Rules of Differentiation: 10 Applications, Worked Step by Step

A Bakery Lowering Its Price Week by Week: the Rate Its Takings Change When Both Factors Move

methodWrite the Takings as a Product of Two Changing Quantities, Differentiate Each Factor in Turn by the Product Rule, and Set the Result to Zero for the Week the Takings Are Steady

A bakery lowers the price of its sourdough loaves week by week. In week x the price is p = 4 − 0.1x dollars a loaf and the number sold is q = 200 + 20x loaves. The takings for the week are R = pq dollars. (a) At what rate are the takings changing in week 4? (b) In which week are the takings momentarily steady?

The takings are the price times the number sold, and both change from week to week: R = pq.

Let x be the number of the week. The price is p = 4 − 0.1x dollars and the number sold is q = 200 + 20x loaves, so the takings are R = pq dollars.

step 1 of 6

The price and the number sold both change from week to week, so the takings are a product of two changing quantities. The product rule differentiates each factor in turn while the other is held as it stands.

Let x be the number of the week. The price is p = 4 − 0.1x dollars and the number sold is q = 200 + 20x loaves, so the takings are R = pq dollars.
Differentiate each factor on its own. Both are linear, so dpdx = −0.1 and dqdx = 20.
Apply the product rule: dRdx = dpdxq + pdqdx = −0.1(200 + 20x) + 20(4 − 0.1x).
Expand and collect the terms: −20 − 2x + 80 − 2x = 60 − 4x dollars a week.
(a) In week 4 the rate is 60 − 4 × 4 = 44, so the takings are rising at $44 a week.
(b) The takings are momentarily steady when 60 − 4x = 0, which gives x = 15. Check: the takings are $1248 in week 14, $1250 in week 15 and $1248 in week 16.

answer(a) The takings are rising at $44 a week; (b) week 15

techniqueThe Product Rule · Differentiating Linear Functions · The Constant Multiple and Sum Rules

examsO-Level

Common pitfalls

Differentiating the two factors and multiplying the results, giving −0.1 × 20 = −2. The derivative of a product is not the product of the derivatives: each factor is differentiated in turn while the other is left as it stands.
Reading a falling price as falling takings. The price drops by $0.10 a week while the number sold rises by 20 loaves, and until week 15 the second effect is the larger of the two.

A Drug in the Blood After an Injection: the Rate of Change of a Concentration Written as a Quotient

methodName the Numerator and the Denominator, Differentiate Each, Assemble the Quotient Rule, and Set the Numerator of the Result to Zero for the Greatest Concentration

After an injection, the concentration of a drug in a patient's blood is C = 20xx² + 4 milligrams per liter, where x is the number of hours since the injection. (a) How fast is the concentration changing after 1 hour? (b) When is the concentration greatest, and what is it then?

The concentration is a quotient: u = 20x over v = x² + 4, and both parts change with time.

Let x be the number of hours since the injection and write the concentration as C = uv, with u = 20x and v = x² + 4.

step 1 of 6

The concentration is one expression divided by another, and both parts change with time. The quotient rule differentiates the numerator and the denominator inside a single formula.

Let x be the number of hours since the injection and write the concentration as C = uv, with u = 20x and v = x² + 4.
Differentiate each part: dudx = 20 and dvdx = 2x.
Apply the quotient rule: dCdx = vdudx − udvdxv² = 20(x² + 4) − 20x(2x)(x² + 4)² = 80 − 20x²(x² + 4)².
(a) After 1 hour, dCdx = 80 − 2025 = 2.4, so the concentration is rising at 2.4 mg per liter each hour.
The concentration stops rising where the numerator is zero: 80 − 20x² = 0, so x² = 4 and x = 2. A time cannot be −2, so that root is rejected.
(b) After 2 hours the concentration is greatest, at 408 = 5 mg per liter. Check: it is 4 mg per liter after 1 hour and about 4.6 after 3 hours, both below 5.

answer(a) It is rising at 2.4 mg per liter each hour; (b) after 2 hours, at 5 mg per liter

techniqueThe Quotient Rule · The Power Rule · Derivative Notation

examsO-Level

Common pitfalls

Differentiating the top and the bottom separately to get 202x. A quotient is not differentiated term by term; the quotient rule keeps the numerator and the denominator together in one expression.
Writing the numerator of the quotient rule as udvdx − vdudx. The order matters, and reversing it changes the sign of the whole derivative.

