2.5 – The Quotient Rule
Learning Objectives
By the end of this section, you should be able to:
- State and apply the Quotient Rule using the notation: \( \frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \)
- Differentiate quotients of functions using the u-v notation
- Recognize when to use the Quotient Rule vs. other rules
- Simplify derivatives obtained using the Quotient Rule
- Apply the Quotient Rule to solve real-world problems
Notation Guide
Continuing our u-v notation from the Product Rule:
- Let \( y = \frac{u}{v} \) where u and v are functions of x
- \( \frac{du}{dx} \) = derivative of numerator (u)
- \( \frac{dv}{dx} \) = derivative of denominator (v)
- The Quotient Rule: \( \frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \)
The Quotient Rule
If \( y = \frac{u}{v} \), then:
\[ \frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \]
In words: The derivative of a quotient equals the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the denominator squared.
🎵 Easy to Remember! 🎵
"Low d-high minus high d-low, over the square of what's below"
Where "low" = denominator, "high" = numerator, "d-high" = derivative of numerator, "d-low" = derivative of denominator
Why We Need the Quotient Rule
Just like with products, the derivative of a quotient is NOT the quotient of the derivatives. For example:
If \( u = x^3 \) and \( v = x^2 \), then:
\[ \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{3x^2}{2x} = \frac{3}{2}x \]
But the actual derivative of \( y = \frac{u}{v} = \frac{x^3}{x^2} = x \) is:
\[ \frac{dy}{dx} = 1 \]
These are different! This shows why we need a special rule for quotients.
Worked Examples
Example 1: Basic Quotient Rule Application
Fundamental ApplicationProblem: Find the derivative of \( y = \frac{x^2 + 3x}{2x - 1} \)
Identify u and v:
Let \( u = x^2 + 3x \) and \( v = 2x - 1 \)
Find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
\[ \frac{du}{dx} = 2x + 3 \]
\[ \frac{dv}{dx} = 2 \]
Apply the Quotient Rule:
\[ \frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \]
\[ \frac{dy}{dx} = \frac{(2x - 1)(2x + 3) - (x^2 + 3x)(2)}{(2x - 1)^2} \]
Expand and simplify numerator:
Numerator: \( (2x - 1)(2x + 3) - 2(x^2 + 3x) \)
\[ = (4x^2 + 6x - 2x - 3) - (2x^2 + 6x) \]
\[ = 4x^2 + 4x - 3 - 2x^2 - 6x \]
\[ = 2x^2 - 2x - 3 \]
Final answer:
\[ \frac{dy}{dx} = \frac{2x^2 - 2x - 3}{(2x - 1)^2} \]
Example 2: Quotient with Radical Functions
Advanced ApplicationProblem: Find the derivative of \( y = \frac{x^2}{\sqrt{x}} \)
Rewrite using exponents and identify u and v:
\[ y = \frac{x^2}{x^{1/2}} = x^{3/2} \]
Let \( u = x^2 \) and \( v = x^{1/2} \)
Find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
\[ \frac{du}{dx} = 2x \]
\[ \frac{dv}{dx} = \frac{1}{2}x^{-1/2} \]
Apply the Quotient Rule:
\[ \frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \]
\[ \frac{dy}{dx} = \frac{(x^{1/2})(2x) - (x^2)\left(\frac{1}{2}x^{-1/2}\right)}{(x^{1/2})^2} \]
Simplify:
Numerator: \( 2x^{3/2} - \frac{1}{2}x^{3/2} = \frac{3}{2}x^{3/2} \)
Denominator: \( x \)
\[ \frac{dy}{dx} = \frac{\frac{3}{2}x^{3/2}}{x} = \frac{3}{2}x^{1/2} \]
Verify with power rule:
Since \( y = x^{3/2} \), using power rule:
\[ \frac{dy}{dx} = \frac{3}{2}x^{1/2} \] ✓
Example 3: Real-World Application
Physics ContextProblem: The position of an object is given by \( s(t) = \frac{t^2 + 1}{t + 2} \) meters. Find its velocity function.
Identify u and v:
Let \( u = t^2 + 1 \) and \( v = t + 2 \)
So \( s(t) = \frac{u}{v} \)
Find \( \frac{du}{dt} \) and \( \frac{dv}{dt} \):
\[ \frac{du}{dt} = 2t \]
\[ \frac{dv}{dt} = 1 \]
Apply the Quotient Rule:
\[ v(t) = \frac{ds}{dt} = \frac{v\frac{du}{dt} - u\frac{dv}{dt}}{v^2} \]
\[ v(t) = \frac{(t + 2)(2t) - (t^2 + 1)(1)}{(t + 2)^2} \]
Simplify:
Numerator: \( 2t^2 + 4t - t^2 - 1 = t^2 + 4t - 1 \)
\[ v(t) = \frac{t^2 + 4t - 1}{(t + 2)^2} \]
Interpretation:
The velocity function is \( v(t) = \frac{t^2 + 4t - 1}{(t + 2)^2} \) m/s. This tells us how fast the object is moving at any time t.
