Derivatives: practice problems with step-by-step solutions

Free Derivatives practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: power rule, chain and product rules, tangents and extrema.

Free practice problemEasy
xyx = −2

Let f(x) = −5x3 + 7x2 − 4x + 9. Find the slope of the graph at x = −2, that is f′(−2).

f(x) = −5x3 + 7x2 − 4x + 9x = −2
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What you practise

  1. Easypower rule
  2. Mediumchain and product rules
  3. Hardtangents and extrema

How to solve Derivatives problems

  1. Identify the structure: sum, product, quotient or composition.
  2. Apply the power rule to each power of x.
  3. Use the product, quotient or chain rule where needed.
  4. Simplify, and evaluate at the point if asked.

Common mistakes

  • Forgetting the inner derivative in the chain rule.
  • Getting the order wrong in the quotient rule.
  • Treating a constant multiple as a product.

Key formulas

Power rule
(xⁿ)′ = n·xⁿ⁻¹
Product rule
(uv)′ = u′v + uv′
Quotient rule
(u/v)′ = (u′v − uv′)/v²
Chain rule
f(g(x))′ = f′(g(x))·g′(x)
Common derivatives
(eˣ)′ = eˣ, (ln x)′ = 1/x, (sin x)′ = cos x, (cos x)′ = −sin x

Learn Derivatives from scratch

Before this: Limits

1What a derivative is

A derivative measures how fast something changes. If a car’s position changes by 50 km every hour, its speed is 50 km/h: the rate of change of position.

On a graph, the derivative at a point is the slope of the curve there: how steeply it rises or falls if you zoom in very close.

That slope is the slope of the tangent, the straight line that just touches the curve at that point.

xy(1, 1)m = 2

For y = x² the tangent at x = 1 has slope 2. We write f′(1) = 2, read “f prime of 1”.

  • Positive derivative: the graph goes up.
  • Negative derivative: the graph goes down.
  • Zero derivative: the graph is flat there, for example at a top or a bottom.
Self-check At a point where the graph of f is going down, the derivative is…
  1. negative
  2. positive
  3. undefined
  4. zero

Answer: negative. The derivative is the slope. A graph that goes down from left to right has a negative slope.

2The basic rules

You do not need limits every time. A few rules cover most functions.

(xⁿ)′ = n · xⁿ⁻¹

This is the power rule: bring the power down in front and lower the power by one. So (x³)′ = 3x² and (x)′ = 1.

  • A constant on its own has derivative 0: it does not change.
  • A constant factor stays: (5x³)′ = 5 · 3x² = 15x².
  • In a sum, take the derivative of every term separately.

Learn these as well: (eˣ)′ = eˣ, (ln x)′ = 1/x, (sin x)′ = cos x and (cos x)′ = −sin x.

Self-check What is the derivative of 4x³ − 2x + 7?
  1. 12x² − 2 + 7
  2. 12x² − 2
  3. 12x³ − 2x
  4. 4x² − 2

Answer: 12x² − 2. Power rule on each term: (4x³)′ = 12x², (−2x)′ = −2 and the constant 7 gives 0.

3Product and chain rules

Two more rules handle functions that are built from other functions.

Product rule, for two functions multiplied together:

(u · v)′ = u′ · v + u · v′

Chain rule, for a function inside another function: take the derivative of the outside, keep the inside as it is, then multiply by the derivative of the inside.

[(3x + 1)⁴]′ = 4(3x + 1)³ · 3 = 12(3x + 1)³

The quotient rule follows from these two: (u/v)′ = (u′v − uv′) / v².

Self-check What is the derivative of (2x + 5)³?
  1. (2x + 5)² · 2
  2. 3(2x + 5)²
  3. 6(2x + 5)²
  4. 6x(2x + 5)²

Answer: 6(2x + 5)². Outside: 3(2x + 5)². Inside: (2x + 5)′ = 2. Multiply: 3 · 2 · (2x + 5)² = 6(2x + 5)². Forgetting the inner 2 is the classic slip.

4The method step by step

  1. Look at the structure: a sum of terms, a product, a quotient or a function inside a function.
  2. Rewrite roots and fractions as powers first: √x = x1/2 and 1/x² = x−2.
  3. Apply the rules from the outside in, one piece at a time.
  4. Simplify the result.
  5. If a value is asked, put the number in only at the end.

To find a top or a bottom of a graph (an extremum), solve f′(x) = 0. The second derivative tells you which one it is: f″ < 0 is a maximum, f″ > 0 a minimum.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

Let f(x) = x³ − 6x² + 9x + 1. Find the slope at x = 0, and the local maximum and minimum.

1

Differentiate term by term

Use the power rule on every term. The constant 1 disappears.

f′(x) = 3x² − 12x + 9
2

The slope at x = 0

f′(0) = 3·0 − 12·0 + 9 = 9

So the tangent at x = 0 rises 9 units for every unit to the right.

