Derivative Quotient Rule

# The Quotient Rule

Like the product rule for differentiation, there is an analogous rule for quotients of functions.

 Theorem 1 (The Quotient Rule): If $f$ and $g$ are differentiable functions, then the derivative of the quotient $\frac{f}{g}$ is $\frac{d}{dx} \left ( \frac{f(x)}{g(x)} \right) = \frac{g(x) \frac{d}{dx} f(x) - f(x) \frac{d}{dx} g(x)}{(g(x))^2}$ provided that $g(x) \neq 0$.
• Proof: Suppose that $f$ and $g$ are differentiable functions, and using the definition of a derivative we get that:
(1)
\begin{align} \frac{d}{dx} \left ( \frac{f(x)}{g(x)} \right) = \lim_{h \to 0} \left ( \frac{\frac{f(x+h)}{g(x + h)} - \frac{f(x)}{g(x)}}{h} \right )\\ \frac{d}{dx} \left ( \frac{f(x)}{g(x)} \right) = \lim_{h \to 0} \left ( \frac{\frac{f(x+h)g(x) - f(x)g(x+h)}{g(x+h)g(x)}}{h} \right )\\ \frac{d}{dx} \left ( \frac{f(x)}{g(x)} \right) = \lim_{h \to 0} \left ( \frac{f(x+h)g(x) - f(x)g(x+h)}{h \cdot g(x+h)g(x)} \right )\\ \end{align}
• Now factor $\frac{1}{g(x+h)g(x)}$ from the denominator and add/subtract $f(x)g(x)$ to the numerator to get that:
(2)
\begin{align} \frac{d}{dx} \left ( \frac{f(x)}{g(x)} \right) = \lim_{h \to 0} \left ( \frac{1}{g(x+h)g(x)} \cdot \frac{f(x+h)g(x) - f(x)g(x+h)}{h} \right ) \\ \: \frac{d}{dx} \left ( \frac{f(x)}{g(x)} \right) = \lim_{h \to 0} \left ( \frac{1}{g(x+h)g(x)} \cdot \frac{f(x+h)g(x) - f(x)g(x) + f(x)g(x) - f(x)g(x+h)}{h} \right ) \\ \frac{d}{dx} \left ( \frac{f(x)}{g(x)} \right) = \lim_{h \to 0} \left ( \frac{1}{g(x+h)g(x)} \cdot \frac{g(x)[f(x+h) - f(x)] + f(x)[g(x) - g(x+h)]}{h} \right ) \\ \frac{d}{dx} \left ( \frac{f(x)}{g(x)} \right) = \lim_{h \to 0} \left ( \frac{1}{g(x+h)g(x)} \cdot \frac{g(x)[f(x+h) - f(x)] - f(x)[g(x+h) - g(x)]}{h} \right ) \\ \frac{d}{dx} \left ( \frac{f(x)}{g(x)} \right) = \lim_{h \to 0} \frac{1}{g(x+h)g(x)} \cdot \lim_{h \to 0} \left( \frac{g(x)[f(x+h) - f(x)] - f(x)[g(x+h) - g(x)]}{h} \right ) \\ \: \frac{d}{dx} \left ( \frac{f(x)}{g(x)} \right) = \lim_{h \to 0} \frac{1}{g(x+h)g(x)} \cdot \left ( \lim_{h \to 0} g(x) \frac{f(x + h) - f(x)}{h} - \lim_{h \to 0} f(x) \frac{g(x + h) - g(x)}{h} \right ) \\ \: \frac{d}{dx} \left ( \frac{f(x)}{g(x)} \right) = \lim_{h \to 0} \frac{1}{g(x+h)g(x)} \cdot \left ( g(x) \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} - f(x) \lim_{h \to 0} \frac{g(x + h) - g(x)}{h} \right )\\ \frac{d}{dx} \left ( \frac{f(x)}{g(x)} \right) = \frac{1}{[g(x)]^2} \cdot \left ( g(x) \frac{d}{dx}f(x) - f(x) \frac{d}{dx}g(x) \right ) \\ \frac{d}{dx} \left ( \frac{f(x)}{g(x)} \right) = \frac{g(x) \frac{d}{dx} f(x) - f(x) \frac{d}{dx} g(x)}{(g(x))^2} \quad \blacksquare \end{align}

Sometimes it is easier to remember this rule using prime notation, that is $\left ( \frac{f}{g} \right)' = \frac{gf' - fg'}{g^2}$.

We will now look at some examples applying the quotient rule.

## Example 1

Calculate the derivative of: $f(x) = \frac{\sin x}{\cos x}$.

We already know that $f(x) = \frac{\sin x}{\cos x} = \tan x$ and that $\frac{d}{dx} \tan x = \sec ^2 x$. Let's use the quotient rule to verify this result with the trigonometric identity that $\cos ^2 x + \sin ^2 x = 1$:

(3)
\begin{align} \frac{d}{dx} f(x) = \frac{h(x) \frac{d}{dx} g(x) - g(x) \frac{d}{dx} h(x)}{(h(x))^2} \\ \frac{d}{dx} f(x) = \frac{\cos x \cos x - \sin x(- \sin x) }{\cos^2 x} \\ \frac{d}{dx} f(x) = \frac{\cos^2 x + \sin^2 x }{\cos^2 x} \\ \frac{d}{dx} f(x) = \frac{1}{\cos^2 x} \\ \frac{d}{dx} f(x) = \sec^2 x \\ \end{align}

## Example 2

Calculate the derivative of: $f(x) = \frac{\sin (x) - x^2}{(2x - 5x^4 + 3)}$.

Applying the quotient rule, we get:

(4)
\begin{align} \frac{d}{dx} f(x) = \frac{h(x) \frac{d}{dx} g(x) - g(x) \frac{d}{dx} h(x)}{(h(x))^2} \\ \frac{d}{dx} f(x) = \frac{(2x - 5x^4 + 3)(\cos (x) - 2x) - (\sin (x) - x^2)(2 - 20x^3)}{(2x - 5x^4 + 3)^2} \\ \end{align}