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Derivatives

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.

Derivatives Updated 4 min read

At a glance

(fg)' = f'g + fg', \qquad \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}
Problems
12, with answers and full solutions
Level
Calculus 1 and AP Calculus AB, Unit 2
You need
The 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
On this page
  1. Problems
  2. Answers
  3. Step-by-step solutions
  4. Common mistakes on these problems
  5. Keep practicing

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 and quotient rule pages, or the product and quotient rules guide.

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

  1. y = \dfrac{x}{x + 1}
  2. y = \dfrac{2x - 1}{x^2 + 3}
  3. y = \dfrac{\sin x}{x}
  4. y = \dfrac{\ln x}{x}
  5. y = \dfrac{e^x}{x^2}

Putting them together

  1. y = x\,e^x\sin x
  2. y = \dfrac{x^2 e^x}{x + 1}
  3. 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 and the quotient rule lesson. When the factors themselves have something inside, like x^2\sin(3x), move on to the chain rule practice problems.

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The quotient rule

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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.

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