The constant and power rules, the sum, difference and constant multiple rules, the product rule with the area proof that explains its two terms, the quotient rule and why its order matters, and the discipline of deciding which rule applies outermost when several are needed at once.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 3 — Derivatives
Differentiation Rules
Objectives
Five outcomes. After this section the difference quotient is never needed again for any function you will actually meet.
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 216-229 — the section these objectives are drawn from
Warm-up
Sections 3.1 and 3.2 computed derivatives from the definition. Three of those computations are worth putting side by side.
Discussion prompt
The derivatives of x, x squared and x cubed came out as 1, 2x and 3x squared. What rule is this pattern pointing at, and would you trust it for x to the tenth?
Hint: Look at what happens to the exponent in each case.
Answer:
\[ x^1 \to 1x^0, \quad x^2 \to 2x^1, \quad x^3 \to 3x^2 \]
In each case the exponent comes down in front and is then reduced by one. The pattern predicts that x to the tenth gives 10 x to the ninth.
A pattern spotted in three cases is a conjecture, not a rule. This section proves it — and proves it for every real exponent, not merely the whole numbers where the pattern was noticed. That gap between spotting and proving is what the section is for.
Concept
Rather than running the difference quotient for each new function, we prove one rule per structural feature: a power, a sum, a constant multiple, a product, a quotient. Any algebraic function is then differentiated by decomposing it into those features.
differentiation rules — Theorems, each proved once from the definition, that give the derivative of a function built in a particular way from simpler ones. Together they replace the difference quotient for all algebraic functions.
\[ \frac{d}{dx}\left[x^{n}\right] = n x^{n-1} \]
This is the same economy as the quadratic formula: complete the square once with letters and never do it again with numbers. Here the definition is run once per structure and never again per function.
Figure (svg): The differentiation rules of this section, each with an instance
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 216-218
Section
Section 1
Concept
The derivative of x to the n is n times x to the n minus 1, for every real exponent. A constant has derivative zero, and a constant factor passes straight through the differentiation untouched.
the power rule — For any real number n, the derivative of x to the power n is n times x to the power n minus 1. It holds for negative and fractional exponents as well as whole ones.
\[ \frac{d}{dx}\left[x^{n}\right] = nx^{n-1}, \qquad \frac{d}{dx}[c] = 0, \qquad \frac{d}{dx}[cf] = cf' \]
The constant rule has an obvious picture: a constant function is a horizontal line, whose slope is zero everywhere. The constant multiple rule says stretching a graph vertically stretches its slopes by the same factor.
Figure (svg): The power rule shown as a pattern emerging from three computed cases
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 216-220 — the basic rules
Picture it
Three computed cases and the rule they point to.
Figure (svg): The power rule shown as a pattern emerging from three computed cases
The three whole-number cases were computed from the definition in Section 3.2. The rule stated for every real exponent goes well beyond them, and that extension is what makes roots and reciprocals differentiable in one line.
Worked example
Example 3.17. Rewrite as a power first, then apply the rule once.
\[ \text{Differentiate } x^{7}, \; \frac{1}{x^{3}}, \; \sqrt{x}. \]
Apply the rule to the first directly
Why: Exponent down, then reduced.
\[ 7 x ^{6} \]
Rewrite the second as a power
Why: A reciprocal is a negative exponent.
\[ 1 / x ^{3} = x ^{-3} \]
Apply the rule
Why: Negative 3 down, then reduced to negative 4.
\[ -3 x ^{-4} = -3 / x ^{4} \]
Rewrite the third as a power
Why: A square root is the one-half power.
\[ \sqrt{x} = x ^{\frac{1}{2}} \]
Apply the rule
Why: One half down, exponent becomes negative one half.
\[ (\frac{1}{2}) x ^{-\frac{1}{2}} = \frac{1}{2 \sqrt{x}} \]
Figure (svg): The solution to Worked example powers of every kind shown as a ladder of expressions, one row per legal move
\[ 7x^6, \qquad -\frac{3}{x^4}, \qquad \frac{1}{2\sqrt{x}} \]
Verify: compare the last two with Section 3.1's definition computations
Why: Section 3.1 computed the derivative of the square root from the definition and got one over twice the root — matching exactly, in one line instead of five. The reciprocal's derivative came out negative there too, and the negative sign here is produced automatically by the exponent coming down. Rewriting as a power BEFORE differentiating is the whole technique: the rule cannot be applied to a root or a fraction written in its usual form.
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 218-219
Matching
Rewrite as a power, then apply the rule.
Match the pairs
Why: The middle two are the ones that need rewriting first: the rule applies to a power, and neither a fraction nor a root looks like one until it is rewritten. The last is the horizontal line whose slope is zero everywhere.