3

Set the derivative to zero

3x² − 12x + 9 = 0 ⇒ x² − 4x + 3 = 0
(x − 1)(x − 3) = 0 ⇒ x = 1 or x = 3
4

Which is the maximum?

f″(x) = 6x − 12

f″(1) = −6 is negative, so x = 1 is a maximum. f″(3) = 6 is positive, so x = 3 is a minimum.

5

Find the values

f(1) = 1 − 6 + 9 + 1 = 5
f(3) = 27 − 54 + 27 + 1 = 1
xy(1, 5)(3, 1)

The graph confirms it: flat at both points, a top at (1, 5) and a bottom at (3, 1).

Answer

f′(0) = 9. Local maximum at (1, 5), local minimum at (3, 1).

6Common mistakes

  • Forgetting the derivative of the inside in the chain rule.
  • Writing (u · v)′ = u′ · v′. The product rule has two terms.
  • Getting the order wrong in the quotient rule: it is u′v − uv′ on top, with a minus.
  • Putting the number in before differentiating. Differentiate first, then substitute.
  • Treating a constant multiple such as 5x³ as a product. It is simpler: the 5 just stays.
Self-check What is the derivative of √x?
  1. 1/(2√x)
  2. 1/√x
  3. 2√x
  4. ½x

Answer: 1/(2√x). √x = x1/2, so the derivative is ½x−1/2 = 1/(2√x). Rewriting the root as a power first makes it easy.

7How exam questions look

  • Easy: a polynomial. Use the power rule and find the slope at a given x.
  • Medium: the chain or product rule, for example (2x + 5)⁵ or x²·e−x, evaluated at a point.
  • Hard: tangents and extrema. Solve f′(x) = 0 and decide which point is the maximum.

A quick check for slope questions: look at the graph in your head. If it is going up there, your answer should be positive.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
xyx = 1

Let f(x) = −x5 + 3x2 − 7x + 7. Find the slope of the graph at x = 1, that is f′(1).

f(x) = −x5 + 3x2 − 7x + 7x = 1

Find f′(1)

Show hints
  1. Differentiate term by term with the power rule (xⁿ)′ = n·xⁿ⁻¹. A constant disappears.
  2. f′(x) = −5x4 + 6x − 7.
  3. Now substitute x = 1.
Show the answer and the step-by-step solution

Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.

Differentiate term by term

f′(x) = −5x4 + 6x − 7

Substitute

f′(1) = (−5)·14 + 6·1 + (−7) = −6

Answer

f′(1) = −6
Example 2Medium
xyx = 1

Let f(x) = x3·e−x. Find f′(1).

f(x) = x3·e−xx = 1

Find f′(1)

Show hints
  1. Product rule with u = x3 and v = e−x.
  2. u′ = 3x2, v′ = −1·e−x.
  3. f′(x) = e−x·(3x2 + (−1)·x3).
Show the answer and the step-by-step solution

Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.

Product rule

f′(x) = u′v + uv′ = e−x·(3x2 + (−1)·x3)

Substitute

f′(1) = 0.3679·(3 + (−1)) = 0.7358

Answer

f′(1) = 0.7358
Example 3Hard
xy

Let f(x) = x3 − 13.5x2 + 54x + 6. Find the x-coordinate x of the local minimum.

f(x) = x3 − 13.5x2 + 54x + 6

Find x

Show hints
  1. At a local maximum or minimum the derivative is zero.
  2. f′(x) = 3x2 − 27x + 54 = 0.
  3. For a cubic with a positive x³ term, the maximum comes first (smaller x) and the minimum second.
Show the answer and the step-by-step solution

Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.

Derivative

f′(x) = 3x2 − 27x + 54 = 3·(x2 − 9x + 18)

Solve f′(x) = 0

x = 3 or x = 6

Which one is which

f″(x) = 6x − 27: negative at a maximum, positive at a minimum.

f″(3) = −9, f″(6) = 9
x = 6

Answer

x = 6

Questions and answers

What is Derivatives?

A derivative measures how fast something changes. If a car’s position changes by 50 km every hour, its speed is 50 km/h: the rate of change of position.

On a graph, the derivative at a point is the slope of the curve there: how steeply it rises or falls if you zoom in very close.

How do I solve Derivatives problems step by step?
  1. Identify the structure: sum, product, quotient or composition.
  2. Apply the power rule to each power of x.
  3. Use the product, quotient or chain rule where needed.
  4. Simplify, and evaluate at the point if asked.
What are common mistakes in Derivatives?
  • Forgetting the inner derivative in the chain rule.
  • Getting the order wrong in the quotient rule.
  • Treating a constant multiple as a product.
What do the Derivatives exercises cover?
  • Easy: power rule
  • Medium: chain and product rules
  • Hard: tangents and extrema

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

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Printable worksheet

Derivatives worksheet with answers
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