A Weather Balloon Being Inflated: the Volume and the Surface Area Growing Through a Radius That Grows

methodLink the Two Rates with the Chain Rule, Differentiate the Volume with Respect to the Radius, and Multiply by the Rate at Which the Radius Itself Grows

A weather balloon is inflated so that it stays spherical and its radius grows at a steady 0.5 cm per second. Its volume is V = 43π r³ and its surface area is A = 4π r², with r in centimeters. Let x be the number of seconds. (a) How fast is the volume growing when the radius is 6 cm? (b) How fast is the surface area growing at that same moment?

The radius grows steadily, so drdx = 0.5 cm per second. The dashed circle is the balloon a moment later.

Let x be the number of seconds. The radius grows steadily, so drdx = 0.5 centimeters per second.

step 1 of 6

The volume depends on the radius and the radius depends on the time, so the two rates multiply. That is the chain rule, and it is what turns a rate per centimeter into a rate per second.

Let x be the number of seconds. The radius grows steadily, so drdx = 0.5 centimeters per second.
The chain rule links the two rates: dVdx = dVdr × drdx.
Differentiate the volume with respect to the radius by the power rule: V = 43π r³ gives dVdr = 4π r², which is 144π when r = 6.
(a) dVdx = 144π × 0.5 = 72π ≈ 226 cubic centimeters per second.
For the surface area, A = 4π r² gives dAdr = 8π r, which is 48π when r = 6.
(b) dAdx = 48π × 0.5 = 24π ≈ 75.4 square centimeters per second. Check: in a tenth of a second the radius goes from 6 to 6.05 cm and the volume grows by about 22.8 cubic centimeters.

answer(a) 72π ≈ 226 cubic centimeters per second; (b) 24π ≈ 75.4 square centimeters per second

techniqueThe Chain Rule · The Power Rule · Derivative Notation

examsO-Level

Common pitfalls

Giving dVdr = 144π as the answer to part (a). That is the rate for each centimeter of radius, not for each second; it must still be multiplied by drdx.
Differentiating 43π r³ as though π were a variable. Here π is a constant multiplier and stays where it is, while only r³ is differentiated.

A Reservoir Through a Dry Spell and Then Rain: a Derivative from First Principles, and a Day That Has None

methodForm the Difference Quotient, Simplify It and Let the Step Tend to Zero; at the Join, Take the Quotient from Each Side and Compare the Two Limits

A reservoir is measured every day. For the first 6 days its depth is D = 120 − 0.5x² centimeters, where x is the number of days since the measurements began. Rain then sets in, and from day 6 onwards the depth is D = 102 + 5(x − 6) centimeters. (a) Differentiate the dry-spell depth from first principles and give the rate on day 4. (b) Show that the depth has no derivative on day 6.

For the first six days the depth is f(x) = 120 − 0.5x², and the dot is day 4.

Let f(x) = 120 − 0.5x² be the depth during the dry spell. Then f(x + h) − f(x) = −0.5[(x + h)² − x²] = −xh − 0.5h².

step 1 of 6

The first piece is differentiated straight from the definition: form the difference quotient, simplify it, and let the step tend to zero. At the day the rain starts the two pieces meet at a corner, and there the quotient tends to a different number from each side.

Let f(x) = 120 − 0.5x² be the depth during the dry spell. Then f(x + h) − f(x) = −0.5[(x + h)² − x²] = −xh − 0.5h².
Divide by the step: f(x + h) − f(x)h = −x − 0.5h.
Let h tend to zero: f'(x) = −x. At x = 4 the quotient is −4.05 for h = 0.1, −4.005 for h = 0.01 and −4.0005 for h = 0.001, closing in on −4.
(a) On day 4 the depth is falling at 4 centimeters a day, so dDdx = −4 there.
On day 6 the dry-spell piece gives a left-hand quotient tending to f'(6) = −6, while the rain line D = 102 + 5(x − 6) is straight and gives a right-hand quotient of exactly 5.
(b) The two one-sided rates, −6 and 5 centimeters a day, are different, so there is no derivative on day 6: the graph has a corner. The depth itself is unbroken, because both pieces give 102 centimeters there.

answer(a) f'(x) = −x, so the depth is falling at −4 centimeters a day on day 4; (b) the one-sided rates are −6 and 5 centimeters a day, so there is no derivative on day 6

techniqueDifferentiating From First Principles · Where a Derivative Fails to Exist · The Derivative As a Limit

examsO-Level

Common pitfalls

Putting h = 0 into the difference quotient before it is simplified. That gives 00, which is no number at all; the h in the denominator must be canceled first.
Saying the depth jumps on day 6 because the derivative fails there. A corner and a jump are different: the depth is 102 centimeters from both sides, so it is unbroken, and only its rate disagrees.

