Can I expand a product instead of using the product rule?

For polynomials, yes: expanding and differentiating term by term gives the same answer and makes a useful check. It stops being practical once the factors are not polynomials, which is why the rule is worth being fluent in.

Do I need to simplify my derivative?

Enough to make it usable. If the question then asks you to evaluate at a point, find critical points or compare against given options, simplifying first is almost always faster than not.

How do I know which rule to use first?

Ask what the outermost operation is. If the whole expression is one thing divided by another, start with the quotient rule; if it is two things multiplied, start with the product rule; if it is a function applied to an expression, that is the chain rule, which has its own topic.

MATH 100 derivatives practice

The differentiation rules are short enough to write on one line each. What makes them difficult in practice is structural: deciding which rule the expression in front of you actually needs, and then keeping track of the order in which they nest when a function is a quotient with a product in its numerator.

Almost all lost marks in this topic are bookkeeping rather than calculus. A dropped inner derivative, a reversed numerator in the quotient rule, or a substitution made before differentiating rather than after will each turn a correct method into a wrong answer.

These questions work through the power, sum, product and quotient rules individually and then in combination, and use them for what they are for: tangent lines, rates of change and second derivatives. Exponential and trigonometric functions have a topic of their own that follows this one, and the chain rule and logarithms have the one after.

Start practising Browse all 23 questions

In this topic

Taught: Before the midterm
Questions: 23
Total marks: 60
Suggested time: 98 min
Difficulty: Introductory to Capstone
Show the 4 hardest

Key results

The power rule: $\frac{d}{dx}x^{n} = n\,x^{n - 1}$
For any constant $n$ when $x > 0$ , and for any integer $n$ when $x \neq 0$ .
Sums and constant multiples: $\frac{d}{dx}\bigl(a\,f(x) + b\,g(x)\bigr) = a\,f'(x) + b\,g'(x)$
For constants $a$ and $b$ , when $f$ and $g$ are differentiable at $x$ .
The product rule: $\frac{d}{dx}\bigl(f(x)\,g(x)\bigr) = f'(x)\,g(x) + f(x)\,g'(x)$
When $f$ and $g$ are differentiable at $x$ .
The quotient rule: $\frac{d}{dx}\,\frac{f(x)}{g(x)} = \frac{f'(x)\,g(x) - f(x)\,g'(x)}{g(x)^{2}}$
When $f$ and $g$ are differentiable at $x$ and $g(x) \neq 0$ .
The tangent line: $y = f(a) + f'(a)(x - a)$
When $f$ is differentiable at $a$ .
Velocity and acceleration: $v(t) = s'(t), \qquad a(t) = v'(t) = s''(t)$
For position $s(t)$ on a line; the speed $|v|$ is increasing when $v$ and $a$ have the same sign and decreasing when their signs differ.

Read the theoryUBC Math 100 §4.1 Arithmetic of derivativesUBC Math 100 §3.4 Higher order derivativesCLP-1 §2.4 Arithmetic of derivativesCLP-1 §2.6 Using the arithmetic of derivativesCLP-1 §2.14 Higher order derivatives

Worked example

Standard3 marks · 4 min

Let $f(x) = \dfrac{3x^{2} + 5}{2 - x}$ . Find $f'(-1)$ .

Strategy

Apply the quotient rule carefully, keeping the derivative of the denominator (which is −1) explicit, and only then substitute x = −1.

Working

$u = 3x^{2} + 5,\ u' = 6x;\qquad v = 2 - x,\ v' = -1$
Name the parts.
$f'(x) = \frac{6x(2 - x) - (3x^{2} + 5)(-1)}{(2 - x)^{2}} = \frac{-3x^{2} + 12x + 5}{(2 - x)^{2}}$
The −1 turns the subtraction into an addition; then expand and collect.
$f'(-1) = \frac{-3 - 12 + 5}{\bigl(2 - (-1)\bigr)^{2}} = \frac{-10}{9}$
Substitute. The denominator is 3² = 9.

Answer

f'(-1) = -\frac{10}{9}

Check

f(−0.99) ≈ 2.6556 and f(−1.01) ≈ 2.6778, so f falls about 0.022 over a run of 0.02, a slope near −1.11.

Where this goes wrong

Treating the derivative of 2 − x as +1, which gives (−18 − 8)/9 = −26/9. The denominator decreases as x increases, so v′ = −1.

