The Product Rule for Differentiation

# The Product Rule for Differentiation

Recall from The Derivative of a Function page that if $f$ is a function defined on the open interval $(a, b)$ and if $c \in (a, b)$ then $f$ is said to be differentiable at $c$ if the following limit (called the derivative of $f$ at $c$) exists:

(1)\begin{align} \quad f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} = \lim_{h \to 0} \frac{f(c + h) - f(c)}{h} \end{align}

The process by which $f'$ is obtained from $f$ is called differentiation.

We will now look at a nice theorem called the product rule for differentiation which says that if $f$ and $g$ are both differentiable at $c$ then the product function $fg$ is differentiable at $c$ and $(fg)'(c) = f(c)g'(c) + f'(c)g(c)$.

Theorem 1: Let $f$ and $g$ be functions defined on the open interval $(a, b)$ and let $c \in (a, b)$. If $f$ and $g$ are both differentiable at $c$ then $fg$ is differentiable at $c$ and $(fg)'(c) = f(c)g'(c) + f'(c)g(c)$. |

**Proof:**Let $f$ and $g$ both be differentiable at $c$. Then $f'(c)$ and $g'(c)$ both exist. Now:

\begin{align} \quad (fg)'(c) &= \lim_{x \to c} \frac{(fg)(x) - (fg)(c)}{x - c} \\ \quad &= \lim_{x \to c} \frac{f(x)g(x) - f(c)g(c)}{x - c} \\ \quad &= \lim_{x \to c} \frac{f(x)g(x) - f(x)g(c) + f(x)g(c) - f(c)g(c)}{x - c} \\ \quad &= \lim_{x \to c} \left ( f(x) \frac{g(x) - g(c)}{x - c} + g(c) \frac{f(x) - f(c)}{x - c} \right ) \\ \quad &= \lim_{x \to c} f(x) \cdot \lim_{x \to c} \frac{g(x) - g(c)}{x - c} + g(c) \cdot \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \\ \quad &= f(c)g'(c) + g(c)f'(c) \end{align}

- So $(fg)'(c)$ exists, so $fg$ is differentiable at $c$ and moreover, $(fg)'(c) = f(c)g'(c) + f'(c)g(c)$. $\blacksquare$