# Product and Quotient Rules: Formulas, Mnemonics and Examples

> Learn the product rule and quotient rule step by step, with easy mnemonics, when to simplify first, and fully worked derivative examples.

Source: https://practicecalculus.com/blog/product-and-quotient-rules
Updated: 2026-08-15

## Key takeaways

- The product rule is $(uv)' = u'v + uv'$: differentiate one factor at a time and add the results.
- The quotient rule is $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$, remembered as "low d-high minus high d-low, over the square of what is below."
- The derivative of a product is not the product of the derivatives, and the same goes for quotients.
- Simplify first when you can: dividing out a monomial or rewriting $\frac{5}{x^2}$ as $5x^{-2}$ often removes the need for the quotient rule entirely.

You know how to differentiate $x^3$ and you know how to differentiate $\sin x$. But what is the derivative of $x^3 \sin x$, or of $\frac{x^2+1}{x-1}$? It is tempting to differentiate each piece and multiply or divide the results, and that gives the wrong answer every time. The **product rule** and the **quotient rule** are the correct tools, and once you have the patterns down they take seconds.

## What are the product and quotient rules?

The **product rule** is the formula for the derivative of two functions multiplied together. The **quotient rule** is the formula for the derivative of one function divided by another. Both are shortcuts that come from the [limit definition of the derivative](/blog/limit-definition-of-the-derivative), so you never have to go back to limits for a product or a fraction.

Write $u$ and $v$ for two differentiable functions of $x$, and $u'$ and $v'$ for their derivatives.

**Product rule:**

$$\frac{d}{dx}\big(uv\big) = u'v + uv'$$

**Quotient rule:**

$$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$$

Notice that neither rule is "multiply the derivatives" or "divide the derivatives." For example, $\frac{d}{dx}(x \cdot x) = \frac{d}{dx}x^2 = 2x$, but multiplying the derivatives would give $1 \cdot 1 = 1$. The rules exist precisely because the naive guess fails.

## Mnemonics that make them stick

For the product rule, say: **"derivative of the first times the second, plus the first times the derivative of the second."** Since addition does not care about order, you can also write it as $uv' + u'v$. Either way, each term has exactly one derivative in it.

For the quotient rule, the classic rhyme is:

> **Tip:** "Low d-high minus high d-low, over the square of what's below." Here "high" is the numerator $u$, "low" is the denominator $v$, and "d" means "the derivative of." So: $v \cdot u'$ minus $u \cdot v'$, all over $v^2$.

The order in the quotient rule matters because of the minus sign. Always start with the derivative of the **top**. If you swap the two terms, your answer will have the wrong sign.

## Why the product rule works

Picture a rectangle with sides $u$ and $v$, so its area is $uv$. Now nudge $x$ a little. The side $u$ grows by a small amount $\Delta u$ and the side $v$ grows by $\Delta v$. The new area adds three pieces: a strip of size $\Delta u \cdot v$, a strip of size $u \cdot \Delta v$, and a tiny corner $\Delta u \cdot \Delta v$.

Divide by $\Delta x$ and let $\Delta x \to 0$. The two strips become $u'v$ and $uv'$. The corner shrinks to zero much faster than the strips, so it disappears. What is left is exactly $u'v + uv'$. The area changes because *both* sides change, and each side contributes its own strip.

## Worked example 1: a polynomial times a trig function

Differentiate $y = x^3 \sin x$.

**Step 1. Name the factors.** $u = x^3$ and $v = \sin x$.

**Step 2. Differentiate each factor.** $u' = 3x^2$ and $v' = \cos x$.

**Step 3. Plug into $u'v + uv'$.**

$$y' = 3x^2 \sin x + x^3 \cos x$$

**Step 4. Factor if it helps.** Pulling out $x^2$ gives $y' = x^2(3\sin x + x\cos x)$. Both forms are correct.

## Worked example 2: checking the rule on a product of polynomials

Differentiate $y = (2x+1)(x^2-3)$.

**Step 1.** $u = 2x + 1$, so $u' = 2$. And $v = x^2 - 3$, so $v' = 2x$.

**Step 2. Apply the product rule.**

$$y' = 2(x^2 - 3) + (2x+1)(2x)$$

**Step 3. Simplify.**

$$y' = 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6$$

**Check by expanding first.** $(2x+1)(x^2-3) = 2x^3 + x^2 - 6x - 3$, whose derivative is $6x^2 + 2x - 6$. Same answer. For two polynomials, expanding and using the [power rule](/blog/power-rule) is often just as fast. The product rule earns its keep when the factors cannot be multiplied out, like $x^3 \sin x$ or $x^2 e^x$.