On the paper

Name each rule as you use it, write the derivative out in full before substituting, and simplify only as far as the question asks.

Your turn

Standard3 marks · 4 min

The table gives values of two differentiable functions $f$ and $g$ and of their derivatives. Let $Q(x) = \dfrac{f(x)}{g(x)}$ . Find $Q'(3)$ .

\begin{array}{c|cccc} x & f(x) & f'(x) & g(x) & g'(x) \\ \hline 0 & 1 & -3 & 2 & 7 \\ 3 & 4 & 2 & -2 & 5 \end{array}

Answer this question

What you should be able to do

Apply the power rule to any exponent
Including negative and fractional powers, which is why rewriting a root or a reciprocal as a power before differentiating is usually the first move.
Use the product rule without dropping a term
The derivative of a product is $u'v + uv'$ . It is never the product of the derivatives, and both cross terms are needed every time.
Get the quotient rule the right way round
The numerator is $u'v - uv'$ , starting with the derivative of the top. Reversing it produces exactly the negative of the correct answer, which is easy to miss.
Read a tangent line off a derivative
The slope at a point is the derivative there, and the line through the point with that slope is the tangent. Finding where a tangent passes through a given point, or is shared by two curves, is the same idea run backwards.
Recognise a limit as a derivative
A limit of the form $\lim_{x \to a}\dfrac{f(x) - f(a)}{x - a}$ is $f'(a)$ in disguise. Spotting that turns an ugly limit into one line of the power rule.
Differentiate first, substitute second
A derivative at a point is the derivative function evaluated there. Substituting the point early turns your function into a constant and the derivative into zero.

Where marks are usually lost

Multiplying the derivatives of a product
Writing $(uv)'$ as $u'v'$ is the single most common error in the topic. Expanding the product first and differentiating term by term is a good way to check yourself.
Differentiating a quotient by differentiating top and bottom
The derivative of $\dfrac{u}{v}$ is not $\dfrac{u'}{v'}$ . If the denominator is a single power of $x$ , rewriting the quotient as a product with a negative exponent is often the cleanest route.
Reversing the quotient rule numerator
$uv' - u'v$ gives the negative of the right answer. If your result has the correct shape but the wrong sign, check this first.
Confusing speed with velocity
Velocity is the derivative of position and carries a sign; speed is its size. An object is speeding up when velocity and acceleration have the same sign, not simply when acceleration is positive.
Evaluating before differentiating
$f(1)$ and $f'(1)$ are different questions. Differentiate the whole function, then substitute.

Practice questions

5 of the 23, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.