When to Use the Quotient Rule
Use the Quotient Rule when you have:
- A quotient of two functions: \( y = \frac{u}{v} \)
- A rational function (fraction with polynomials)
- Any function written as a fraction where both numerator and denominator depend on x
When NOT to use the Quotient Rule:
- When the denominator is a constant (use Constant Multiple Rule)
- When you can simplify the fraction to a single power of x
- For products of functions (use Product Rule)
Common Mistakes to Avoid
| Mistake | Why It's Wrong | Correct Approach |
|---|---|---|
| \( \frac{d}{dx}[\frac{u}{v}] = \frac{\frac{du}{dx}}{\frac{dv}{dx}} \) | Derivative of quotient ≠ Quotient of derivatives | Use Quotient Rule: \( \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \) |
| Forgetting the minus sign | Quotient Rule has subtraction in numerator | Remember: "Low d-high MINUS high d-low" |
| Forgetting to square the denominator | Denominator must be squared in Quotient Rule | Always divide by \( v^2 \) |
| Not simplifying the result | Leaves answer in unnecessarily complex form | Factor and cancel common terms when possible |
Quick Concept Check
Question 1: If \( y = \frac{u}{v} \), what is \( \frac{dy}{dx} \)?
Question 2: Find \( \frac{dy}{dx} \) for \( y = \frac{x^2 + 1}{2x - 1} \)
Problem-Solving Strategy
- Identify if you have a quotient of two functions: \( y = \frac{u}{v} \)
- Label the functions as u (numerator) and v (denominator)
- Find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) separately
- Apply the Quotient Rule: \( \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \)
- Simplify the numerator and look for common factors
- Verify your answer makes sense
AP Exam Tip
On the AP Calculus exam, Quotient Rule problems often appear in both multiple-choice and free-response sections. Remember to:
- Show all steps when using the Quotient Rule
- Clearly identify u and v in free-response questions
- Simplify your final answer when possible
- Watch for the minus sign - this is a common error
- Check if the function can be simplified before differentiating
Practice Problems
Problem 1: Basic Quotient
Find \( \frac{dy}{dx} \) for \( y = \frac{x^3}{x^2 + 1} \)
Let \( u = x^3 \), \( v = x^2 + 1 \)
\[ \frac{du}{dx} = 3x^2 \], \( \frac{dv}{dx} = 2x \)
\[ \frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} = \frac{(x^2 + 1)(3x^2) - (x^3)(2x)}{(x^2 + 1)^2} \]
\[ = \frac{3x^4 + 3x^2 - 2x^4}{(x^2 + 1)^2} = \frac{x^4 + 3x^2}{(x^2 + 1)^2} \]
Problem 2: With Constants
Differentiate \( y = \frac{3x^2 - 2x}{x^3 + 4} \)
Let \( u = 3x^2 - 2x \), \( v = x^3 + 4 \)
\[ \frac{du}{dx} = 6x - 2 \], \( \frac{dv}{dx} = 3x^2 \)
\[ \frac{dy}{dx} = \frac{(x^3 + 4)(6x - 2) - (3x^2 - 2x)(3x^2)}{(x^3 + 4)^2} \]
\[ = \frac{6x^4 - 2x^3 + 24x - 8 - 9x^4 + 6x^3}{(x^3 + 4)^2} \]
\[ = \frac{-3x^4 + 4x^3 + 24x - 8}{(x^3 + 4)^2} \]
Problem 3: Complex Rational Function
Find \( \frac{dy}{dx} \) for \( y = \frac{x^2 - 4}{x^2 + 4} \)
Let \( u = x^2 - 4 \), \( v = x^2 + 4 \)
\[ \frac{du}{dx} = 2x \], \( \frac{dv}{dx} = 2x \)
\[ \frac{dy}{dx} = \frac{(x^2 + 4)(2x) - (x^2 - 4)(2x)}{(x^2 + 4)^2} \]
\[ = \frac{2x^3 + 8x - 2x^3 + 8x}{(x^2 + 4)^2} \]
\[ = \frac{16x}{(x^2 + 4)^2} \]
Note: The numerator simplified nicely because u and v were similar functions.