You've learned that derivatives work nicely with sums (the derivative of a sum is the sum of derivatives). But what about quotients? Unfortunately, $\left(\frac{f}{g}\right)' \neq \frac{f'}{g'}$: the derivative of a quotient is NOT the quotient of derivatives.
The Quotient Rule is the tool that handles fractions properly. It's slightly more complex than the Product Rule, but follows a memorable pattern: "low d-high minus high d-low, all over the square of what's below."
| Property | Value |
|---|---|
| Concept | Differentiation Formulas |
| Chapter | 2.3 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
If $f$ and $g$ are both differentiable and $g(x) \neq 0$, then:
$$\boxed{\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{[g(x)]^2}}$$
In prime notation: $$\left(\frac{f}{g}\right)' = \frac{gf' - fg'}{g^2}$$
"Low d-high minus high d-low, all over the square of what's below"
Where:
So: $\frac{\text{Low} \cdot \text{d-High} - \text{High} \cdot \text{d-Low}}{(\text{Low})^2}$
g · f' - f · g'
Quotient' = ─────────────────────
g²
↑ ↑
bottom top (denominator)·(derivative of numerator)
stays stays MINUS
same same (numerator)·(derivative of denominator)
─────────────────────────────────────────
(denominator)²| Product Rule | Quotient Rule |
|---|---|
| $(fg)' = fg' + gf'$ | $\left(\frac{f}{g}\right)' = \frac{gf' - fg'}{g^2}$ |
| Addition of two terms | Subtraction (order matters!) |
| No denominator | Divide by $g^2$ |
Warning: The order of subtraction matters! It's $gf' - fg'$, not $fg' - gf'$.
We can derive it from the Product Rule. Write $\frac{f}{g} = f \cdot g^{-1}$.
By the Product Rule: $$\frac{d}{dx}(f \cdot g^{-1}) = f \cdot (g^{-1})' + g^{-1} \cdot f'$$
Using the Chain Rule (preview): $(g^{-1})' = -g^{-2} \cdot g' = -\frac{g'}{g^2}$
$$= f \cdot \left(-\frac{g'}{g^2}\right) + \frac{f'}{g} = -\frac{fg'}{g^2} + \frac{f'}{g} = \frac{-fg' + gf'}{g^2} = \frac{gf' - fg'}{g^2}$$
Use the Quotient Rule when:
Don't use it when:
Example of simplifying first:
For $F(x) = \frac{3x^2 + 2\sqrt{x}}{x}$, don't use the Quotient Rule!
Instead, divide term by term: $$F(x) = 3x + 2x^{-1/2}$$
Then: $F'(x) = 3 - x^{-3/2}$
Much easier than applying the Quotient Rule to the original expression.
Find $\frac{d}{dx}\left(\frac{x}{x+1}\right)$.
Differentiate $y = \frac{x^2 - 1}{x^2 + 1}$.
Find $g'(t)$ if $g(t) = \frac{2t - 5}{3t + 7}$.
Find the equation of the tangent line to $y = \frac{x^2}{1 + x}$ at the point $(1, \frac{1}{2})$.
Use the Quotient Rule to prove that $\frac{d}{dx}(x^{-n}) = -nx^{-n-1}$ for positive integers $n$.
A student claims that since $fg' + gf' = gf' + fg'$ (by commutativity of addition), the order in the Quotient Rule shouldn't matter either. Why is this wrong?
For $f(x) = \frac{x^4 - 2x^2 + 1}{x^2}$, would you use the Quotient Rule or simplify first? Justify your choice, then find $f'(x)$.
Think of division as "keeping a ratio":
When you have $\frac{f}{g}$, the rate of change depends on two competing effects:
The $g^2$ in the denominator reflects that the "base level" of the ratio matters. A small absolute change in a small ratio is relatively more significant.
| Rule | Formula | When to Use |
|---|---|---|
| Constant | $\frac{d}{dx}(c) = 0$ | Any constant |
| Power | $\frac{d}{dx}(x^n) = nx^{n-1}$ | Powers of $x$ |
| Constant Multiple | $(cf)' = cf'$ | Constant times function |
| Sum | $(f + g)' = f' + g'$ | Adding functions |
| Difference | $(f - g)' = f' - g'$ | Subtracting functions |
| Product | $(fg)' = fg' + gf'$ | Multiplying functions |
| Quotient | $\left(\frac{f}{g}\right)' = \frac{gf' - fg'}{g^2}$ | Dividing functions |
| Previous | Up | Next |
|---|---|---|
| Product Rule | Skills Index | Trig Derivatives |
Last updated: 2026-01-22