Worked example
Checkpoint 3.17. The sum rule lets each term go separately.
\[ \text{Differentiate } f(x) = 4x^{5} - 3x^{2} + 7x - 9. \]
Apply the sum rule to split the terms
Why: Each is differentiated on its own.
Differentiate the first
Why: Constant multiple, then power rule.
\[ 4(5 x ^{4}) = 20 x ^{4} \]
Differentiate the second
Why: The minus travels with the term.
\[ -3(2 x) = -6 x \]
Differentiate the third
Why: The derivative of x is 1.
\[ 7 \]
Differentiate the constant
Why: Zero.
\[ 0 \]
Figure (svg): The solution to Worked example a polynomial termwise shown as a ladder of expressions, one row per legal move
\[ f'(x) = 20x^4 - 6x + 7 \]
Verify: check the degree and the constant's fate
Why: The original has degree 5 and the derivative degree 4, as every differentiation must lower it by exactly one. The constant negative 9 has vanished entirely, which is right: shifting a graph up or down changes no slope, so the derivative cannot see it. Both are structural checks worth running on any polynomial derivative, and either failing points to an arithmetic slip.
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 220-220
Trap
\[ y = 2^{x} \]
Bring the exponent down and reduce it
Why: The student treats it like a power of x.
\[ y' = x \cdot 2^{x-1} \quad \text{(wrong)} \]
The power rule is about x raised to a constant. Here the base is constant and the exponent varies, which is a different family entirely.
\[ \frac{d}{dx}\left[x^{n}\right] = nx^{n-1} \quad \text{but} \quad \frac{d}{dx}\left[2^{x}\right] = 2^{x}\ln 2 \]
Check which of the base and the exponent is the variable
Why: Variable base with constant exponent is the power rule; constant base with variable exponent is an exponential.
The exponential's derivative is proved in Section 3.9 and looks nothing like the power rule. Mixing them is a genuine error rather than a slip: x squared and 2 to the x are different families of function with different calculus, and the notation makes them look deceptively similar.
Fill the middle
The reciprocal cube, expressed as a power before the rule is applied.
Fill in the blanks
\frac-4___ = x^___ \;\Longrightarrow\; \frac______ = -3x^___}
Why: Negative 3 minus 1 is negative 4, so the derivative is negative 3 x to the negative 4, or negative 3 over x to the fourth. Reducing a negative exponent makes it more negative, which is where sign errors creep in.
Sorting
Look at what is variable.
Sort into buckets
Sort each expression.
Pi cubed catches people because it looks like a power. It is a number, roughly 31, and its derivative is zero — a reminder to ask what is actually varying before reaching for any rule.
Prediction
Commit before reasoning.
Predict first
The pattern was spotted for n = 1, 2, 3. For which n does the rule actually hold?
Correct: Every real number.
\[ \frac{d}{dx}\left[x^{\pi}\right] = \pi x^{\pi - 1} \quad \text{is a legitimate application} \]
Why: The rule holds for every real exponent, which is why one line differentiates the square root as the one-half power and the reciprocal as the negative-one power. The whole-number proof is the easiest and comes first, but the general statement is what makes the rule useful — without it, roots and reciprocals would each need their own difference quotient. The full proof for irrational exponents needs logarithmic differentiation, which Section 3.9 supplies.
Section
Section 2
Concept
The derivative of a product is the derivative of the first times the second, plus the first times the derivative of the second. It is emphatically not the product of the derivatives.
the product rule — For differentiable f and g, the derivative of their product is f prime times g plus f times g prime. Each term differentiates one factor and leaves the other alone.
\[ \frac{d}{dx}\left[f(x)g(x)\right] = f'(x)g(x) + f(x)g'(x) \]
The area picture explains the shape. A product is a rectangle; growing both sides adds two long strips and one tiny corner, and the corner is second order and vanishes in the limit. Two strips, two terms.
Figure (svg): The product rule as an area picture: two rectangles growing, with the change in area split into pieces
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 220-224 — the product rule
Picture it
The product as an area, grown slightly.
Figure (svg): The product rule as an area picture: two rectangles growing, with the change in area split into pieces
The corner rectangle has both sides small, so it shrinks far faster than the strips do and contributes nothing in the limit. That is why two terms survive and not three.
