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Applications of Derivatives
Peaks, valleys, bends and speeds: what slopes tell you.
You know how to find a derivative. Now put it to work. The sign of f′ tells you where a graph rises and falls, so you can find its highest and lowest points. The sign of f″ tells you which way it bends. The same ideas size a box for the biggest volume, track a sliding ladder, and turn a position formula into velocity and acceleration.
On this page: tangent lines and linear approximation, the first and second derivative tests with a sign-chart explorer, an optimization slider, related rates, motion, the Mean Value Theorem, six worked examples, nine practice problems and an eight-question quiz.
Section ITangent lines and linear approximation
The tangent line at x = a goes through the point (a, f(a)) with slope f′(a):
y = f(a) + f′(a)(x − a)Zoom in on a smooth curve and it looks like its tangent line. So near a, the tangent line is a good, easy stand-in for the curve. Used this way it is called the linearization L(x), and f(x) ≈ L(x) is a linear approximation.
For example, √4.1 ≈ L(4.1) = 2 + 0.1/4 = 2.025. The true value is 2.024846, so the estimate is off by about 0.00015. The farther you go from a, the worse the estimate gets: L(9) = 3.25, but √9 = 3.
Section IIIncreasing, decreasing and the first derivative test
On an interval where f′(x) > 0, f is increasing: the graph rises left to right. Where f′(x) < 0, f is decreasing. A critical number is an x in the domain where f′(x) = 0 or f′(x) does not exist. Local maxima and minima can only happen at critical numbers.
- If f′ changes from + to − at c, then f(c) is a local maximum.
- If f′ changes from − to + at c, then f(c) is a local minimum.
- If f′ does not change sign, there is no local extreme value at c.
A critical number is only a candidate. f(x) = x³ has f′(0) = 0, but it rises on both sides, so x = 0 is neither a max nor a min.
To find the absolute (overall) maximum and minimum on a closed interval a ≤ x ≤ b, compare the values of f at every critical number inside the interval and at both endpoints. The Extreme Value Theorem promises that a continuous function on a closed interval has both.
Section IIIConcavity and the second derivative test
The second derivative f″ is the derivative of f′. Where f″(x) > 0, the slopes are increasing and the graph is concave up, shaped like a cup that holds water. Where f″(x) < 0, it is concave down, like an upside-down cup. A point where the concavity changes is an inflection point.
- If f″(c) < 0, then f(c) is a local maximum (concave down: a peak).
- If f″(c) > 0, then f(c) is a local minimum (concave up: a valley).
- If f″(c) = 0, the test says nothing. Go back to the first derivative test.
Careful: f″(c) = 0 does not by itself make an inflection point. f(x) = x⁴ has f″(0) = 0, but it is concave up on both sides.
Try the interactive version at jugami.com/calculus/applications/
Section IVSign-chart explorer: f, f′ and f″ together
Pick a function and move x with the slider. The top graph shows f with its tangent line. The three strips below are sign charts for f, f′ and f″ on the same scale: + or “rising” where the function is positive, − or “falling” where it is negative. Black bars mark where a sign changes. On the graph, filled gold dots mark points where f′ = 0, and open red dots mark where f″ changes sign, the inflection points.
Section VOptimization: the biggest box
Optimization means finding the best value: the largest area, the least cost, the shortest time. The plan is always the same:
- Draw a picture and name the variables.
- Write the quantity to optimize as a function of one variable, and find its domain.
- Find the critical numbers, and check them and the endpoints.
- Answer the question that was asked, with units.
The classic problem: from a 12-inch square sheet of card, cut a square of side x from each corner and fold up the sides to make an open box. The base is 12 − 2x on each side, so the volume is
V(x) = x(12 − 2x)², for 0 ≤ x ≤ 6With calculus:
V′(x) = (12 − 2x)² − 4x(12 − 2x) = (12 − 2x)(12 − 6x)This is 0 at x = 6 (no box at all) and x = 2. Checking the endpoints, V(0) = V(6) = 0, so the maximum is V(2) = 2 · 8² = 128 cubic inches.
Section VIRelated rates: the sliding ladder
In a related-rates problem, several quantities change over time and are tied together by an equation. Differentiate the equation with respect to time t and the rates become tied together too.
A 10 ft ladder leans against a wall. Its foot slides away from the wall at 2 ft/s. How fast is the top sliding down when the foot is 6 ft from the wall?
The key habit: substitute the numbers for this instant only after you differentiate. If you plug in x = 6 first, it becomes a constant and its rate disappears.
Section VIIMotion: position, velocity, acceleration
If s(t) is the position of an object moving along a line, then its velocity is v(t) = s′(t) and its acceleration is a(t) = v′(t) = s″(t). Velocity is positive when the object moves right (or up) and negative when it moves left (or down); speed is |v(t)|. The object is speeding up when v and a have the same sign, and slowing down when they have opposite signs.
Take s(t) = t³ − 6t² + 9t metres, for 0 ≤ t ≤ 4 seconds. Then v(t) = 3t² − 12t + 9 = 3(t − 1)(t − 3) and a(t) = 6t − 12.
| Time | Velocity | Acceleration | What it does |
|---|---|---|---|
| 0 < t < 1 | + | − | Moves right, slowing down |
| 1 < t < 2 | − | − | Moves left, speeding up |
| 2 < t < 3 | − | + | Moves left, slowing down |
| 3 < t < 4 | + | + | Moves right, speeding up |
It finishes at s(4) = 4, so its displacement is 4 m. But it travelled 4 m out, 4 m back and 4 m out again: a total distance of 12 m.