Introductory1 mark · 2 min
Let $f(x) = 3x^{4} - 5x^{2} + 7$ . Find $f'(x)$ .
Strategy
The derivative of a sum is the sum of the derivatives, so take each term on its own with the power rule.
Working
$\frac{d}{dx}\bigl(3x^{4}\bigr) = 3\cdot 4x^{3} = 12x^{3}, \qquad \frac{d}{dx}\bigl(5x^{2}\bigr) = 5\cdot 2x = 10x$
Bring each exponent down as a factor and lower it by one.
$\frac{d}{dx}(7) = 0$
A constant does not change, so its rate of change is zero.
$f'(x) = 12x^{3} - 10x$
Combine the pieces, keeping the minus sign.
Answer
$f'(x) = 12x^{3} - 10x$
The answer is f prime of x equals 12x cubed minus 10x.
Check
f is even, so f′ must be odd, as 12x³ − 10x is; at x = 1 it gives 2, close to (f(1.01) − f(0.99))/0.02 ≈ 2.001.
Where this goes wrong
Bringing each exponent down but keeping it, which gives 12x⁴ − 10x². The power rule also lowers every exponent by one: x⁴ becomes 4x³.
Try a similar question
Let $f(x) = x^{2}\sqrt{x} - \dfrac{6}{\sqrt[3]{x}}$ for $x > 0$ . Find $f'(x)$ .
Answer this question
Introductory2 marks · 3 min
A student differentiates $f(x) = \dfrac{5x}{x^{2} + 1}$ as shown below. Which line contains the first error?
$\begin{aligned} \text{Line 1:}\quad f'(x) &= \frac{5\,(x^{2} + 1) + 5x\cdot 2x}{(x^{2} + 1)^{2}} \\[6pt] \text{Line 2:}\quad &= \frac{5x^{2} + 5 + 10x^{2}}{(x^{2} + 1)^{2}} \\[6pt] \text{Line 3:}\quad &= \frac{5\,(3x^{2} + 1)}{(x^{2} + 1)^{2}} \end{aligned}$
Strategy
Compare each line against the rule it claims to apply. Lines 2 and 3 are correct algebra on what came before them; the error is in the rule itself.
Working
$u = 5x,\ u' = 5;\qquad v = x^{2} + 1,\ v' = 2x$
Name the parts.
$f'(x) = \frac{u'v - uv'}{v^{2}} = \frac{5(x^{2} + 1) - 5x\cdot 2x}{(x^{2} + 1)^{2}} = \frac{5 - 5x^{2}}{(x^{2} + 1)^{2}}$
The rule subtracts. Line 1 added, and everything after it inherits the mistake.
Answer
$\text{Line 1};\qquad f'(x) = \frac{5(1 - x^{2})}{(x^{2} + 1)^{2}}$
The answer is the error is in line 1; the correct derivative is 5(1 − x²) over (x² + 1)².
Check
f(1) = 2.5 is the largest value of f, so f′(1) must be 0, but line 3 gives 5 there.
Where this goes wrong
Hunting for an arithmetic slip in lines 2 and 3, which are both correct, instead of checking the rule in line 1. When a derivative looks wrong, check the rule before the algebra.
Try a similar question
Let $q(x) = \dfrac{x^{2} + 1}{x - 2}$ . Which expression is $q'(x)$ ?
Answer this question
Standard2 marks · 3 min
Let $h(x) = (2x + 1)(x^{2} - 3)$ . Find $h'(2)$ .
Strategy
This is a product of two polynomials. The product rule handles it directly; expanding first also works and is a useful check.
Working
$u = 2x + 1, \quad v = x^{2} - 3, \quad u' = 2, \quad v' = 2x$
Name the factors and differentiate each one before combining.
$h'(x) = 2\bigl(x^{2} - 3\bigr) + (2x + 1)(2x)$
Apply u′v + uv′. Neither term is optional.
$= 2x^{2} - 6 + 4x^{2} + 2x = 6x^{2} + 2x - 6$
Expand and collect like terms.
$h'(2) = 6(4) + 2(2) - 6 = 22$
Substitute the point into the simplified derivative.
Answer
$h'(2) = 22$
The answer is 22.
Check
h(2) = 5 and h(2.01) ≈ 5.2213, a rise of about 0.22 over 0.01, so a slope near 22.
Where this goes wrong
Writing h′ = u′v′ = 2 · 2x = 4x, which gives 8. The derivative of a product is not the product of the derivatives; both cross terms are needed.
Try a similar question
Let $q(x) = \dfrac{x^{2} + 1}{x - 2}$ . Which expression is $q'(x)$ ?
Answer this question
Challenging4 marks · 7 min
Two distinct lines pass through the point $(1, -6)$ and are tangent to the parabola $y = 2x^{2}$ . Find the slope of the steeper one.
Strategy
Parametrise the unknown point of tangency by its x-coordinate a. The tangent line at a is determined by the derivative; requiring it to pass through the given external point produces an equation whose solutions are the two tangency points.
Working
$\text{tangent at } (a, 2a^{2}):\quad y = 4a(x - a) + 2a^{2} = 4ax - 2a^{2}$
Slope y′(a) = 4a, through the point of tangency.
$-6 = 4a(1) - 2a^{2} \quad\Longrightarrow\quad 2a^{2} - 4a - 6 = 0 \quad\Longrightarrow\quad a^{2} - 2a - 3 = 0$
Substitute the external point (1, −6).
$(a - 3)(a + 1) = 0 \quad\Longrightarrow\quad a = 3 \text{ or } a = -1$
Two tangency points, as promised.
$\text{slopes } 4a = 12 \text{ and } -4;\qquad \text{steeper: } 12$
The line tangent at (3, 18) has slope 12.
Answer
$12 \quad (\text{tangent at } (3, 18): y = 12x - 18)$
The answer is 12.
Check
The line y = 12x − 18 passes through (1, −6), and 2x² − (12x − 18) = 2(x − 3)² has a double root, so it touches the parabola.
Where this goes wrong
Evaluating the derivative at x = 1 and answering 4. The line through (1, −6) with slope 4 is not tangent to the parabola; (1, −6) is not even on it.
Try a similar question
Find an equation of the tangent line to $y = 2x^{3} - x$ at the point where $x = 2$ .
Answer this question
Capstone5 marks · 10 min
There is exactly one line that is tangent to both of the parabolas below, at different points. Find its slope.
$y = x^{2} \qquad\text{and}\qquad y = x^{2} - 4x + 5$
Strategy
A common tangent is the same line written two ways. Write the tangent to each parabola at an unknown point, then equate slopes and equate intercepts.
Working
$\text{first, at } a:\quad y = 2a(x - a) + a^{2} = 2ax - a^{2}$
Derivative 2x; tangent at (a, a²).
$\text{second, at } b:\quad y = (2b - 4)(x - b) + b^{2} - 4b + 5 = (2b - 4)x - b^{2} + 5$
Derivative 2x − 4; tangent at (b, b² − 4b + 5).
$2a = 2b - 4 \quad\Longrightarrow\quad a = b - 2$
Equal slopes.
$-a^{2} = -b^{2} + 5 \quad\Longrightarrow\quad b^{2} - a^{2} = 5 \quad\Longrightarrow\quad (b - a)(b + a) = 5$
Equal intercepts, factored as a difference of squares.
$2(a + b) = 5 \quad\Longrightarrow\quad b = \tfrac{9}{4},\ a = \tfrac{1}{4},\ \text{slope } 2a = \tfrac{1}{2}$
Using b − a = 2. The line is y = x/2 − 1/16, touching at x = 1/4 and x = 9/4.
Answer
$\text{slope } \frac{1}{2} \quad\bigl(y = \frac{1}{2}x - \frac{1}{16}\bigr)$
The answer is one half.
Check
x² − (x/2 − 1/16) = (x − 1/4)² and x² − 4x + 5 − (x/2 − 1/16) = (x − 9/4)², so the line touches each parabola once.
Where this goes wrong
Assuming the line touches both parabolas at the same x, which forces a = b and no solution. The two points of tangency are different, which is why two unknowns are needed.
Try a similar question
Two distinct lines pass through the point $(1, -6)$ and are tangent to the parabola $y = 2x^{2}$ . Find the slope of the steeper one.
Answer this question

