Mathematics guide, Unit 2, step 12 of 33Previous step Next step

Step 12: Power, Product and Quotient Rules

By the EconTutorials team. 10-minute read with 6 practice questions.

SyllabusAP 2.5–2.10 · 9709 P1 1.7, P3 3.4 · IB SL 5.6

In 30 seconds

Power rule: $ \dfrac{d}{dx}(x^n) = nx^{n-1} $ for any real $ n $. Rewrite roots and fractions as powers first. Constants differentiate to 0, and sums and multiples go term by term.
Learn four more: $ \sin x \to \cos x $, $ \cos x \to -\sin x $ (radians only), $ e^x \to e^x $ and $ \ln x \to \dfrac{1}{x} $.
Product rule $ (uv)’ = uv’ + vu’ $ and quotient rule $ \left(\dfrac{u}{v}\right)’ = \dfrac{vu’-uv’}{v^2} $. The quotient rule then gives $ \dfrac{d}{dx}(\tan x) = \sec^2 x $.
Exam point: the derivative of a product is not the product of the derivatives. List $ u $, $ v $, $ u’ $, $ v’ $ before you assemble either rule; most lost marks are sign or order slips.

Video lesson coming soonA short animated lesson for this step will appear here.

The power rule, and sums and multiples

In Step 11 every derivative took a page of algebra. Doing that for every function would be exhausting, so mathematicians did the algebra once, in general, and turned the results into rules. Look at what first principles gives:

$ f(x) $	$ x^2 $	$ x^3 $	$ x^{-1} $	$ x^{1/2} $
$ f'(x) $	$ 2x $	$ 3x^2 $	$ -x^{-2} $	$ \tfrac{1}{2}x^{-1/2} $

The pattern is the same every time: bring the power down in front, then reduce the power by one.

\[ \frac{d}{dx}\left(x^n\right) = nx^{n-1} \qquad \frac{d}{dx}\big(kf(x)\big) = kf'(x) \qquad \frac{d}{dx}\big(f \pm g\big) = f’ \pm g’ \]

Why does the power come down? For a whole number $ n $, the binomial theorem gives $ (x+h)^n = x^n + nx^{n-1}h + (\text{terms in } h^2 \text{ and higher}) $. Subtract $ x^n $, divide by $ h $, and every term except $ nx^{n-1} $ still contains an $ h $, so it vanishes in the limit. The rule holds for negative and fractional $ n $ too, as Step 13 shows. A constant has a flat graph, so its derivative is 0; and since limits respect sums and multiples, so do derivatives.

The rule needs a power of $ x $, so the real skill is rewriting. Turn $ \sqrt{x} $ into $ x^{1/2} $, $ \dfrac{1}{x^3} $ into $ x^{-3} $, and split a fraction whose numerator is a sum into separate terms.

Worked example

Differentiate (a) $ y = 5x^4-\dfrac{3}{x} + 4\sqrt{x}-9 $ (b) $ y = \dfrac{x^3-4}{x^2} $.

Step 1(a) Rewrite as powers: $ y = 5x^4-3x^{-1} + 4x^{1/2}-9 $.

Step 2Term by term: $ \dfrac{dy}{dx} = 20x^3 + 3x^{-2} + 2x^{-1/2} = 20x^3 + \dfrac{3}{x^2} + \dfrac{2}{\sqrt{x}} $. The constant $ -9 $ disappears.

Step 3(b) Divide each term of the numerator by $ x^2 $: $ y = x-4x^{-2} $.

Step 4So $ \dfrac{dy}{dx} = 1 + 8x^{-3} = 1 + \dfrac{8}{x^3} $.

In (a), watch the signs: $ -3x^{-1} $ differentiates to $ (-1)(-3)x^{-2} = +3x^{-2} $.

Common mistakeDifferentiating a denominator as though it were a numerator. $ \dfrac{d}{dx}\left(\dfrac{1}{x^2}\right) $ is not $ \dfrac{1}{2x} $. Rewrite it as $ x^{-2} $ first, giving $ -2x^{-3} = -\dfrac{2}{x^3} $. Splitting a fraction is fine when the numerator is a sum, as in (b) above, but $ \dfrac{1}{x^2 + 1} $ cannot be split; that needs the chain or quotient rule.

An economics example: marginal cost from a formula. If $ TC = 500 + 12Q + 0.03Q^2 $ dollars, then $ MC = \dfrac{d(TC)}{dQ} = 12 + 0.06Q $. At $ Q = 100 $, marginal cost is $18 per unit; at $ Q = 200 $ it is $24. The power rule gives the marginal cost at every output in one line, and the fixed cost of $500, a constant, drops out exactly as it did from first principles.

