# Product and Quotient Rule Practice Problems with Answers

> 12 product rule and quotient rule practice problems with answers and full step-by-step solutions, including triple products and a table problem.

Source: https://practicecalculus.com/practice-problems/product-and-quotient-rules
Updated: 2026-11-06

$$ (fg)' = f'g + fg', \qquad \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2} $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 1 and AP Calculus AB, Unit 2
- **You need:** The [power rule](/practice-problems/power-rule) and the derivatives of $e^x$, $\ln x$, $\sin x$ and $\cos x$
- **Main idea:** Product: first times derivative of second plus second times derivative of first. Quotient: low d-high minus high d-low, over low squared
- **Practice with new numbers:** [Product rule lesson](/learn/calculus-1/derivative-rules/product-rule)

## Key takeaways

- In the quotient rule the order matters: $f'g$ comes first, then subtract $fg'$. Swapping them flips the sign of the answer.
- Simplify the top of a quotient rule answer by factoring out common pieces such as $e^x$ or a power of $x$.
- For three factors, take the derivative of one factor at a time and add the three terms.

These **product and quotient rule practice problems** start with two-factor products, move to quotients, and finish with a triple product, a product inside a quotient and a problem from a table of values. Try each one first. The answers come right after the list, then the full solutions.

The two rules:

$$(fg)' = f'g + fg' \qquad\qquad \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$$

For a refresher, see the [product rule](/rules/product-rule) and [quotient rule](/rules/quotient-rule) pages, or the [product and quotient rules guide](/blog/product-and-quotient-rules).

## Problems

**Product rule**

1. $y = x^2 e^x$
2. $y = (x^2 + 1)(3x - 2)$
3. $y = \sqrt{x}\,\ln x$
4. $y = e^x\cos x$

**Quotient rule**

5. $y = \dfrac{x}{x + 1}$
6. $y = \dfrac{2x - 1}{x^2 + 3}$
7. $y = \dfrac{\sin x}{x}$
8. $y = \dfrac{\ln x}{x}$
9. $y = \dfrac{e^x}{x^2}$

**Putting them together**

10. $y = x\,e^x\sin x$
11. $y = \dfrac{x^2 e^x}{x + 1}$
12. Given $f(2) = 3$, $f'(2) = -1$, $g(2) = 4$ and $g'(2) = 5$, find $(fg)'(2)$ and $\left(\dfrac{f}{g}\right)'(2)$.

## Answers

| Problem | Answer |
| --- | --- |
| 1 | $(x^2 + 2x)\,e^x$ |
| 2 | $9x^2 - 4x + 3$ |
| 3 | $\dfrac{\ln x + 2}{2\sqrt{x}}$ |
| 4 | $e^x(\cos x - \sin x)$ |
| 5 | $\dfrac{1}{(x + 1)^2}$ |
| 6 | $\dfrac{-2x^2 + 2x + 6}{(x^2 + 3)^2}$ |
| 7 | $\dfrac{x\cos x - \sin x}{x^2}$ |
| 8 | $\dfrac{1 - \ln x}{x^2}$ |
| 9 | $\dfrac{e^x(x - 2)}{x^3}$ |
| 10 | $e^x(\sin x + x\sin x + x\cos x)$ |
| 11 | $\dfrac{x e^x(x^2 + 2x + 2)}{(x + 1)^2}$ |
| 12 | $(fg)'(2) = 11$, $\left(\frac{f}{g}\right)'(2) = -\dfrac{19}{16}$ |

## Step-by-step solutions

**1.** Let $f = x^2$ and $g = e^x$, so $f' = 2x$ and $g' = e^x$:

$$y' = 2x\cdot e^x + x^2\cdot e^x = (x^2 + 2x)\,e^x$$

**2.** Let $f = x^2 + 1$ and $g = 3x - 2$:

$$y' = 2x(3x - 2) + (x^2 + 1)(3) = 6x^2 - 4x + 3x^2 + 3 = 9x^2 - 4x + 3$$

You can check by expanding first: $y = 3x^3 - 2x^2 + 3x - 2$ gives the same derivative.

