Calculus
Intermediate

Quotient Rule

Derivative of a quotient of two functions.

Formula

(fg)′=f′g−fg′g2\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}

Variables

f, gFunctions

Example

d/dx (x/(x+1)) = ((x+1) − x)/(x+1)² = 1/(x+1)².

How it works

The quotient rule tells you how to differentiate a ratio of two differentiable functions when both the top and bottom depend on the variable. It comes from combining the product rule with the fact that a quotient can be rewritten as a product with a reciprocal, and the denominator must stay in the result because changes in both functions affect the slope. The minus sign appears because an increase in the denominator tends to decrease the quotient.

When to use it

Use this rule whenever you need the derivative of one function divided by another, such as rates involving ratios, normalized quantities, or expressions like position over time-varying scale factors. It is especially useful in physics, economics, and calculus problems where both numerator and denominator change with x.

Worked example, step by step

Find the derivative of h(x) = (x^2 + 3x)/(2x - 1). Let f(x) = x^2 + 3x and g(x) = 2x - 1. Then f'(x) = 2x + 3 and g'(x) = 2. Apply the quotient rule: h'(x) = [f'g - fg'] / g^2. So h'(x) = [(2x + 3)(2x - 1) - (x^2 + 3x)(2)] / (2x - 1)^2. Expand the numerator: (2x + 3)(2x - 1) = 4x^2 + 4x - 3. Subtract the second product: 4x^2 + 4x - 3 - 2x^2 - 6x = 2x^2 - 2x - 3. Final answer: h'(x) = (2x^2 - 2x - 3)/(2x - 1)^2.

Common mistakes to avoid

  • Differentiating the top and bottom separately and dividing the results
  • Forgetting the negative sign on the fg' term
  • Using g instead of g^2 in the denominator

Did You Know?

Students memorize the quotient rule with "low d-high minus high d-low, over low squared".

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026