Worked example
Example 3.21. Label the factors before differentiating anything.
\[ \text{Differentiate } y = (x^{2}+3)(2x - 1). \]
Name the two factors
Why: Keeping them separate prevents confusion.
\[ f = x ^{2} + 3, g = 2 x - 1 \]
Differentiate each
Why: Power rule termwise.
\[ f' = 2 x, g' = 2 \]
Apply the rule
Why: First derivative times second, plus first times second derivative.
\[ (2 x) (2 x - 1) + (x ^{2} + 3) (2) \]
Expand
Why: Both products.
\[ 4 x ^{2} - 2 x + 2 x ^{2} + 6 \]
Collect
Why: Like terms.
\[ 6 x ^{2} - 2 x + 6 \]
Figure (svg): The solution to Worked example applying the product rule shown as a ladder of expressions, one row per legal move
\[ y' = 6x^2 - 2x + 6 \]
Verify: expand first and differentiate to check
Why: Multiplying out gives 2x cubed minus x squared plus 6x minus 3, whose derivative termwise is 6x squared minus 2x plus 6 — matching exactly. That the two routes agree is a complete check, and for a product of polynomials expanding first is often quicker. The product rule earns its place when the factors cannot be multiplied out, which happens constantly once trigonometric and exponential factors appear.
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 222-223
Fill the middle
The two factors from the worked example, each differentiated.
Fill in the blanks
y' = (2x)(2x-1) + (x^2+3)(2)
Why: The derivative of the second factor is 2. The rule pairs each factor's derivative with the OTHER factor undifferentiated, which is what produces two terms rather than one.
Worked example
Checkpoint 3.21. The trick is adding and subtracting the same thing.
\[ \text{Prove that } (fg)' = f'g + fg'. \]
Write the difference quotient for the product
Why: The definition.
\[ \frac{f(x + h) g(x + h) - f(x) g(x)}{h} \]
Add and subtract a bridging term
Why: The step that makes it work.
Group into two quotients
Why: Each isolating one factor's change.
\[ g(x + h) (f(x + h) - f(x)) / h + f(x) (g(x + h) - g(x)) / h \]
Take limits of each piece
Why: Using continuity of g for the first factor.
\[ g(x) f'(x) + f(x) g'(x) \]
State the rule
Why: Two terms, as the picture predicted.
\[ (f g)' = f' g + f g' \]
Figure (svg): The solution to Worked example proving the rule shown as a ladder of expressions, one row per legal move
\[ (fg)' = f'g + fg' \]
Verify: identify where continuity was needed
Why: In step four the factor g of x plus h had to approach g of x, which requires g to be continuous — and it is, because Section 3.2 proved differentiability implies continuity. So the hypothesis that both functions are differentiable is doing double duty. The add-and-subtract move in step two is the entire idea, and it is the same device that will prove the quotient rule and the chain rule.
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 221-222
Trap
\[ f = x^2, \; g = x^3 \;\Longrightarrow\; f' = 2x, \; g' = 3x^2 \]
Multiply the two derivatives
Why: The student assumes differentiation distributes over multiplication.
\[ (fg)' = (2x)(3x^2) = 6x^3 \quad \text{(wrong)} \]
The product is x to the fifth, whose derivative is 5x to the fourth. The guess is not even the right degree.
\[ (fg)' = f'g + fg' = 2x\cdot x^3 + x^2\cdot 3x^2 = 2x^4 + 3x^4 = 5x^4 \]
Use the rule: differentiate one factor at a time and add
Why: Each term leaves one factor untouched, which is what the area picture shows.
The degree check refutes the guess instantly: multiplying derivatives lowers the degree twice, while differentiating the product should lower it once. Whenever the rule feels uncertain, testing it on two powers takes five seconds and settles it — and the same test works for the quotient rule.
Two truths and a lie
All three are about products.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it is the most tempting wrong guess in the chapter. Testing on x squared times x cubed refutes it in one line: the product is x to the fifth with derivative 5x to the fourth, while the guess gives 6x cubed — wrong value and wrong degree. Differentiation distributes over addition but not over multiplication.
Ranking
The standard argument.
Put in order
Why: Step b is the whole trick and looks unmotivated until step c reveals what it was for. Step d is where differentiability is used twice: once for each quotient, and once more to guarantee the continuity that lets the undifferentiated factor pass to its limit.
Prediction
Commit before reasoning.
Predict first
In the area picture, why does the small corner rectangle contribute nothing to the derivative?
Correct: Both its sides are small, so after dividing by h it still tends to zero.
\[ \frac{\Delta f \cdot \Delta g}{h} \approx \frac{(f'h)(g'h)}{h} = f'g'h \;\longrightarrow\; 0 \]
Why: Each strip has one small side, so dividing its area by h leaves something finite. The corner has TWO small sides, so its area is of order h squared, and dividing by h still leaves something of order h — which vanishes in the limit. Nothing is being approximated or discarded by hand; the limit does it. This second-order-vanishing argument recurs throughout calculus and is the same reason the h squared term disappeared in every difference quotient of Section 3.1.
