp = \frac{du}{dx} v + u \frac{dv}{dx} + 0 \cdot \frac{dv}{dx}\\
\amp = \frac{du}{dx} v + u \frac{dv}{dx}
\end{align*}