**3.** Let $f = \sqrt{x}$ and $g = \ln x$, so $f' = \frac{1}{2\sqrt{x}}$ and $g' = \frac{1}{x}$:

$$y' = \frac{\ln x}{2\sqrt{x}} + \frac{\sqrt{x}}{x} = \frac{\ln x}{2\sqrt{x}} + \frac{2}{2\sqrt{x}} = \frac{\ln x + 2}{2\sqrt{x}}$$

Here $\frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}} = \frac{2}{2\sqrt{x}}$.

**4.** The derivative of $\cos x$ is $-\sin x$:

$$y' = e^x\cos x + e^x(-\sin x) = e^x(\cos x - \sin x)$$

**5.** Top $x$, bottom $x + 1$, and both derivatives are $1$:

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

**6.** Top derivative $2$, bottom derivative $2x$:

$$y' = \frac{2(x^2 + 3) - (2x - 1)(2x)}{(x^2 + 3)^2} = \frac{2x^2 + 6 - 4x^2 + 2x}{(x^2 + 3)^2} = \frac{-2x^2 + 2x + 6}{(x^2 + 3)^2}$$

**7.** Top derivative $\cos x$, bottom derivative $1$:

$$y' = \frac{x\cos x - \sin x\cdot 1}{x^2} = \frac{x\cos x - \sin x}{x^2}$$

**8.** Top derivative $\frac{1}{x}$, bottom derivative $1$:

$$y' = \frac{\frac{1}{x}\cdot x - \ln x\cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$$

**9.** Top derivative $e^x$, bottom derivative $2x$:

$$y' = \frac{e^x x^2 - e^x\cdot 2x}{x^4} = \frac{x e^x(x - 2)}{x^4} = \frac{e^x(x - 2)}{x^3}$$

**10.** Three factors, so three terms. Differentiate $x$, then $e^x$, then $\sin x$, one at a time:

$$y' = 1\cdot e^x\sin x + x\cdot e^x\sin x + x e^x\cdot\cos x = e^x(\sin x + x\sin x + x\cos x)$$

**11.** The top is a product. From problem 1, its derivative is $(x^2 + 2x)e^x$. Now use the quotient rule:

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

Factor $e^x$ from the top and expand: $(x^2 + 2x)(x + 1) - x^2 = x^3 + 2x^2 + 2x = x(x^2 + 2x + 2)$. So

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

**12.** Plug the values into each rule:

$$(fg)'(2) = f'(2)g(2) + f(2)g'(2) = (-1)(4) + (3)(5) = 11$$

$$\left(\frac{f}{g}\right)'(2) = \frac{f'(2)g(2) - f(2)g'(2)}{g(2)^2} = \frac{(-1)(4) - (3)(5)}{16} = -\frac{19}{16}$$

## Common mistakes on these problems

- **Multiplying the derivatives.** $(x^2e^x)' \ne 2x\cdot e^x$. The product rule always gives two terms.
- **Subtracting in the wrong order.** In problem 6, writing $(2x - 1)(2x) - 2(x^2 + 3)$ on top gives the right size but the wrong sign.
- **Forgetting to square the bottom.** The quotient rule denominator is $g^2$, not $g$.
- **Expanding when factoring is easier.** In problem 9, factor $xe^x$ out of the top so you can cancel an $x$.

## Keep practicing

These 12 problems use fixed numbers. For unlimited practice with new functions and instant feedback, try the [product rule lesson](/learn/calculus-1/derivative-rules/product-rule) and the [quotient rule lesson](/learn/calculus-1/derivative-rules/quotient-rule). When the factors themselves have something inside, like $x^2\sin(3x)$, move on to the [chain rule practice problems](/practice-problems/chain-rule).

## Frequently asked questions

### When do you use the product rule vs the quotient rule?

Use the product rule when two functions are multiplied, like $x^2e^x$, and the quotient rule when one is divided by another, like $\frac{\sin x}{x}$. If the bottom is a single power of $x$, you can often rewrite and use the power rule instead.

### How do you remember the quotient rule?

Say "low d-high minus high d-low, over the square of what's below". Low is the bottom $g$, high is the top $f$, and d means derivative: $\frac{g f' - f g'}{g^2}$.

### What is the product rule for three functions?

$(fgh)' = f'gh + fg'h + fgh'$. Differentiate one factor at a time, keep the other two, and add the three results. Problem 10 below uses it.