# The Product Rule

Theorem:

If $f$ and $g$ are functions that are differentiable at $x$, then the derivative of their product exists and is given by

$$\frac{d}{dx} \left[ f\,(x) g(x) \right] = f\,(x) g'(x) + f\,'(x) g(x)$$

Proof:

After appealing to the limit definition of the derivative, we add a well-chosen value of zero, $f\,(x+h)g(x) - f\,(x+h)g(x)$, to the numerator:

$$\begin{array}{rcl} \frac{d}{dx} \left[ f\,(x) g(x) \right] & = & \lim_{h \rightarrow 0} \ \frac{ f\,(x+h) g(x+h) - f\,(x) g(x)}{h}\\\\ & = & \lim_{h \rightarrow 0} \ \frac{ f\,(x+h) g(x+h) - f\,(x+h) g(x) + f\,(x+h) g(x) - f\,(x) g(x)}{h} \end{array}$$

We then manipulate this expression to reveal the same limits that appear in the definitions of $f\,'(x)$ and $g'(x)$.

$$\begin{array}{rcl} \frac{d}{dx} \left[ f\,(x) g(x) \right] & = & \lim_{h \rightarrow 0} \ \frac{ f\,(x+h) [g(x+h) - g(x)] + [f\,(x+h) - f\,(x)] g(x)}{h}\\\\ & = & \lim_{h \rightarrow 0} \ f\,(x+h) \cdot \frac{ g(x+h) - g(x) }{h} + \lim_{h \rightarrow 0} \ \frac{f\,(x+h) - f\,(x)}{h} \cdot g(x)\\\\ & = & \lim_{h \rightarrow 0} \ f\,(x+h) \cdot \lim_{h \rightarrow 0} \ \frac{ g(x+h) - g(x) }{h} + \lim_{h \rightarrow 0} \ \frac{f\,(x+h) - f\,(x)}{h} \cdot \lim_{h \rightarrow 0} \ g(x) \end{array}$$

Replacing these limits with $f\,'(x)$ and $g'(x)$, and noting the last limit is of a constant, we have

$$\frac{d}{dx} \left[ f\,(x) g(x) \right] = \lim_{h \rightarrow 0} \ f\,(x+h) \cdot g'(x) + f\,'(x) g(x)$$

To simplify the last limit, we note that if $f$ is differentiable at $x$, it must also be continuous there. As such,

$$\lim_{h \rightarrow 0} \ f\,(x+h) = \lim_{u \rightarrow x} \ f\,(u) = f(x)$$

Making this replacement, we arrive at our desired result:

$$\frac{d}{dx} \left[ f\,(x) g(x) \right] = f\,(x) g'(x) + f\,'(x) g(x)$$ QED.