Sine, cosine, eˣ and ln x

Four more derivatives complete the basic toolkit. With $ x $ in radians:

\[ \frac{d}{dx}(\sin x) = \cos x \qquad \frac{d}{dx}(\cos x) = -\sin x \]

These come from first principles using the compound angle formula $ \sin(x+h) = \sin x\cos h + \cos x\sin h $ and the two limits $ \dfrac{\sin h}{h} \to 1 $ and $ \dfrac{1-\cos h}{h} \to 0 $ from Step 9. You can also see the result in Figure 1. The sine curve is flat at its peaks and troughs, exactly where $ \cos x = 0 $. It is steepest as it crosses the axis, with gradient $ +1 $ at $ x = 0 $ and $ -1 $ at $ x = \pi $, exactly where $ \cos x = \pm 1 $.

y = sin x and its gradient function y = cos x

y = sin xy = cos x (the gradient of sin x)tangent to sin x at the origin: gradient 1

Figure 1. Where sin x is flat (x = π/2, 3π/2) the cos x curve crosses zero; where sin x is steepest (x = 0, π, 2π) the cos x curve is at ±1. The gradient of sin x at each x is cos x.

Common mistakeUsing degrees. The limit $ \dfrac{\sin h}{h} \to 1 $ is true only when $ h $ is in radians, so $ \dfrac{d}{dx}(\sin x) = \cos x $ is a radian fact. In degrees the derivative would be $ \tfrac{\pi}{180}\cos x $. Set your calculator to radians for all calculus.

Worked example

Find the values of $ x $ in $ 0 \le x < 2\pi $ at which the curve $ y = 2\sin x-x $ has gradient zero.

Step 1Differentiate: $ \dfrac{dy}{dx} = 2\cos x-1 $.

Step 2Set it to zero: $ \cos x = \tfrac{1}{2} $.

Step 3In the interval, $ x = \dfrac{\pi}{3} $ or $ x = 2\pi-\dfrac{\pi}{3} = \dfrac{5\pi}{3} $.

These are the turning points of the curve, a topic that returns in Step 18.

Now exponentials. Try $ a^x $ from first principles: $ \dfrac{a^{x+h}-a^x}{h} = a^x \cdot \dfrac{a^h-1}{h} $. The fraction $ \dfrac{a^h-1}{h} $ does not involve $ x $, so the gradient of $ a^x $ is always a fixed multiple of $ a^x $ itself. That multiple is the gradient at $ x = 0 $: about 0.693 for $ a = 2 $ and about 1.099 for $ a = 3 $. Somewhere between 2 and 3 is the base for which the multiple is exactly 1. That base is $ e = 2.71828\ldots $, and it makes $ e^x $ its own derivative (Figure 2).

y = 2ˣ, eˣ and 3ˣ near the point (0, 1)

y = 2ˣ: gradient at (0, 1) ≈ 0.693y = eˣ: gradient at (0, 1) = 1y = 3ˣ: gradient at (0, 1) ≈ 1.099tangent to eˣ: y = x + 1

Figure 2. All three exponentials pass through (0, 1). Only for the base e = 2.718… is the gradient there exactly 1, and then the gradient at every point equals the height: d/dx(eˣ) = eˣ.

The natural logarithm $ \ln x $ is the inverse of $ e^x $, so its graph is the reflection of $ y = e^x $ in the line $ y = x $. Reflection swaps rise and run, so each gradient becomes its reciprocal. The point $ (b, e^b) $ on $ y = e^x $ has gradient $ e^b $; it reflects to $ (e^b, b) $ on $ y = \ln x $, where the gradient is $ \dfrac{1}{e^b} $. Since that point has $ x = e^b $, the gradient of $ \ln x $ is $ \dfrac{1}{x} $.

\[ \frac{d}{dx}\left(e^x\right) = e^x \qquad \frac{d}{dx}(\ln x) = \frac{1}{x} \quad (x > 0) \]

Worked example

(a) Find the gradient of $ y = 4e^x-3\ln x + x $ at $ x = 1 $. (b) Differentiate $ \ln(2x) $ and $ \ln\sqrt{x} $.

Step 1(a) $ \dfrac{dy}{dx} = 4e^x-\dfrac{3}{x} + 1 $.

Step 2At $ x = 1 $: $ 4e-3 + 1 = 4e-2 \approx 8.87 $.

Step 3(b) Use the log laws before differentiating. $ \ln(2x) = \ln 2 + \ln x $, and $ \ln 2 $ is a constant, so the derivative is $ \dfrac{1}{x} $.

