The Sum Difference Rules for Differentiation

# The Sum Difference Rules 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 some nice theorems which say that if $f$ and $g$ are both differentiable at some point $c$ then the sum function $f + g$ and difference function $f - g$ are both differentiable at $c$ and moreover, $(f + g)'(c) = f'(c) + g'(c)$ and $(f - g)'(c) = f'(c) - g'(c)$.

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 $f + g$ is differentiable at $c$ and $(f + g)'(c) = f'(c) + g'(c)$. |

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

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

- So $(f + g)'(c)$ exists and $(f + g)'(c) = f'(c) + g'(c)$. $\blacksquare$

Theorem 2: 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 $f - g$ is differentiable at $c$ and $(f - g)'(c) = f'(c) - g'(c)$. |

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

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

- So $(f - g)'(c)$ exists and $(f - g)'(c) = f'(c) - g'(c)$. $\blacksquare$