Browse all 23 questions

Common questions

Can I expand a product instead of using the product rule?: For polynomials, yes: expanding and differentiating term by term gives the same answer and makes a useful check. It stops being practical once the factors are not polynomials, which is why the rule is worth being fluent in.
Do I need to simplify my derivative?: Enough to make it usable. If the question then asks you to evaluate at a point, find critical points or compare against given options, simplifying first is almost always faster than not.
How do I know which rule to use first?: Ask what the outermost operation is. If the whole expression is one thing divided by another, start with the quotient rule; if it is two things multiplied, start with the product rule; if it is a function applied to an expression, that is the chain rule, which has its own topic.

Needs
The definition of the derivative
Secants and tangents, average and instantaneous velocity, the difference quotient and differentiability.
27 questions
Leads to
Exponential and trigonometric functions
Derivatives of eˣ, aˣ, sine, cosine and tangent, and the limits behind them.
22 questions
Leads to
The chain rule and logarithms
Compositions, nested rules, the natural logarithm and logarithmic differentiation.
30 questions
Implicit differentiation and inverse trigonometric functions
Differentiating relations, tangent lines to curves, and the derivatives of inverse and inverse trigonometric functions.
30 questions

Key results

Worked example

Strategy

Working

Answer

Check

Where this goes wrong

On the paper

Your turn

What you should be able to do

Apply the power rule to any exponent

Use the product rule without dropping a term

Get the quotient rule the right way round

Read a tangent line off a derivative

Recognise a limit as a derivative

Differentiate first, substitute second

Where marks are usually lost

Multiplying the derivatives of a product

Differentiating a quotient by differentiating top and bottom

Reversing the quotient rule numerator

Confusing speed with velocity

Evaluating before differentiating

Practice questions

Strategy

Working

Answer

Check

Where this goes wrong

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Try a similar question

Common questions

Related topics

The definition of the derivative

Exponential and trigonometric functions

The chain rule and logarithms

Implicit differentiation and inverse trigonometric functions