Section
Section 3
Concept
The derivative of a quotient is the bottom times the derivative of the top, minus the top times the derivative of the bottom, all over the bottom squared. Because the numerator is a difference, swapping the terms changes every answer's sign.
the quotient rule — For differentiable f and g with g not zero, the derivative of f over g is f prime g minus f g prime, all divided by g squared.
\[ \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{\left[g(x)\right]^{2}} \]
The order is the only difficult thing about the rule, and one known case settles it permanently. The reciprocal of x must have a negative derivative, since its graph falls; whichever order produces that is the right one.
Figure (svg): The quotient rule with its order emphasised, and the memory device for which term comes first
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 224-227 — the quotient rule
Picture it
The rule with its order emphasised.
Figure (svg): The quotient rule with its order emphasised, and the memory device for which term comes first
The squared denominator is easy to remember and the numerator's order is not, so the check is worth running whenever the rule is used after a gap: differentiate one over x and confirm the answer is negative.
Worked example
Example 3.23. Label top and bottom before starting.
\[ \text{Differentiate } y = \frac{3x + 1}{4x - 3}. \]
Name the parts
Why: Top and bottom, kept separate.
\[ f = 3 x + 1, g = 4 x - 3 \]
Differentiate each
Why: Both linear.
\[ f' = 3, g' = 4 \]
Apply the rule
Why: Bottom times top prime, minus top times bottom prime.
\[ (3(4 x - 3) - (3 x + 1) (4)) / (4 x - 3) ^{2} \]
Expand the numerator
Why: Carefully, keeping the minus outside the second product.
\[ (12 x - 9 - 12 x - 4) / (4 x - 3) ^{2} \]
Collect
Why: The x terms cancel.
\[ -13 / (4 x - 3) ^{2} \]
Figure (svg): The solution to Worked example applying the quotient rule shown as a ladder of expressions, one row per legal move
\[ y' = \frac{-13}{(4x-3)^2} \]
Verify: check the sign against the graph
Why: The derivative is negative for every x, since the numerator is negative and the denominator is a square. That says the function is decreasing on each of its two branches, which the graph confirms — this is a hyperbola falling on both sides of its asymptote. Had the two terms been swapped the answer would have been positive 13 over the square, and the graph would have refuted it immediately. Note also that the x terms cancelling in step five is typical for a quotient of linear functions.
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 225-226
Fill the middle
The quotient from the worked example, with both parts differentiated.
Fill in the blanks
y' = \frac2___}}}
Why: The denominator is the bottom squared. The numerator puts the derivative of the top first, which is what makes the answer come out negative — as a decreasing function requires.
Worked example
Checkpoint 3.23. Use a derivative you already know.
\[ \text{Differentiate } y = \frac{1}{x} \text{ with the quotient rule and check.} \]
Name the parts
Why: The top is the constant 1.
\[ f = 1, g = x \]
Differentiate each
Why: A constant and the identity.
\[ f' = 0, g' = 1 \]
Apply the rule in the correct order
Why: Bottom times top prime, minus top times bottom prime.
\[ (0 \cdot x - 1 \cdot 1) / x ^{2} \]
Simplify
Why: The first term vanishes.
\[ -1 / x ^{2} \]
Compare with Section 3.1
Why: That computed it from the definition.
Figure (svg): The solution to Worked example checking the order with a known case shown as a ladder of expressions, one row per legal move
\[ \frac{d}{dx}\left[\frac1x\right] = -\frac{1}{x^2} \]
Verify: see what the wrong order would have given
Why: Swapping the numerator's terms gives 1 times 1 minus 0 times x, which is positive 1 over x squared — a positive derivative for a decreasing function, which is impossible. So this single case pins the order down permanently. It is worth doing deliberately whenever the rule is used after a break, because the order is the only thing about the quotient rule that is genuinely easy to forget.
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 226-227
Error analysis
A student differentiates a quotient.
Annotate
On: \( \frac{d}{dx}\left[\frac{3x+1}{4x-3}\right] = \frac{(3x+1)(4) - 3(4x-3)}{(4x-3)^2} = \frac{13}{(4x-3)^2} \)
The graph settles it: this function decreases on both branches, so its derivative must be negative everywhere. Checking the sign against the shape of the graph catches a swapped order every time, and it costs nothing.
Sorting
Read the outermost structure.
Sort into buckets
Sort each expression by the rule that applies FIRST.
Both the fourth and the fifth can be avoided entirely by simplifying first: x squared times the root of x is x to the five-halves, and the last splits into x plus one over x. Looking for a simplification before reaching for a rule is worth ten seconds every time.