Step 4$ \ln\sqrt{x} = \tfrac{1}{2}\ln x $, so its derivative is $ \dfrac{1}{2x} $.

Leave exact answers such as $ 4e-2 $ in terms of $ e $ unless a decimal is asked for.

The product rule

The derivative of a product is not the product of the derivatives. Test it: $ x \cdot x = x^2 $ has derivative $ 2x $, but the derivatives of the two factors multiply to $ 1 \times 1 = 1 $.

\[ y = uv \quad \Longrightarrow \quad \frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} \]

Picture $ uv $ as the area of a rectangle with sides $ u $ and $ v $. When $ x $ increases slightly, $ u $ grows by $ \Delta u $ and $ v $ by $ \Delta v $. The extra area is a strip $ u\,\Delta v $ along one side, a strip $ v\,\Delta u $ along the other, and a tiny corner $ \Delta u\,\Delta v $. Divide by $ \Delta x $ and let it shrink: the corner term vanishes, and the two strips give $ uv’ + vu’ $. Each side grows while the other is held fixed.

Worked example

Differentiate (a) $ y = x^3\cos x $ (b) $ y = (x^2 + 1)e^x $.

Step 1(a) $ u = x^3,\ v = \cos x $, so $ u’ = 3x^2,\ v’ = -\sin x $.

Step 2$ \dfrac{dy}{dx} = x^3(-\sin x) + 3x^2\cos x = x^2(3\cos x-x\sin x) $.

Step 3(b) $ u = x^2 + 1,\ v = e^x $, so $ u’ = 2x,\ v’ = e^x $.

Step 4$ \dfrac{dy}{dx} = (x^2 + 1)e^x + 2xe^x = e^x(x^2 + 2x + 1) = e^x(x + 1)^2 $.

Factorising at the end pays off: in (b) it shows at once that the gradient is never negative and is zero only at $ x = -1 $.

An economics example: marginal revenue. Total revenue is price times quantity, $ TR = P \cdot Q $. On a downward-sloping demand curve the price itself depends on $ Q $, so this is a product, and

\[ MR = \frac{d(TR)}{dQ} = P + Q\,\frac{dP}{dQ}. \]

Selling one more unit brings in its price $ P $, but to sell it the firm must lower the price slightly on all $ Q $ units, which costs $ Q\frac{dP}{dQ} $. Since $ \frac{dP}{dQ} < 0 $, marginal revenue is below price. With demand $ P = 90-3\sqrt{Q} $, $ \frac{dP}{dQ} = -\frac{3}{2\sqrt{Q}} $, so $ MR = 90-3\sqrt{Q}-\tfrac{3}{2}\sqrt{Q} = 90-4.5\sqrt{Q} $. At $ Q = 36 $, price is $72 but marginal revenue is only $63 (Figure 3).

Demand P = 90 − 3√Q and marginal revenue MR = 90 − 4.5√Q $

price (demand), P = 90 − 3√Qmarginal revenue, MR = 90 − 4.5√Q

Figure 3. Marginal revenue lies below price at every positive quantity. At Q = 36 the 37th unit sells for about $72, but revenue rises by only about $63 because the price cut applies to all 36 earlier units too.

The quotient rule, and tan and sec

For one function divided by another:

\[ y = \frac{u}{v} \quad \Longrightarrow \quad \frac{dy}{dx} = \frac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2} \]

It is the product rule in disguise. If $ y = \frac{u}{v} $ then $ u = yv $, so $ u’ = yv’ + vy’ $, and solving for $ y’ $ gives $ \dfrac{u’-\frac{u}{v}v’}{v} = \dfrac{vu’-uv’}{v^2} $. The minus sign makes sense: if the denominator grows, the fraction shrinks. That is also why the order matters. A useful chant is “bottom times derivative of top, minus top times derivative of bottom, over bottom squared”.

Worked example

Differentiate (a) $ y = \dfrac{2x + 1}{x-3} $ (b) $ y = \dfrac{e^x}{x} $.

Step 1(a) $ u = 2x + 1,\ u’ = 2,\ v = x-3,\ v’ = 1 $.

Step 2$ \dfrac{dy}{dx} = \dfrac{(x-3)(2)-(2x + 1)(1)}{(x-3)^2} = \dfrac{-7}{(x-3)^2} $.

Step 3(b) $ u = e^x,\ u’ = e^x,\ v = x,\ v’ = 1 $: $ \dfrac{dy}{dx} = \dfrac{xe^x-e^x}{x^2} = \dfrac{e^x(x-1)}{x^2} $.

In (a) the gradient is negative for every $ x \neq 3 $: the curve falls on both sides of its vertical asymptote.

