Quotient Rule
In short: Quotient rule differentiates a fraction whose numerator and denominator both depend on x, using the bottom’s square as the new denominator and a difference of two cross products on top.
AP Calculus ABCalculusAP Calculus AB Unit 2
What is Quotient Rule?
The quotient rule is the correct recipe for a fraction made of two changing expressions, which cannot be differentiated by dividing the derivatives.
If \(u\) is the numerator, \(v\) the denominator and \(v \neq 0\):
\[\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u^{\prime}v – u\,v^{\prime}}{v^2}\]
Order matters in the top: the subtraction is not commutative. A memory aid says “low d-high minus high d-low, over the square of what’s below.”
How to spot it
- The function is a ratio of two expressions in x that does not simplify to a single power.
- A rate such as concentration, average cost or speed is given as one formula divided by another.
- You are asked where the graph of a rational expression levels off, which means finding a zero of the new numerator.
- Turning the fraction into a multiplication with a reciprocal power would be messier than working directly.
The method
- Label the top \(u\) and the bottom \(v\), and differentiate both.
- Write the numerator \(u^{\prime}v – u\,v^{\prime}\) in exactly that order.
- Place it over \(v^2\); there is no need to multiply out the squared denominator.
- Simplify the numerator; to find horizontal tangents, set that numerator equal to 0.
Quotient Rule: solved example
Problem
The concentration of a medicine in a patient’s blood is \(C(t) = \frac{5t}{t^2 + 4}\) milligrams per liter, \(t\) hours after the dose. Find the rate of change of the concentration at \(t = 1\).
Solution
Here \(u = 5t\), \(u^{\prime} = 5\), \(v = t^2 + 4\), \(v^{\prime} = 2t\).
\(C^{\prime}(t) = \frac{5(t^2 + 4) – 5t(2t)}{(t^2 + 4)^2} = \frac{20 – 5t^2}{(t^2 + 4)^2}\).
At \(t = 1\): \(\frac{20 – 5}{25} = \frac{15}{25}\). The concentration is rising at 0.6 milligram per liter per hour.
Try it
Question 1
For the dose model above, when does the blood level reach its highest point for \(t > 0\)?
Show the answer
Set \(20 – 5t^2 = 0\): \(t^2 = 4\), so \(t = 2\) hours. The peak value is \(C(2) = \frac{10}{8} = 1.25\) mg/L.
Question 2
Differentiate \(y = \frac{x}{x + 3}\).
Show the answer
\(y^{\prime} = \frac{1(x + 3) – x(1)}{(x + 3)^2} = \frac{3}{(x + 3)^2}\).
Quotient Rule vs. Product Rule
The quotient rule and its cousin both involve two functions and two derivatives, but the quotient rule has a minus sign and a squared denominator, so swapping the two terms in the numerator flips the sign of your answer. If you remember only the product rule, you can also treat \(\frac{u}{v}\) as \(u \cdot v^{-1}\), at the cost of using the chain rule.
Keep going
Lessons
Worksheets and quizzes
- Product Rule, Quotient Rule and Trig Derivatives – Worksheet
- Product Rule, Quotient Rule and Trig Derivatives – Quiz
Related words
Whole dictionary: A–Z · AP Calculus AB · Calculus