Two truths and a lie
All three are about quotients.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and the same test that refuted the product guess refutes it. Take x to the fifth over x squared, which is x cubed with derivative 3x squared; the guess gives 5x to the fourth over 2x, which is 2.5x cubed — a different function. Differentiation distributes over addition and subtraction only.
Prediction
Commit before reasoning.
Predict first
You cannot remember whether f'g or fg' comes first. What is the fastest way to settle it?
Correct: Differentiate one over x both ways and keep the order giving negative one over x squared.
\[ \frac{0\cdot x - 1\cdot 1}{x^2} = -\frac{1}{x^2} \;\checkmark \qquad \frac{1\cdot 1 - 0\cdot x}{x^2} = +\frac{1}{x^2} \;\times \]
Why: The reciprocal's derivative is known independently from Section 3.1 and its sign is forced by the graph being decreasing. Running the rule both ways takes fifteen seconds and one of them gives a positive answer, which is immediately impossible. Picking the simpler answer is no guide at all, since both orders give equally simple expressions. Using the product rule on the reciprocal power is a legitimate alternative but does not settle the quotient rule's order.
Section
Section 4
Concept
When an expression needs several rules, identify what the whole thing is — a sum, a product or a quotient — and apply that rule first. The inner pieces are then handled by whatever rules they need.
outermost structure — The single operation that combines the largest pieces of an expression. Identifying it determines which rule to apply first, and each resulting piece is then differentiated by the same procedure.
\[ \text{sum} \to \text{termwise}; \quad \text{product} \to \text{two terms}; \quad \text{quotient} \to \text{the fraction} \]
Before applying any rule at all, look for a simplification. A quotient by a single power splits into a sum; a product of powers collapses into one power. Both save considerable work.
Figure (svg): An expression needing three rules, with the order of application marked
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 227-229 — combining differentiation rules
Picture it
One expression needing three rules.
Figure (svg): An expression needing three rules, with the order of application marked
Reading the structure from the outside in is what makes a complicated expression tractable. And the note at the bottom is worth taking seriously: this particular one collapses under simplification and never needs the quotient rule at all.
Worked example
Example 3.25. Structure first, algebra second.
\[ \text{Differentiate } y = \frac{(x^2+1)(x-2)}{x}. \]
Look for a simplification first
Why: Expand the numerator and divide by x.
\[ \frac{x ^{3} - 2 x ^{2} + x - 2}{x} \]
Split the division termwise
Why: Each term over x.
\[ x ^{2} - 2 x + 1 - \frac{2}{x} \]
Rewrite the last term as a power
Why: So the power rule applies.
\[ x ^{2} - 2 x + 1 - 2 x ^{-1} \]
Differentiate termwise
Why: Sum rule and power rule.
\[ 2 x - 2 + 0 + 2 x ^{-2} \]
Tidy
Why: Rewrite the negative power.
\[ 2 x - 2 + 2 / x ^{2} \]
Figure (svg): The solution to Worked example a product inside a quotient shown as a ladder of expressions, one row per legal move
\[ y' = 2x - 2 + \frac{2}{x^2} \]
Verify: confirm the quotient rule would agree
Why: Applying the quotient rule directly, with the product rule inside for the numerator, is several lines longer and gives the same answer — as it must. The lesson is that simplifying first turned a three-rule problem into a one-rule problem. Checking one value settles it numerically: at x equal to 1 the simplified derivative gives 2 minus 2 plus 2, which is 2, and a numerical difference quotient there gives about 2.000.
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 227-228
Ranking
Differentiating a complicated expression.
Put in order
Why: Step a is skipped most often and saves the most work — a quotient by a monomial never needs the quotient rule. Step e catches sign errors, which are the commonest failure once the quotient rule is involved.
Worked example
Checkpoint 3.25. The first real payoff of the rules.
\[ \text{Find every point where } f(x) = x^3 - 3x \text{ has a horizontal tangent.} \]
Differentiate
Why: Termwise with the power rule.
\[ f'(x) = 3 x ^{2} - 3 \]
Set the derivative to zero
Why: A horizontal tangent has zero slope.
\[ 3 x ^{2} - 3 = 0 \]
Solve
Why: Divide by 3 and take roots.
\[ x = 1\text{ and } x = -1 \]
Find the corresponding heights
Why: Evaluate the original.
\[ f(-1) = 2, f(1) = -2 \]
State the points
Why: Both coordinates.
\[ (-1, 2)\text{ and } (1, -2) \]
Figure (svg): A curve with a horizontal tangent located by setting the derivative to zero
\[ (-1, 2) \text{ and } (1, -2) \]
Verify: check against the graph from Section 3.2
Why: That section sketched this cubic and its derivative, and the derivative's zeros sat directly beneath the turning points at negative 1 and 1 — matching exactly. Note that the question asked for POINTS, so both coordinates are needed: the derivative supplies the inputs and the original function supplies the heights. Reporting only the x-values answers half the question, which is the standard omission here.