A Crank Pin Turning on a Wheel: the Speed at Which It Rises, and the Instant It Is at Rest

methodDifferentiate the Sine to a Cosine, Read the Rate at the Stated Angle, and Solve Cosine Equals Zero for the Moment the Pin Stops Rising

A crank wheel of radius 10 cm turns at a steady 1 radian per second. The height of the crank pin above the center of the wheel is h = 10sin x centimeters, where x is the number of seconds after the pin passes the level of the center. (a) How fast is the pin rising after π3 seconds? (b) At what time is the pin first momentarily at rest, and how high is it then?

The pin is 10sin x cm above the center, and x is both the seconds and the angle turned in radians.

Let x be the number of seconds. The wheel turns at 1 radian per second, so x is also the angle turned, in radians, and h = 10sin x centimeters.

step 1 of 5

The height is a constant multiple of a sine, so it differentiates to the same multiple of a cosine. Where that cosine is zero the pin has stopped rising and has not yet begun to fall.

Let x be the number of seconds. The wheel turns at 1 radian per second, so x is also the angle turned, in radians, and h = 10sin x centimeters.
Differentiate: the derivative of sin x is cos x, and the constant multiple stays, so dhdx = 10cos x centimeters per second.
(a) At x = π3, cosπ3 = 12, so dhdx = 10 × 12 = 5: the pin is rising at 5 centimeters per second.
The pin is momentarily at rest when 10cos x = 0. The first time this happens is x = π2 ≈ 1.57 seconds.
(b) There h = 10sinπ2 = 10 centimeters, the top of the pin's travel. Check: at x = π3 the pin is only 10sinπ3 ≈ 8.66 centimeters up and still rising.

answer(a) It is rising at 5 centimeters per second; (b) at x = π2 ≈ 1.57 seconds, 10 centimeters above the center

techniqueDifferentiating Trigonometric Functions · The Constant Multiple and Sum Rules

examsO-Level

Common pitfalls

Working in degrees and writing cos 60^° into a rate per second. The derivatives of sin x and cos x hold only when x is in radians, so the angle and the time must both be read in radians here.
Reading the greatest height as the moment of the greatest rate. The pin rises fastest as it passes the center, where cos x = 1, and it is at rest at the very top.

A Lighthouse Beam Sweeping Along a Sea Wall: the Speed of the Lit Spot Far from the Perpendicular

methodDifferentiate the Tangent to a Secant Squared for the Meters per Radian, Then Multiply by the Turning Rate of the Lamp for the Meters per Second

A lighthouse stands 300 m from a long straight sea wall, and its lamp turns at a steady 0.2 radians per second. The beam lights a point on the wall y meters from the foot of the perpendicular from the lighthouse, where y = 300tanθ and θ is the angle between the beam and that perpendicular. (a) Find dydθ when θ = π4. (b) How fast is the lit spot moving along the wall at that moment?

The lit spot is y = 300tanθ meters from the foot of the perpendicular, where θ is the beam's angle.

Let θ be the angle between the beam and the perpendicular to the wall. The lit spot is y = 300tanθ meters from the foot of that perpendicular.

step 1 of 5

The distance along the wall is a tangent of the beam's angle, so it differentiates to a secant squared. Multiplying that by the rate at which the lamp turns changes meters per radian into meters per second.

Let θ be the angle between the beam and the perpendicular to the wall. The lit spot is y = 300tanθ meters from the foot of that perpendicular.
Differentiate: ddθ(tanθ) = sec²θ, so dydθ = 300sec²θ meters per radian.
(a) At θ = π4, cosπ4 = 1√2, so sec²π4 = 2 and dydθ = 300 × 2 = 600 meters per radian.
Let x be the number of seconds. The lamp turns at dθdx = 0.2 radians per second, and the chain rule gives dydx = dydθ × dθdx.
(b) dydx = 600 × 0.2 = 120 meters per second. Check: a hundredth of a radian past π4 the spot has moved about 6 m, and the lamp takes 0.05 seconds to turn that far.

answer(a) 600 meters per radian; (b) the lit spot is moving at 120 meters per second

techniqueDifferentiating tan x and the Reciprocal Ratios · The Quotient Rule · The Chain Rule

examsO-Level

Common pitfalls

Differentiating tanθ as sinθcosθ term by term, giving cosθ−sinθ. The quotient rule gives cos²θ + sin²θcos²θ = sec²θ instead.
Reading the 0.2 as a speed in meters per second. It is the turning rate of the lamp, in radians per second, and it becomes a speed along the wall only after it is multiplied by dydθ.

