The Quotient Rule: Formula, Proof, and Examples

The quotient rule in calculus is used to differentiate the quotient (division) of two or more functions.

For a quotient of two functions, the quotient rule is equal to the denominator multiplied by the derivative of the numerator minus the numerator multiplied by the derivative of the denominator all divided by the square of the denominator.

It's perhaps easier to learn the formula!

Before learning how to differentiate using the quotient rule, it is helpful to know how to differentiate $ e^x, ln(x), sin(x), cos(x), $ and $ tan(x)$.

Knowledge of the product rule is also beneficial.

Quotient Rule Formula

Here is the quotient rule formula:

For $ y = \frac{u}{v} $,

$ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} $

Using function notation, we can also write the product rule like this:

$$ \frac{d}{dx} [\frac{f(x)} {g(x)}] = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{g^2(x)} $$

A proof of the quotient rule formula can be found at the bottom of this page.

Quotient Rule Examples

Example 1:

Differentiate $ y = \frac{x^2 - 3}{2x - 1} $ with respect to $ x $.

Solution:

Let $ u = x^2 - 3 $ and let $ v = 2x - 1 $

Then $ \frac{du}{dx} = 2x $ and $ \frac{dv}{dx} = 2 $

Substitute into the quotient rule:

$ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} $

$ \frac{dy}{dx} = \frac{(2x - 1) \cdot 2x - (x^2 - 3) \cdot 2}{(2x - 1)^2} $

$ \frac{dy}{dx} = \frac{2x(2x - 1) - 2(x^2 - 3)}{(2x - 1)^2} $

$ \frac{dy}{dx} = \frac{4x^2 - 2x - 2x^2 + 6}{(2x - 1)^2} $

$ \frac{dy}{dx} = \frac{2x^2 - 2x + 6}{(2x - 1)^2} $

Example 2:

Differentiate $ y = \frac{e^{3x}}{x^2} $ with respect to $ x $.

Solution:

Let $ u = e^{3x} $ and let $ v = x^2 $

Then $ \frac{du}{dx} = 3e^{3x} $ and $ \frac{dv}{dx} = 2x $

Substitute into the quotient rule:

$ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} $

$ \frac{dy}{dx} = \frac{x^2 \cdot 3e^{3x} - e^{3x} \cdot 2x}{x^4} $

$ \frac{dy}{dx} = \frac{e^{3x}(3x^2 - 2x)}{x^4} $

$ \frac{dy}{dx} = \frac{e^{3x}(3x - 2)}{x^3} $

Example 3:

Differentiate $ y = \frac{\cos 2x}{e^{2x + 1}} $ with respect to $ x $.

Solution:

Let $ u = \cos 2x $ and $ v = e^{2x + 1} $.

Then, $ \frac{du}{dx} = -2 \sin 2x $

and $ \frac{dv}{dx} = 2e^{2x + 1} $ (using the chain rule)

Using the quotient rule, we have:

\[ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \] \[ = \frac{e^{2x + 1} \cdot (-2 \sin 2x) - \cos 2x \cdot 2e^{2x + 1}}{(e^{2x + 1})^2} \] \[ = \frac{-2e^{2x + 1} (\sin 2x + \cos 2x)}{e^{2x + 1} \cdot e^{2x + 1}} \] \[ = -\frac{2(\sin 2x + \cos 2x)}{e^{2x + 1}} \]

When to Use the Product Rule or Quotient Rule

By rewriting the denominator of the quotient as a term with a negative power, we can change the quotient into a product.

For example, consider the function $ f(x) = \frac{2x^3}{x^2 + 1} $.,

We can rewrite it as:

\[ f(x) = 2x^3 \cdot (x^2 + 1)^{-1} \]

Rewriting the quotient as a product means that we can also use the product rule to differentiate a quotient.

Whether you use the quotient rule or the product rule is a matter of personal preference. The product rule is simpler to use, but you'll need to rewrite the expression with a negative power first. Choose the option that you think you're less likely to make a mistake.

Product Rule vs Quotient Rule Example:

Differentiate $ y = \frac{x^2 + 3}{x^4} $ with respect to $ x $ using both the quotient rule and the product rule.

Solution (using the Quotient Rule):

Let $ u = x^2 + 3 $ and $ v = x^4 $.

Then, $ \frac{du}{dx} = 2x $ and $ \frac{dv}{dx} = 4x^3 $.