## Worked example 3: finding where a product has a flat tangent

Find where $y = x^2 e^x$ has a horizontal tangent line.

**Step 1.** $u = x^2$, $u' = 2x$, $v = e^x$, $v' = e^x$.

**Step 2. Apply the product rule.**

$$y' = 2x e^x + x^2 e^x$$

**Step 3. Factor.** Both terms share $x e^x$:

$$y' = x e^x (x + 2)$$

**Step 4. Set it equal to zero.** Since $e^x$ is never zero, $y' = 0$ only when $x = 0$ or $x = -2$. The points are $(0, 0)$ and $(-2, 4e^{-2}) \approx (-2, 0.54)$.

```graph
title: 'Graph of y = x² eˣ with horizontal tangents at x = -2 and x = 0'
caption: 'The derivative $x e^x (x+2)$ is zero at $x = -2$ (a local max) and $x = 0$ (a local min).'
x: [-6, 1.5]
y: [-0.5, 3]
functions:
  - expr: x^2 e^x
    label: y = x² eˣ
points:
  - [-2, 0.5413, '(-2, 4/e²)']
  - [0, 0, '(0, 0)']
```

Factoring the derivative is the habit that pays off here. A factored derivative tells you its zeros at a glance.

## Worked example 4: the quotient rule with polynomials

Differentiate $y = \dfrac{x^2 + 1}{x - 1}$.

**Step 1. Name the top and bottom.** High: $u = x^2 + 1$, so $u' = 2x$. Low: $v = x - 1$, so $v' = 1$.

**Step 2. Low d-high minus high d-low.**

$$y' = \frac{(x-1)(2x) - (x^2+1)(1)}{(x-1)^2}$$

**Step 3. Expand the numerator only.**

$$(x-1)(2x) - (x^2 + 1) = 2x^2 - 2x - x^2 - 1 = x^2 - 2x - 1$$

**Step 4. Write the answer.** Leave the denominator squared; expanding it rarely helps.

$$y' = \frac{x^2 - 2x - 1}{(x-1)^2}$$

At $x = 2$ the function equals $\frac{5}{1} = 5$ and the slope is $\frac{4 - 4 - 1}{1} = -1$, so the tangent line there is $y = -x + 7$.

```graph
title: 'Graph of y = (x² + 1)/(x - 1) with its tangent line at x = 2'
caption: 'At $x = 2$ the quotient rule gives slope $-1$, so the tangent line is $y = -x + 7$. The curve has a vertical asymptote at $x = 1$.'
x: [-4, 6]
y: [-10, 14]
functions:
  - expr: (x^2+1)/(x-1)
    label: y = (x² + 1)/(x - 1)
  - expr: -x + 7
    label: tangent at x = 2
    dashed: true
points:
  - [2, 5, '(2, 5)']
vertical:
  - {x: 1, label: 'x = 1'}
```

## Worked example 5: deriving the derivative of tan x

Differentiate $y = \tan x$ by writing it as $\dfrac{\sin x}{\cos x}$.

**Step 1.** High: $u = \sin x$, $u' = \cos x$. Low: $v = \cos x$, $v' = -\sin x$.

**Step 2. Apply the quotient rule.**

$$y' = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$$

**Step 3. Use $\sin^2 x + \cos^2 x = 1$.**

$$y' = \frac{1}{\cos^2 x} = \sec^2 x$$

This is how the derivatives of $\tan x$, $\cot x$, $\sec x$ and $\csc x$ are all found. See [derivatives of trig functions](/blog/derivatives-of-trig-functions) for the full set.

## When to simplify first instead

The quotient rule is reliable but messy. Before reaching for it, ask whether a little algebra makes it unnecessary.

**A constant on the bottom.** $y = \dfrac{x^3 - 5x}{4}$ is just $\tfrac{1}{4}(x^3 - 5x)$, so $y' = \tfrac{1}{4}(3x^2 - 5)$.

**A constant on top.** $y = \dfrac{5}{x^2}$ is $5x^{-2}$, so $y' = -10x^{-3} = -\dfrac{10}{x^3}$. The quotient rule gives the same thing, but with four extra lines.