A Radioactive Tracer That Halves Every Six Hours: the Same Decay Written with Base Two and with Base e

methodDifferentiate a Power of Two by Multiplying by the Natural Logarithm of Two and by the Derivative of the Index, Then Rewrite the Base as e and Check the Two Rates Agree

A hospital's radioactive tracer halves in activity every 6 hours, so its activity x hours after it is made is A = 800 × 2^−x/6 becquerel. (a) At what rate is the activity falling after 6 hours? (b) The same activity can be written A = 800e^−kx. Find k, and check that it gives the same rate.

The activity halves every 6 hours: 800, 400, 200, 100 becquerel at 0, 6, 12 and 18 hours.

Let x be the number of hours. Every 6 hours the index −x6 falls by 1 and the activity halves, so A = 800 × 2^−x/6 becquerel.

step 1 of 6

A power of 2 is differentiated by the rule for a^x: the same power again, multiplied by ln 2 and by the derivative of the index. Writing the decay with base e instead must give the very same rate.

Let x be the number of hours. Every 6 hours the index −x6 falls by 1 and the activity halves, so A = 800 × 2^−x/6 becquerel.
Differentiate a^u as a^u ln a × dudx, with a = 2 and u = −x6, so dudx = −16. This gives dAdx = −ln 26 × 800 × 2^−x/6.
After 6 hours the activity is 800 × 2⁻¹ = 400 becquerel.
(a) dAdx = −400ln 26 ≈ −46.21, so the activity is falling at about 46.21 becquerel an hour.
Every base is e in disguise: 2^−x/6 = e^{−xln 2/6}, so A = 800e^−kx with k = ln 26 ≈ 0.1155 per hour.
(b) Then dAdx = −kA = −0.1155 × 400 ≈ −46.21, the same rate. Check: between 6 and 6.1 hours the activity falls from 400 to about 395.4 becquerel.

answer(a) The activity is falling at about 46.21 becquerel an hour; (b) k = ln 26 ≈ 0.1155 per hour

techniqueDifferentiating aˣ and logₐ x · Differentiating Exponentials and Logarithms · The Number E

examsO-Level

Common pitfalls

Differentiating 2^−x/6 by the power rule, as though the index were fixed and the base were the variable. The variable is in the index here, so the rule for a^x applies and a factor of ln 2 appears.
Taking k to be 16 because the activity halves every 6 hours. The index of e carries a factor of ln 2, so k = ln 26, which is about 0.1155 rather than 0.1667.

A Leaf-Shaped Panel Whose Edge Obeys a Relation That Cannot Be Solved for y

methodDifferentiate Both Sides with Respect to x, Attaching dy/dx Wherever y Is Differentiated, Then Collect the dy/dx Terms and Divide

A metalworker cuts a leaf-shaped panel whose edge follows x³ + y³ = 9xy, with x and y in centimeters, and the cutting head must be set along the tangent to that edge. (a) Find dydx at the point (2, 4). (b) Find it at the point (4, 2), and say what the two answers show about the shape.

The point (2, 4) lies on the edge: 2³ + 4³ = 72 and 9 × 2 × 4 = 72.

Check first that (2, 4) lies on the edge: 2³ + 4³ = 8 + 64 = 72, and 9 × 2 × 4 = 72.

step 1 of 6

The relation cannot be rearranged to give y on its own, so differentiate both sides with respect to x, treating y as a function of x and attaching dydx each time y is differentiated.

Check first that (2, 4) lies on the edge: 2³ + 4³ = 8 + 64 = 72, and 9 × 2 × 4 = 72.
Differentiate both sides with respect to x. The chain rule gives ddx(y³) = 3y²dydx, and the product rule gives ddx(9xy) = 9y + 9xdydx.
So 3x² + 3y²dydx = 9y + 9xdydx. Collect the terms in dydx on one side: (3y² − 9x)dydx = 9y − 3x².
Divide both sides by 3 and then by (y² − 3x): dydx = 3y − x²y² − 3x.
(a) At (2, 4): 12 − 416 − 6 = 810 = 0.8, so the head is set at a gradient of 0.8.
(b) At (4, 2): 6 − 164 − 12 = −10−8 = 1.25. Swapping x and y leaves the relation unchanged, so the edge is its own mirror image in the line y = x, and 0.8 × 1.25 = 1.

answer(a) dydx = 0.8 at (2, 4); (b) dydx = 1.25 at (4, 2), and the two gradients multiply to 1 because the edge is symmetrical about y = x

techniqueImplicit Differentiation · The Chain Rule · The Product Rule

examsH2

Common pitfalls

Differentiating y³ as 3y² and stopping there. Since y is itself a function of x, the chain rule attaches dydx, and without it the equation cannot be solved for the gradient.
Differentiating 9xy as 9dydx. It is a product of 9x and y, so the product rule gives 9y + 9xdydx.