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 228-229
Trap
\[ y = \frac{x^2 + 1}{x} \]
Apply the quotient rule immediately
Why: The student sees a fraction and reaches for the rule.
\[ y' = \frac{(2x)(x) - (x^2+1)(1)}{x^2} = \frac{x^2 - 1}{x^2} \]
The answer is correct, but it took four lines and an expansion that could have been avoided entirely.
\[ y = x + \frac1x = x + x^{-1} \;\Longrightarrow\; y' = 1 - x^{-2} = 1 - \frac{1}{x^2} \]
Divide through first, then use the power rule
Why: A quotient by a single power always splits termwise.
The two answers agree — x squared minus 1 over x squared is 1 minus one over x squared — so nothing was lost, only time. The habit worth forming is a ten-second look for simplification before any rule is applied, since a quotient by a monomial and a product of powers both collapse.
Fill the middle
The cubic from the worked example, differentiated and set to zero.
Fill in the blanks
3x^2 - 3 = 0 \;\Longrightarrow\; x = \pm 1
Why: Dividing by 3 gives x squared equal to 1, so x is plus or minus 1. The heights come from the original function: 2 and negative 2 respectively, giving the two points.
Matching
Simplify where you can.
Match the pairs
Why: Three of the four collapse and only the last genuinely needs its rule, because its denominator is not a monomial. That ratio is typical of a first course, which is why looking for the simplification is worth doing every time.
Prediction
Commit before reasoning.
Predict first
Solving f'(x) = 0 gives x = -1 and x = 1. What have you found?
Correct: The inputs where the tangent is horizontal — the heights still need computing from f.
\[ f'(x) = 0 \;\Longrightarrow\; x = \pm 1; \quad \text{then } f(-1) = 2, \; f(1) = -2 \]
Why: The derivative's zeros are inputs, not points, so the question is only half answered until the original function supplies the heights. And a horizontal tangent need not be a maximum or minimum, as the cubic at the origin showed in Section 3.2 — that requires the derivative to change sign, which is Section 4.5's business. The zeros of f itself are an entirely different set, found by solving f equals zero rather than f prime equals zero.
Section
Section 5
Concept
The product rule extends to three factors by applying it twice, differentiating one factor at a time. The rules can also be applied repeatedly to obtain higher derivatives, and the quotient rule specialises neatly to a reciprocal.
the extended product rule — For three differentiable factors, the derivative is the sum of three terms, each differentiating exactly one factor and leaving the other two alone. The pattern continues for any number of factors.
\[ (fgh)' = f'gh + fg'h + fgh' \]
The pattern is worth noticing rather than memorising: one term per factor, each differentiating that factor alone. It is the same structure as the two-factor case and it follows from applying that case twice.
Figure (svg): The differentiation rules of this section, each with an instance
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 223-229 — extending the rules
Picture it
Six rules, and what each covers.
Figure (svg): The differentiation rules of this section, each with an instance
Every algebraic function is built from powers by sums, products and quotients, so these six rules differentiate all of them. What is missing is composition, which Section 3.6 supplies, and the transcendental families, which Sections 3.5 and 3.9 supply.
Worked example
Apply the two-factor rule twice, or use the pattern.
\[ \text{Differentiate } y = x(x+1)(x+2). \]
Use the three-factor pattern
Why: One term per factor.
\[ 1(x + 1) (x + 2) + x(1) (x + 2) + x(x + 1) (1) \]
Expand each term
Why: Three quadratics.
\[ (x ^{2} + 3 x + 2) + (x ^{2} + 2 x) + (x ^{2} + x) \]
Collect
Why: Add the like terms.
\[ 3 x ^{2} + 6 x + 2 \]
Note the alternative
Why: Expanding first also works.
\[ y = x ^{3} + 3 x ^{2} + 2 x \]
Figure (svg): The solution to Worked example a product of three factors shown as a ladder of expressions, one row per legal move
\[ y' = 3x^2 + 6x + 2 \]
Verify: differentiate the expanded form
Why: Expanding gives x cubed plus 3x squared plus 2x, whose derivative termwise is 3x squared plus 6x plus 2 — matching exactly. For a product of three polynomials, expanding first is usually faster; the extended rule earns its place when the factors cannot be multiplied out. The structural check is that each of the three terms should have exactly one factor differentiated, and here each does.