Substitute into the quotient rule:

\[ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \] \[ = \frac{x^4 \cdot 2x - (x^2 + 3) \cdot 4x^3}{(x^4)^2} \] \[ = \frac{2x^5 - 4x^5 - 12x^3}{x^8} \] \[ = \frac{-2x^5 - 12x^3}{x^8} \] \[ = -\frac{2x^5}{x^8} - \frac{12x^3}{x^8} \] \[ = -\frac{2}{x^3} - \frac{12}{x^5} \]

Solution (using the Product Rule):

Rewrite $ y $ as $ y = (x^2 + 3) \cdot x^{-4} $.

Let $ u = x^2 + 3 $ and $ v = x^{-4} $.

Then, $ \frac{du}{dx} = 2x $ and $ \frac{dv}{dx} = -4x^{-5} $.

Substitute into the product rule:

\[ \frac{dy}{dx} = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} \] \[ = (x^2 + 3) \cdot (-4x^{-5}) + x^{-4} \cdot 2x \] \[ = -4(x^2 + 3)x^{-5} + 2x \cdot x^{-4} \] \[ = -\frac{4(x^2 + 3)}{x^5} + \frac{2}{x^3} \] \[ = -\frac{4x^2 + 12}{x^5} + \frac{2}{x^3} \] \[ = -\frac{2}{x^3} - \frac{12}{x^5} \]

Thus, in both methods, we find that:

\[ \frac{dg}{dx} = -\frac{2}{x^3} - \frac{12}{x^5} \]

Quotient Rule Questions in Context

The quotient rule can be used in problems such as finding the equations of tangents and normals to a curve and in identifying stationary points and their nature.

Example 4:

Find the coordinates of the points on the curve $ y = \frac{(x - 1)^2}{2x + 5} $ where the tangent is parallel to the x-axis.

Solution:

A tangent is parallel to the x-axis when the derivative $ \frac{dy}{dx} = 0 $.

Let $ u = (x - 1)^2 $ and $ v = 2x + 5 $.

Then, $ \frac{du}{dx} = 2(x - 1) $ and $ \frac{dv}{dx} = 2 $.

Using the quotient rule, we have:

\[ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \] \[ = \frac{(2x + 5) \cdot 2(x - 1) - (x - 1)^2 \cdot 2}{(2x + 5)^2} \] \[ = \frac{2(x - 1)[(2x + 5) - (x - 1)]}{(2x + 5)^2} \] \[ = \frac{2(x - 1)(x + 6)}{(2x + 5)^2} \]

Setting $ \frac{dy}{dx} = 0 $ gives:

\[ 2(x - 1)(x + 6) = 0 \] (as the denominator cannot be equal to zero)

Solving for $ x $:

\[ x = 1 \quad \text{or} \quad x = -6 \]

Substitute these values back into the original function to find the corresponding $ y $-coordinates:

\[ y = \frac{(1 - 1)^2}{2(1) + 5} = 0 \] \[ y = \frac{(-6 - 1)^2}{2(-6) + 5} = \frac{49}{-7} = -7 \]

Therefore, the points where the tangent is parallel to the x-axis are:

\[ (-6, -7) \quad \text{and} \quad (1, 0) \]

Example 5:

A curve has equation $ y = \frac{\sin 2x}{e^{2x}} $ for $ 0 \leq x \leq \frac{\pi}{2} $.

a) Find the exact value for the $ x $-coordinate of the stationary point of this curve.

b) Determine the nature of this stationary point.

Solution:

a) To find the stationary points, we need to differentiate $ y $ with respect to $ x $ and set $ \frac{dy}{dx} = 0 $.

Let $ u = \sin 2x $ and $ v = e^{2x} $.

Then, $ \frac{du}{dx} = 2 \cos 2x $ and $ \frac{dv}{dx} = 2e^{2x} $.