Exam tipWrite $ u $, $ v $, $ u’ $ and $ v’ $ in a small list before assembling the rule, as in the worked examples. Simplify the numerator, but leave the denominator as $ v^2 $, for example $ (x-3)^2 $; expanding it gains nothing and makes the next step (solving $ \frac{dy}{dx} = 0 $) harder.

The quotient rule unlocks the remaining trigonometric functions. Write $ \tan x = \dfrac{\sin x}{\cos x} $, so $ u = \sin x $ and $ v = \cos x $. The numerator of the quotient rule is $ \cos x \cdot \cos x-\sin x \cdot (-\sin x) = \cos^2 x + \sin^2 x = 1 $, so

\[ \frac{d}{dx}(\tan x) = \frac{1}{\cos^2 x} = \sec^2 x. \]

In the same way, $ \sec x = \dfrac{1}{\cos x} $ gives $ \dfrac{0-1\cdot(-\sin x)}{\cos^2 x} = \sec x\tan x $.

$ f(x) $	$ \tan x $	$ \sec x $	$ \operatorname{cosec} x $	$ \cot x $
$ f'(x) $	$ \sec^2 x $	$ \sec x\tan x $	$ -\operatorname{cosec} x\cot x $	$ -\operatorname{cosec}^2 x $

A memory aid: the “co” functions (cos, cosec, cot) are the ones whose derivatives carry a minus sign. AP writes cosec as csc.

Worked example

(a) Differentiate $ y = x\tan x $. (b) Find the exact gradient of $ y = \sec x $ at $ x = \dfrac{\pi}{3} $.

Step 1(a) Product rule with $ u = x,\ v = \tan x $: $ \dfrac{dy}{dx} = x\sec^2 x + \tan x $.

Step 2(b) $ \dfrac{dy}{dx} = \sec x\tan x $. At $ \frac{\pi}{3} $, $ \cos\frac{\pi}{3} = \frac{1}{2} $, so $ \sec\frac{\pi}{3} = 2 $, and $ \tan\frac{\pi}{3} = \sqrt{3} $.

Step 3The gradient is $ 2\sqrt{3} $.

Exact trigonometric values (Step 7) come up constantly in derivative questions on non-calculator papers.

Exam tipKnow your syllabus. AP Calculus topics 2.5 to 2.10 cover everything on this page, including the derivatives of $ \tan $, $ \cot $, $ \sec $ and $ \csc $. Cambridge 9709 Paper 1 needs only the power, sum and multiple rules; $ e^x $, $ \ln x $, $ \sin x $, $ \cos x $, $ \tan x $, products and quotients belong to Papers 2 and 3, and sec, cosec and cot are found there by the quotient rule. IB AA SL covers $ x^n $, $ \sin x $, $ \cos x $, $ e^x $, $ \ln x $ and the product and quotient rules; $ \tan x $, $ \sec x $, $ \operatorname{cosec} x $ and $ \cot x $ are HL.

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. Differentiate $ y = 6x^5-2x^3 + 7 $. Show worked answer

Power rule, term by term.
$ 6x^5 \to 30x^4 $, $ -2x^3 \to -6x^2 $, and the constant 7 gives 0.
$ \dfrac{dy}{dx} = 30x^4-6x^2 $.

EasyQ2. Differentiate $ y = \dfrac{3}{x^2} + \sqrt{x} $. Show worked answer

Rewrite: $ y = 3x^{-2} + x^{1/2} $.
$ \dfrac{dy}{dx} = -6x^{-3} + \tfrac{1}{2}x^{-1/2} $.
$ = -\dfrac{6}{x^3} + \dfrac{1}{2\sqrt{x}} $.

MediumQ3. Differentiate $ y = x^2\ln x $, for $ x > 0 $. Show worked answer

Product rule: $ u = x^2,\ u’ = 2x,\ v = \ln x,\ v’ = \dfrac{1}{x} $.
$ \dfrac{dy}{dx} = x^2 \cdot \dfrac{1}{x} + 2x\ln x $.
$ = x + 2x\ln x = x(1 + 2\ln x) $.

MediumQ4. Show that the derivative of $ y = \dfrac{\sin x}{1 + \cos x} $ simplifies to $ \dfrac{1}{1 + \cos x} $. Show worked answer

Quotient rule: $ u = \sin x,\ u’ = \cos x,\ v = 1 + \cos x,\ v’ = -\sin x $.
$ \dfrac{dy}{dx} = \dfrac{(1 + \cos x)\cos x-\sin x(-\sin x)}{(1 + \cos x)^2} $.
Numerator: $ \cos x + \cos^2 x + \sin^2 x = \cos x + 1 $, using $ \sin^2 x + \cos^2 x = 1 $.
So $ \dfrac{dy}{dx} = \dfrac{1 + \cos x}{(1 + \cos x)^2} = \dfrac{1}{1 + \cos x} $.