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 223-224
Fill the middle
The reciprocal, already differentiated once.
Fill in the blanks
y' = -x^2 \;\Longrightarrow\; y'' = ___x^___
Why: The exponent negative 2 comes down and multiplies the existing negative 1, giving positive 2. Two negatives meeting is the whole content of this step, and it is where the sign is usually lost.
Worked example
Differentiate twice, using whatever rule each stage needs.
\[ \text{Find } y'' \text{ for } y = \frac{1}{x}. \]
Rewrite as a power
Why: So the power rule applies.
\[ y = x ^{-1} \]
Differentiate once
Why: Exponent down, reduced.
\[ y' = -x ^{-2} \]
Differentiate again
Why: Negative 2 comes down and multiplies the existing negative.
\[ y'' = 2 x ^{-3} \]
Rewrite
Why: As a fraction.
\[ y'' = 2 / x ^{3} \]
Figure (svg): The solution to Worked example a second derivative with the rules shown as a ladder of expressions, one row per legal move
\[ y'' = \frac{2}{x^{3}} \]
Verify: check the signs on each branch
Why: For positive x the second derivative is positive, so the graph bends upward there; for negative x it is negative and the graph bends downward. That matches the hyperbola's shape exactly — the two branches curve in opposite directions. Note the double negative in step three: negative 1 times negative 2 gives positive 2, and losing that sign is the standard error when differentiating negative powers repeatedly.
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 229-229
Error analysis
A student differentiates a reciprocal twice.
Annotate
On: \( y = x^{-1} \;\Longrightarrow\; y' = -x^{-2} \;\Longrightarrow\; y'' = -2x^{-3} \)
The graph settles it: for positive x the hyperbola bends upward, so its second derivative must be positive there. Whenever a negative power is differentiated twice, two negatives meet, and writing the multiplication out rather than doing it mentally prevents the loss.
Matching
One term per factor, each differentiating that factor alone.
Match the pairs
Why: The pattern is exactly one term per factor, which follows from applying the two-factor rule repeatedly. Noticing the pattern is far better than memorising the three-factor case, because it generalises and the memorised version does not.
Sorting
Ask whether the expression collapses.
Sort into buckets
Sort each.
The last one is borderline: distributing the root gives x to the five-halves plus x to the one-half, which is easier than the product rule. Even where a rule is available, the distributed form is often shorter, which is why the ten-second look is worth it.
Prediction
Commit before reasoning.
Predict first
Differentiating a product of four functions gives how many terms?
Correct: Four — one per factor.
\[ (fghk)' = f'ghk + fg'hk + fgh'k + fghk' \]
Why: The pattern is one term per factor, each differentiating that factor and leaving the other three alone. It follows from applying the two-factor rule three times, and it is why memorising the pattern beats memorising any particular case. The same structure explains why the product rule has two terms and not one: there are two independent ways the product can change, and each gets its own term.
Comparison
Fill the blanks. The last two are the ones that are not what you would guess.
Comparison matrix
| Structure | Rule | Watch out for |
|---|---|---|
| A power of x | n x^(n-1) | rewrite roots and reciprocals as powers first |
| A sum | differentiate termwise | constants vanish |
| A product | f'g + fg' | it is NOT f'g' |
| A quotient | (f'g - fg')/g^2 | the order: swapping flips the sign |
The two guessable rules are the two that hold. Differentiation distributes over addition, and it does not distribute over multiplication or division — which is exactly why those two need proofs.
Pattern
Given any algebraic expression to differentiate.
Steps one and two together resolve most first-course problems without any product or quotient rule at all. The habit of looking before reaching is worth more than fluency with the rules themselves.
Stewart, Calculus: Early Transcendentals 8e, §3.1 Derivatives of Polynomials and Exponential Functions §3.1, pp. 172-182
Check
The power rule. Rewrite first.
Check your understanding
Differentiate 1/x^3.
Answer: A
Why: Rewrite as x^(-3); the exponent comes down and reduces to -4, giving -3x^(-4).
Check
The product rule. Two terms.
Check your understanding
Differentiate (x^2+3)(2x-1).
Answer: A
Why: (2x)(2x-1) + (x^2+3)(2) expands to 4x^2 - 2x + 2x^2 + 6.
Check
The quotient rule. Order matters.
Check your understanding
Differentiate (3x+1)/(4x-3).
Answer: A
Why: The numerator is 3(4x-3) - (3x+1)(4) = -13, over the bottom squared.
Real world
A company's total cost of producing x units is C(x) dollars. Economists define the average cost as the total divided by the number of units, and they care about where that average is smallest.