**A single power of $x$ on the bottom.** Divide term by term:

$$\frac{x^3 + 2x}{x} = x^2 + 2 \quad (x \neq 0), \qquad \text{so } y' = 2x$$

Similarly, $\dfrac{x^2 + \sqrt{x}}{x} = x + x^{-1/2}$, with derivative $1 - \tfrac{1}{2}x^{-3/2}$.

**A product you can expand.** As in example 2, multiplying out two short polynomials and using the power rule avoids the product rule entirely.

> **Tip:** if the denominator is a single term, divide. If it is a sum like $x - 1$ or $x^2 + 4$, use the quotient rule.

## Products of three functions

The product rule extends naturally. For three factors, differentiate one at a time and add:

$$(uvw)' = u'vw + uv'w + uvw'$$

For example, with $y = x e^x \sin x$:

$$y' = e^x \sin x + x e^x \sin x + x e^x \cos x$$

Each term has exactly one factor differentiated.

## Common mistakes with the product and quotient rules

1. **Multiplying the derivatives.** $\frac{d}{dx}(x^2 \sin x) \neq 2x \cos x$. The correct answer is $2x\sin x + x^2 \cos x$.
2. **Reversing the quotient rule numerator.** Writing $uv' - u'v$ flips the sign of the whole answer. Start with the derivative of the top.
3. **Forgetting to square the denominator.** The bottom is $v^2$, not $v$.
4. **Dropping parentheses.** In $(x-1)(2x) - (x^2+1)(1)$, the minus sign applies to all of $x^2 + 1$. Losing the parentheses turns $-1$ into $+1$.
5. **Using the quotient rule on a constant denominator.** It works, but it is slow and creates chances for errors. Pull the constant out.
6. **Expanding the denominator.** $(x-1)^2$ is easier to read, and easier to use for finding where the derivative is undefined, when left factored.

## Quick reference

| Situation                       | Rule                                   | Example                                                    |
| ------------------------------- | -------------------------------------- | ---------------------------------------------------------- |
| $u \cdot v$                     | $u'v + uv'$                            | $(x^2 \sin x)' = 2x\sin x + x^2\cos x$                     |
| $u \cdot v \cdot w$             | $u'vw + uv'w + uvw'$                   | $(x e^x \sin x)'$ has three terms                          |
| $\frac{u}{v}$                   | $\frac{u'v - uv'}{v^2}$                | $\left(\frac{x}{x+1}\right)' = \frac{1}{(x+1)^2}$          |
| $\frac{u}{c}$, $c$ constant     | $\frac{u'}{c}$                         | $\left(\frac{x^3}{4}\right)' = \frac{3x^2}{4}$             |
| $\frac{c}{x^n}$                 | rewrite as $cx^{-n}$                   | $\left(\frac{5}{x^2}\right)' = -\frac{10}{x^3}$            |
| $\frac{\text{poly}}{x^n}$       | divide term by term first              | $\left(\frac{x^3+2x}{x}\right)' = 2x$                      |

When an inside function appears, such as $\sin(3x)$ inside a product, combine these rules with the [chain rule](/blog/chain-rule).

## Practice until the pattern is automatic

The product and quotient rules are mechanical, which is good news: a few dozen problems is enough to make them second nature. Work through the interactive lesson below. Each answer is checked instantly, and if you swap the terms in a quotient or forget a factor, you will see exactly where it went wrong before moving on.

## Frequently asked questions

### What is the product rule in calculus?

The product rule tells you how to differentiate two functions multiplied together. If $y = u \cdot v$, then $y' = u'v + uv'$: the derivative of the first times the second, plus the first times the derivative of the second.

### What is the quotient rule formula?

If $y = \frac{u}{v}$, then $y' = \frac{u'v - uv'}{v^2}$. The order in the numerator matters because of the minus sign, so always start with the derivative of the top.

### What is the mnemonic for the quotient rule?

The classic one is "low d-high minus high d-low, over the square of what's below." Low is the denominator, high is the numerator, and d means "the derivative of."

### Can I use the product rule instead of the quotient rule?

Yes. Rewrite $\frac{u}{v}$ as $u \cdot v^{-1}$ and use the product rule together with the chain rule. You will get the same answer, although it usually needs a little more simplifying at the end.

### When should I not use the quotient rule?

Skip it when the denominator is a constant or a single power of $x$. For $\frac{x^2+3}{4}$ just pull out $\frac{1}{4}$, and for $\frac{x^3+2x}{x}$ divide first to get $x^2 + 2$. Both are much faster than the quotient rule.