Many students assume the derivative of two functions multiplied together is just the product of their derivatives — and the same for division. It isn’t. Rates of change interact with each other when functions multiply or divide, so calculus gives us two dedicated tools: the Product Rule and the Quotient Rule.
In this tutorial you will master both formulas, learn a memory trick that makes the quotient rule stick, and work through three fully solved examples.
What Is the Product Rule?
When two functions of x are multiplied together, differentiate like this:
d/dx[f(x) · g(x)] = f'(x) · g(x) + f(x) · g'(x)
In plain words: (derivative of the first × second) + (first × derivative of the second). Because the terms are added, their order doesn’t matter — the commutative property keeps you safe.
Before differentiating combined expressions, make sure single-term differentiation is automatic. Brush up in our tutorial on how to use the power rule for derivatives. For analyzing behavior near boundary points, review our guide on evaluating calculus limits step-by-step. You can also verify derivative expansions using the CalcSolver computation tool.
Worked Example 1: Applying the Product Rule
Find the derivative of:
h(x) = (x^2 + 3)(2x – 5)
Step 1: Identify f(x) and g(x)
- f(x) = x² + 3
- g(x) = 2x − 5
Step 2: Differentiate each piece separately
- f'(x) = 2x
- g'(x) = 2
Step 3: Plug into the product rule
h'(x) = f'(x)· g(x) + f(x)· g'(x)
h'(x) = (2x)(2x – 5) + (x^2 + 3)(2)
Step 4: Expand and combine like terms
- First group: 2x(2x − 5) = 4x² − 10x
- Second group: 2(x² + 3) = 2x² + 6
- Combine: (4x² − 10x) + (2x² + 6) = 6x² − 10x + 6
Final derivative: h'(x) = 6x² − 10x + 6.
What Is the Quotient Rule?
When one function is divided by another, use:
d/dx[f(x)/g(x)] = (f'(x)· g(x) – f(x)· g'(x))/([g(x)]^2)
The Easy Mnemonic: “Low d-High”
“Low d-High minus High d-Low, over the square of what’s below.”
Low = the denominator g(x). d-High = the derivative of the numerator f'(x). High = the numerator f(x). d-Low = the derivative of the denominator g'(x).
Because the numerator uses subtraction, order matters: always start with (denominator × derivative of numerator). Flip the order and every sign in the numerator reverses.
Worked Example 2: Applying the Quotient Rule
Find the derivative of:
y = (3x – 1)/(x^2 + 2)
Step 1: Identify High and Low
- High: f(x) = 3x − 1 → d-High: f'(x) = 3
- Low: g(x) = x² + 2 → d-Low: g'(x) = 2x
Step 2: Apply the formula
y’ = ((Low)(d-High) – (High)(d-Low))/((Low)^2)
y’ = ((x^2 + 2)(3) – (3x – 1)(2x))/((x^2 + 2)^2)
Step 3: Expand the numerator
- (x² + 2)(3) = 3x² + 6
- (3x − 1)(2x) = 6x² − 2x
- Subtract: (3x² + 6) − (6x² − 2x) = 3x² + 6 − 6x² + 2x = −3x² + 2x + 6
Step 4: Write the final quotient form
Keep the denominator factored — expanding it only creates clutter:
y’ = (-3x^2 + 2x + 6)/((x^2 + 2)^2)
Worked Example 3: Quotient Rule with Simplification
Differentiate y = (2x^3 – 4)/(x).
- High: f(x) = 2x³ − 4 → f'(x) = 6x²
- Low: g(x) = x → g'(x) = 1
y’ = ((6x^2)(x) – (2x^3 – 4)(1))/(x^2) = (6x^3 – 2x^3 + 4)/(x^2) = (4x^3 + 4)/(x^2) = (4(x^3 + 1))/(x^2)
Watch the double negative in the middle: subtracting (2x³ − 4) means subtracting 2x³ and adding 4. That’s where most sign errors creep in.
3 Common Mistakes to Avoid
- Differentiating top and bottom directly: d/dx [f/g] is not f’/g’. Division always requires the full quotient formula.
- Reversing the subtraction: starting with (High)(d-Low) flips every sign in the numerator. Always begin with (Low)(d-High).
- Expanding the denominator too early: keep forms like (x² + 2)² factored. An expanded denominator hides common factors and makes simplification much harder.
Sanity check: after a quotient rule problem, expand the numerator fully and look for a factor that might cancel with the denominator — Example 3 above collapsed nicely into 4(x³ + 1)/x² because we kept the denominator untouched.