Using the quotient rule, we have:

\[ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \] \[ = \frac{e^{2x} \cdot 2 \cos 2x - \sin 2x \cdot 2e^{2x}}{(e^{2x})^2} \] \[ = \frac{2e^{2x} (\cos 2x - \sin 2x)}{e^{4x}} \] \[ = \frac{2 (\cos 2x - \sin 2x)}{e^{2x}} \]

Setting $ \frac{dy}{dx} = 0 $ gives:

\[ \cos 2x - \sin 2x = 0 \] \[ \cos 2x = \sin 2x \] \[ \frac{\sin 2x}{\cos 2x} = 1 \] \[ \tan 2x = 1 \]

This implies $ 2x = \frac{\pi}{4} $, so:

\[ x = \frac{\pi}{8} \]

Therefore, the exact value for the $ x $-coordinate of the stationary point is:

\[ x = \frac{\pi}{8} \]

b) To determine the nature of this stationary point, we need to find the second derivative $ \frac{d^2y}{dx^2} $ and evaluate it at $ x = \frac{\pi}{8} $.

Recall that:

\[ \frac{dy}{dx} = \frac{2 (\cos 2x - \sin 2x)}{e^{2x}} \]

We differentiate $ \frac{dy}{dx} $ with respect to $ x $ again, using the quotient rule:

\[ \frac{d^2y}{dx^2} = \frac{e^{2x} \cdot \frac{d}{dx} (2 (\cos 2x - \sin 2x)) - 2 (\cos 2x - \sin 2x) \cdot \frac{d}{dx} (e^{2x})}{(e^{2x})^2} \]

Calculating each derivative separately:

\[ \frac{d}{dx} (2 (\cos 2x - \sin 2x)) = -4 \sin 2x - 4 \cos 2x = -4 (\sin 2x + \cos 2x) \]

and

\[ \frac{d}{dx} (e^{2x}) = 2e^{2x} \]

Substitute these into $ \frac{d^2y}{dx^2} $:

\[ \frac{d^2y}{dx^2} = \frac{e^{2x} \cdot (-4 (\sin 2x + \cos 2x)) - 2 (\cos 2x - \sin 2x) \cdot 2e^{2x}}{e^{4x}} \] \[ = \frac{-4 (\sin 2x + \cos 2x) - 4 (\cos 2x - \sin 2x)}{e^{2x}} \] \[ = \frac{-8 \cos 2x }{e^{2x}} \]

Evaluating at $ x = \frac{\pi}{8} $:

\[ \frac{d^2y}{dx^2} \bigg|_{x = \frac{\pi}{8}} \approx -2.58 \, (\text{to 3 significant figures}) \]

Since $ \frac{d^2y}{dx^2} < 0 $ at $ x = \frac{\pi}{8} $, this point is a maximum.

Quotient Rule Practice Questions

Question 1

Differentiate $ \frac{2x + 3}{x - 4} $ with respect to $ x $

Quotient Rule Proof

The quotient rule states that for two differentiable functions $ f(x) $ and $ g(x) $ (with $ g(x) \neq 0 $), the derivative of their quotient is given by: $$ \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \frac{f'(x) g(x) - f(x) g'(x)}{[g(x)]^2} $$ We begin by using the definition of the derivative: $$ \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \lim_{\Delta x \to 0} \frac{\frac{f(x + \Delta x)}{g(x + \Delta x)} - \frac{f(x)}{g(x)}}{\Delta x} $$ To simplify, we find a common denominator: $$ = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) g(x) - f(x) g(x + \Delta x)}{\Delta x \cdot g(x) \cdot g(x + \Delta x)} $$ Next, we can add and subtract $ f(x) g(x) $ in the numerator: $$ = \lim_{\Delta x \to 0} \frac{[f(x + \Delta x) - f(x)] g(x) + f(x) [g(x) - g(x + \Delta x)]}{\Delta x \cdot g(x) \cdot g(x + \Delta x)} $$ Now, separate the terms in the limit: $$ = \lim_{\Delta x \to 0} \left( \frac{f(x + \Delta x) - f(x)}{\Delta x} \cdot \frac{g(x)}{g(x) \cdot g(x + \Delta x)} - \frac{f(x)}{g(x) \cdot g(x + \Delta x)} \cdot \frac{g(x + \Delta x) - g(x)}{\Delta x} \right) $$ As $ \Delta x \to 0 $, the first term becomes $ f'(x) \cdot \frac{g(x)}{[g(x)]^2} = \frac{f'(x) g(x)}{[g(x)]^2} $, and the second term becomes $ f(x) \cdot \frac{-g'(x)}{[g(x)]^2} = -\frac{f(x) g'(x)}{[g(x)]^2} $. So, we have: $$ \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \frac{f'(x) g(x) - f(x) g'(x)}{[g(x)]^2} $$ This completes the proof of the quotient rule.