The Quotient Rule (College Board AP® Calculus AB)

Study Guide

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Written by: Jamie Wood

Reviewed by: Dan Finlay

Derivatives of quotients

How do I differentiate one function divided by another?

The derivative of the quotient of two functions can be found using the quotient rule
The quotient rule states that
- If $f (x) = \frac{g (x)}{h (x)}$ ,
- then $f^{'} (x) = \frac{g^{'} (x) \cdot h (x) - g (x) \cdot h^{'} (x)}{{(h (x))}^{2}}$
This is also commonly written as
- If $y = \frac{u}{v}$ ,
- then $\frac{d y}{d x} = \frac{\frac{d u}{d x} \cdot v - u \cdot \frac{d v}{d x}}{v^{2}}$
- Or in a more concise form: $y^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}$
Any quotient rule problem can alternatively be solved as a product rule problem by writing $f (x) = \frac{g (x)}{h (x)}$ as $f (x) = g (x) \cdot {(h (x))}^{- 1}$
- The product rule and the chain rule can then be applied

Examiner Tips and Tricks

Notice that the numerator is the same as the product rule but with a subtraction instead (provided you write the $u^{'} v$ term first!).

Product rule: If $y = u \cdot v$ , then $y^{'} = u^{'} v + u v^{'}$ .

Worked Example

Find the derivative of the following functions.

(a) $f (x) = \frac{4 x + 3}{2 x + 5}$

Answer:

Assign $u$ and $v$ to each function

$u = 4 x + 3$

$v = 2 x + 5$

Find the derivatives of $u$ and $v$

$u^{'} = 4$

$v^{'} = 2$

Apply the quotient rule, $y^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}$

$y^{'} = \frac{4 (2 x + 5) - (4 x + 3) \cdot 2}{{(2 x + 5)}^{2}}$

Simplify

$y^{'} = \frac{(8 x + 20) - (8 x + 6)}{{(2 x + 5)}^{2}} = \frac{14}{{(2 x + 5)}^{2}}$

$f^{'} (x) = \frac{14}{{(2 x + 5)}^{2}}$

(b) $g (x) = \frac{\sin 3 x}{e^{4 x}}$

Answer:

Assign $u$ and $v$ to each function

$u = \sin 3 x$

$v = e^{4 x}$

Find the derivatives of $u$ and $v$

$u^{'} = 3 \cos 3 x$

$v^{'} = 4 e^{4 x}$

Apply the quotient rule, $y^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}$

$y^{'} = \frac{3 \cos 3 x \cdot e^{4 x} - \sin 3 x \cdot 4 e^{4 x}}{{(e^{4 x})}^{2}}$

In the numerator, the $e^{4 x}$ term can be factored out

$y^{'} = \frac{e^{4 x} (3 \cos 3 x - 4 \sin 3 x)}{{(e^{4 x})}^{2}}$

Simplify the powers of $e^{4 x}$

$g^{'} (x) = \frac{3 \cos 3 x - 4 \sin 3 x}{e^{4 x}}$

Last updated: 5 August 2024

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