MediumQ5. For $ y = e^x\sin x $, find the gradient at $ x = 0 $, and the values of $ x $ in $ 0 \le x < 2\pi $ where the gradient is zero. Show worked answer

Product rule: $ \dfrac{dy}{dx} = e^x\cos x + e^x\sin x = e^x(\sin x + \cos x) $.
At $ x = 0 $: $ e^0(0 + 1) = 1 $.
$ e^x $ is never zero, so we need $ \sin x + \cos x = 0 $, that is $ \tan x = -1 $.
In the interval: $ x = \dfrac{3\pi}{4} $ or $ x = \dfrac{7\pi}{4} $.

ExamQ6. The curve $ C $ has equation $ y = \dfrac{\ln x}{x^2} $, for $ x > 0 $. (a) Show that $ \dfrac{dy}{dx} = \dfrac{1-2\ln x}{x^3} $. [3] (b) Find the exact coordinates of the point on $ C $ where the gradient is zero. [3] (c) Find the equation of the tangent to $ C $ at the point where $ x = 1 $. [2] Show worked answer

(a) Quotient rule with $ u = \ln x,\ u’ = \dfrac{1}{x},\ v = x^2,\ v’ = 2x $. [M1]
$ \dfrac{dy}{dx} = \dfrac{x^2 \cdot \frac{1}{x}-2x\ln x}{x^4} = \dfrac{x-2x\ln x}{x^4} $. [A1]
Divide top and bottom by $ x $: $ \dfrac{1-2\ln x}{x^3} $, as required. [A1]
(b) $ 1-2\ln x = 0 $, so $ \ln x = \tfrac{1}{2} $ and $ x = e^{1/2} = \sqrt{e} $. [M1 A1]
$ y = \dfrac{\frac{1}{2}}{(\sqrt{e})^2} = \dfrac{1}{2e} $. The point is $ \left(\sqrt{e},\ \dfrac{1}{2e}\right) $. [A1]
(c) At $ x = 1 $: $ y = \dfrac{\ln 1}{1} = 0 $ and $ \dfrac{dy}{dx} = \dfrac{1-0}{1} = 1 $. [M1]
Tangent: $ y-0 = 1(x-1) $, so $ y = x-1 $. [A1]
Total: 8 marks.

Check yourself

1. What is $ \dfrac{d}{dx}\left(\dfrac{1}{x^3}\right) $?

$ \dfrac{1}{x^3} = x^{-3} $, and the power rule gives $ -3x^{-4} = -\dfrac{3}{x^4} $. The power goes down from $ -3 $ to $ -4 $, not up to $ -2 $.

2. With $ x $ in radians, what is $ \dfrac{d}{dx}(\cos x) $?

$ \cos x $ starts at its maximum and falls, so its gradient just after $ x = 0 $ is negative: $ \dfrac{d}{dx}(\cos x) = -\sin x $.

3. What is $ \dfrac{d}{dx}\left(xe^x\right) $?

Product rule: $ x \cdot e^x + 1 \cdot e^x = (x + 1)e^x $. Multiplying the two derivatives, $ 1 \times e^x $, is the classic mistake.

Your score: 0 out of 3

The short version

$ \dfrac{d}{dx}(x^n) = nx^{n-1} $. Rewrite roots and fractions as powers first; constants vanish; sums and multiples go term by term.
In radians, $ \sin x \to \cos x $ and $ \cos x \to -\sin x $. Also $ e^x \to e^x $ and $ \ln x \to \dfrac{1}{x} $.
Product rule $ (uv)’ = uv’ + vu’ $; quotient rule $ \left(\dfrac{u}{v}\right)’ = \dfrac{vu’-uv’}{v^2} $.
$ \tan x \to \sec^2 x $, $ \sec x \to \sec x\tan x $; the “co” functions get a minus sign.
In economics, $ MC = \dfrac{d(TC)}{dQ} $ ignores fixed cost, and the product rule gives $ MR = P + Q\dfrac{dP}{dQ} < P $.
The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.

Step 12: Power, Product and Quotient Rules

The power rule, and sums and multiples

Sine, cosine, eˣ and ln x

The product rule

The quotient rule, and tan and sec

Practise: 6 questions

Check yourself

The short version

Further reading on EconTutorials

Contact Info

Quick Links

Success!