The Quotient Rule For Differentiation

# The Quotient 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 quotient rule for differentiation which says that if $f$ and $g$ are both differentiable at $c$ and $g(c) \neq 0$ then the quotient function $\frac{f}{g}$ is differentiable at $c$ and $\displaystyle{\left ( \frac{f}{g} \right ) '(c) = \frac{g(c)f'(c) - g'(c)f(c)}{[g(c)]^2}}$.

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 $c$ and $g(c) \neq 0$ then $\left ( \frac{f}{g} \right )$ is differentiable at $c$ and $\displaystyle{\left ( \frac{f}{g} \right )'(c) = \frac{g(c)f'(c) - g'(c)f(c)}{[g(c)]^2}}$. |

**Proof:**Let $f$ and $g$ both be differentiable at $c$ and suppose that $g(c) \neq 0$. Now:

\begin{align} \quad \left ( \frac{f}{g} \right )' (c) &= \lim_{x \to c} \frac{\frac{f(x)}{g(x)} - \frac{f(c)}{g(c)}}{x - c} \\ \quad &= \lim_{x \to c} \frac{\frac{f(x)g(c) - f(c)g(x)}{g(x)g(c)}}{x - c} \\ \quad &= \lim_{x \to c} \frac{f(x)g(c) - f(c)g(x)}{g(x)g(c) [x - c]} \\ \quad &= \lim_{x \to c} \left ( \frac{f(x)g(c) - f(c)g(c) + f(c)g(c) - f(c)g(x)}{g(x)g(c) [ x - c]} \right ) \\ \quad &= \lim_{x \to c} \left ( \frac{g(c)}{g(x)g(c)} \frac{f(x) - f(c)}{x - c} \right ) + \lim_{x \to c} \left ( \frac{f(c)}{g(x)g(c)} \frac{g(c) - g(x)}{x - c}\right ) \\ \quad &= \frac{g(c)}{[g(c)]^2} f'(c) - \frac{f(c)}{[g(c)]^2} g'(c) \\ \quad &= \frac{g(c)f'(c) - g'(c)f(c)}{[g(c)]^2} \end{align}

- Since $g(c) \neq 0$ and since $f'(c)$ and $g'(c)$ exists we see that $\left ( \frac{f}{g} \right )'$ exists, so $\frac{f}{g}$ is differentiable at $c$ and $\displaystyle{\left ( \frac{f}{g} \right ) ' (c) = \frac{g(c)f'(c) - g'(c)f(c)}{[g(c)]^2}}$. $\blacksquare$