Discussion prompt
Write the average cost function, differentiate it, and show that it has a horizontal tangent exactly where the average cost equals the marginal cost.
Hint: Average cost is a quotient, so this is the quotient rule with an interpretation attached.
Answer:
\[ A(x) = \frac{C(x)}{x} \]
Differentiating with the quotient rule, taking the top as C and the bottom as x:
\[ A'(x) = \frac{C'(x)\cdot x - C(x)\cdot 1}{x^{2}} = \frac{xC'(x) - C(x)}{x^{2}} \]
Setting this to zero, the denominator cannot vanish for positive production, so the numerator must:
\[ xC'(x) = C(x) \;\Longleftrightarrow\; C'(x) = \frac{C(x)}{x} = A(x) \]
So the average cost has a horizontal tangent exactly where marginal cost equals average cost — which is one of the standard results of microeconomics, and it falls straight out of the quotient rule with no economics at all.
The interpretation is worth having: while the marginal cost of the next unit is below the current average, making one more unit pulls the average down; once it rises above, the average is pushed up. The turning point is where they cross. That reasoning is exactly the first derivative test of Section 4.5, arrived at from the algebra rather than the graph.
Commit first
Answer, then rate your confidence honestly.
Predict first
What is the derivative of x squared times x cubed?
Correct: 5x to the fourth, by either route.
\[ 2x\cdot x^3 + x^2\cdot 3x^2 = 2x^4 + 3x^4 = 5x^4 = \frac{d}{dx}\left[x^5\right] \]
Why: Collapsing first gives x to the fifth, whose derivative is 5x to the fourth. The product rule gives 2x times x cubed plus x squared times 3x squared, which is 2x to the fourth plus 3x to the fourth, also 5x to the fourth. The first option is the product-of-derivatives guess and is refuted by the degree alone. The third is the product undifferentiated, and the fourth adds the derivatives, which is the sum rule misapplied to a product.
Explain it
They insist the derivative of a product should be the product of the derivatives, because that is how the sum rule works.
Discussion prompt
In four sentences or fewer, convince them with a test they can run themselves.
Hint: Pick two functions whose product they can differentiate another way.
Answer:
Take x squared times x cubed. They can multiply first to get x to the fifth, and they already know its derivative is 5x to the fourth. Now ask them to multiply the derivatives: 2x times 3x squared is 6x cubed.
Those are not the same, and they are not even the same degree — differentiating should lower the degree once, and multiplying the derivatives lowered it twice. The sum rule works because differentiation is linear, and multiplication is not a linear operation, which is exactly why the product needs its own rule with two terms.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For rewriting, do it as a separate first line every time rather than in your head. For the product rule, label the two factors and their derivatives before combining anything. For the quotient rule, differentiate one over x and check you get a negative answer. For structure, ask what the WHOLE expression is before looking at any part of it — and look for a simplification first. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write the six rules of this section in a table, each with one worked instance beside it, and mark the two that are not what you would guess. Below, draw the product rule's area picture: a rectangle with sides f and g, grown by small amounts, with the two strips and the corner shaded differently, and write beneath it why the corner contributes nothing. In the middle, differentiate three things in full: the reciprocal cube by rewriting as a power, the product of x squared plus 3 with 2x minus 1 both by the rule and by expanding first, and the quotient of 3x plus 1 by 4x minus 3. Beside the last, write what the answer would have been with the terms swapped and one sentence saying how the graph refutes it. At the bottom, take the cubic x cubed minus 3x, differentiate it, set the derivative to zero, and give both coordinates of each horizontal tangent, sketching the curve with those two tangents drawn in. In a margin, write the one-term-per-factor pattern for a product of n functions.
If your two routes for the product give different answers, expand the rule's version fully before comparing — they must agree, and the discrepancy is almost always an unexpanded bracket rather than a misapplied rule.
Recap
Five things, and together they retire the difference quotient for every algebraic function.
| If you see | Then |
|---|---|
| A root or a reciprocal | Rewrite it as a power first |
| A polynomial | Differentiate termwise; constants vanish |
| A product that expands | Expanding is usually quicker |
| A product that does not expand | Use the rule: two terms |
| A quotient by a monomial | Split termwise; no quotient rule needed |
| A genuine quotient | Bottom times top prime first, over bottom squared |
| A request for horizontal tangents | Set f' = 0, then get the heights from f |
Section 3.4 pauses the rule-building to interpret what has been gained: a derivative as a rate of change in physics, biology and economics, and the vocabulary of velocity, acceleration and marginal quantities that goes with it.
OpenStax Calculus Volume 1, §3.3 Differentiation Rules §3.3, pp. 216-229 — everything on these slides traces back here
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