A Rocket Watched from 800 Meters Away: the Angle of Elevation Opening Up, and the Climb Behind One Radian

methodDifferentiate the Inverse Tangent by Its Rule and the Chain Rule for the Radians per Second, Then Turn the Derivative Over for the Meters per Radian

An observer stands 800 m from a launch pad on level ground. A rocket rises vertically, and when it is h meters up the observer's angle of elevation is θ = tan⁻¹(h800) radians. At the moment the rocket is 600 m up it is rising at 200 m per second. (a) How fast is the angle of elevation increasing then? (b) Find dhdθ at that height.

The observer is 800 m from the pad, so the angle of elevation is θ = tan⁻¹(h800).

Let x be the number of seconds and h the height in meters. The angle of elevation is θ = tan⁻¹(h800) radians.

step 1 of 5

The angle is an inverse tangent of the height, so it is differentiated by the rule for tan⁻¹ together with the chain rule. Turning that derivative over gives how far the rocket must climb for each radian of elevation.

Let x be the number of seconds and h the height in meters. The angle of elevation is θ = tan⁻¹(h800) radians.
Differentiate with respect to h. The rule is ddhtan⁻¹u = 11 + u²dudh, and here u = h800, so dudh = 1800.
At h = 600, u = 0.75 and 1 + u² = 1.5625, so dθdh = 1800 × 1.5625 = 11250 = 0.0008 radians per meter.
(a) The rocket rises at dhdx = 200 m per second, so the chain rule gives dθdx = 0.0008 × 200 = 0.16 radians per second.
(b) The derivative of the inverse function is the reciprocal of the derivative: dhdθ = 10.0008 = 1250 meters per radian. Check: lifting the rocket from 600 m to 601 m raises the angle from 0.6435 to 0.6443 radians.

answer(a) The angle is increasing at 0.16 radians per second; (b) dhdθ = 1250 meters per radian

techniqueDifferentiating sin⁻¹ x and tan⁻¹ x · The Derivative of an Inverse Function · The Chain Rule

examsO-Level

Common pitfalls

Reading tan⁻¹ as 1tan. It is the inverse tangent, the angle whose tangent is h800, and the same warning applies to sin⁻¹ and cos⁻¹.
Stopping at dθdh = 0.0008 for part (a). That is radians per meter of climb; it becomes radians per second only after it is multiplied by the rocket's speed.

A Tram Pulling Away from a Stop: the Acceleration Its Passengers Feel and the Jerk Behind It

methodDifferentiate the Distance Once for the Velocity, Twice for the Acceleration and Three Times for the Jerk, and Read Each in the Units of the Situation

A tram pulls away from a stop. For the first 10 seconds its distance from the stop is s = 0.4x² + 0.02x³ meters, where x is the number of seconds since it started. (a) Find the acceleration after 5 seconds. (b) Find the third derivative, which engineers call the jerk, and say what it tells the passengers.

The distance from the stop is s = 0.4x² + 0.02x³ meters after x seconds.

Let x be the number of seconds since the tram started, so the distance from the stop is s = 0.4x² + 0.02x³ meters.

step 1 of 6

Differentiating the distance once gives the velocity, twice gives the acceleration, and three times gives the jerk: the rate at which the acceleration itself is changing.

Let x be the number of seconds since the tram started, so the distance from the stop is s = 0.4x² + 0.02x³ meters.
Differentiate once, term by term with the power rule, for the velocity: dsdx = 0.8x + 0.06x² meters per second.
Differentiate again for the acceleration: d²sdx² = 0.8 + 0.12x meters per second squared.
(a) After 5 seconds the acceleration is 0.8 + 0.12 × 5 = 1.4 meters per second squared, and the tram is traveling at 4 + 1.5 = 5.5 meters per second.
Differentiate a third time for the jerk: d³sdx³ = 0.12 meters per second cubed.
(b) The jerk is a steady 0.12 meters per second cubed, so the acceleration builds up evenly and the passengers feel no jolt. Check: the acceleration is 1.28 after 4 seconds and 1.52 after 6, a rise of 0.24 over those 2 seconds.

answer(a) 1.4 meters per second squared; (b) the jerk is a steady 0.12 meters per second cubed

techniqueHigher Derivatives · The Power Rule · Derivative Notation

examsO-Level

Common pitfalls

Giving the velocity 5.5 meters per second as the answer to part (a). The acceleration is the second derivative, not the first.
Treating a constant third derivative as no jerk at all. A jerk of 0.12 is a real and steady change in the acceleration; it is zero jerk that would mean the acceleration never changes.