Section VIIIThe Mean Value Theorem, briefly
In words: somewhere, the tangent slope equals the average slope. If you drive 120 miles in 2 hours, your speedometer must read exactly 60 mph at some instant. For f(x) = x² on 1 ≤ x ≤ 3, the average slope is (9 − 1)/2 = 4, and f′(c) = 2c = 4 at c = 2.
Section IXWorked examples
Example 1: a linear approximation
Estimate √4.1 with the tangent line to y = √x at x = 4.
✓ Numerical check: √4.1 = 2.024846 to six decimals, so the error is 0.000154.
Example 2: the first derivative test
Find the local extreme values of f(x) = x³ − 6x² + 9x + 1.
✓ Numerical check: f(0.99) = 4.9997 and f(1.01) = 4.9997, both just below 5.
Example 3: concavity and the second derivative test
For the same f, find the inflection point and confirm Example 2 with f″.
✓ Numerical check: f(2) = 8 − 24 + 18 + 1 = 3; f″(1.9) = −0.6 and f″(2.1) = 0.6.
Example 4: optimization
Find the cut x that gives the biggest open box from a 12-inch square sheet.
✓ Numerical check: V(1.9) = 127.756 and V(2.1) = 127.764, both just under 128.
Example 5: related rates
The 10 ft ladder from Section VI: how fast is the top moving when the foot is 8 ft out, still sliding at 2 ft/s?
✓ Numerical check: from 0.001 s before to 0.001 s after, the height changes at −2.6667 ft/s.
Example 6: motion
For s(t) = t³ − 6t² + 9t, find the velocity and acceleration at t = 2, and decide whether the particle is speeding up.
✓ Numerical check: the speed is 2.97 at t = 1.9, 3 at t = 2 and 2.97 at t = 2.1, so t = 2 is the fastest leftward moment.
Section XCommon mistakes
- “f′(c) = 0, so it’s a max or a min.” Not always: x³ at 0 is neither. Check the sign of f′ on both sides.
- Forgetting the endpoints. On a closed interval the absolute max or min can be at an endpoint, as in practice problem 6.
- “f″(c) = 0 means inflection point.” Only if f″ actually changes sign there. x⁴ at 0 is a counterexample.
- Plugging in numbers too early in related rates. Differentiate first, then substitute the values for that instant.
- Mixing up distance and displacement. When the object turns around, add the distance of each leg.
Section XIPractice problems
Try each one on paper first, then open the answer to check your steps. Every answer ends with a numerical check you can repeat on a calculator.
Use the tangent line to y = ∛x at x = 8 to estimate ∛8.1. Linear approximation
Show answer and steps
y′ = ⅓x−2/3, so y′(8) = 1 over 1282/3 = 4, and ⅓ · ¼ = 1/12L(x) = 2 + x − 8 over 12Tangent line through (8, 2)∛8.1 ≈ 2 + 0.1 over 12 ≈ 2.008333Answer✓ A calculator gives ∛8.1 = 2.008299 to six decimals, so the estimate is off by only about 0.000035.
Find where f(x) = x³ − 12x is increasing and decreasing, and its local extreme values. First derivative test
Show answer and steps
f′(x) = 3x² − 12 = 3(x − 2)(x + 2)Critical numbers x = −2 and x = 2f′ > 0 for x < −2 and x > 2; f′ < 0 for −2 < x < 2Test x = −3, 0, 3: f′ = 15, −12, 15local max f(−2) = 16, local min f(2) = −16f′ changes + to − at −2 and − to + at 2✓ Check: f(−2.01) = 15.9994 and f(−1.99) = 15.9994, both just below 16.
Find the inflection points of f(x) = x⁴ − 6x². Concavity
Show answer and steps
f′(x) = 4x³ − 12x, f″(x) = 12x² − 12Differentiate twicef″(x) = 0 at x = −1 and x = 112(x − 1)(x + 1) = 0concave up for |x| > 1, down for −1 < x < 1f″(0) = −12, f″(2) = 36inflection points (−1, −5) and (1, −5)f″ changes sign at both✓ Check: f(1) = 1 − 6 = −5, and f″(0.9) = −2.28 while f″(1.1) = 2.52, so the sign really changes.
Classify the critical points of f(x) = 2x³ − 3x² − 12x. Second derivative test
Show answer and steps
f′(x) = 6x² − 6x − 12 = 6(x − 2)(x + 1)Critical numbers x = −1 and x = 2f″(x) = 12x − 6Second derivativef″(−1) = −18 < 0: local max f(−1) = 7Concave down, so a peakf″(2) = 18 > 0: local min f(2) = −20Concave up, so a valley✓ Check: f(−1) = −2 − 3 + 12 = 7 and f(2) = 16 − 12 − 24 = −20.
A farmer has 200 m of fence for a rectangular pen along a straight river; the river side needs no fence. What is the largest area? Optimization
Show answer and steps
A = x(200 − 2x)x is each side touching the river; the far side is 200 − 2xA′(x) = 200 − 4x = 0, so x = 50Only critical number in 0 < x < 100A″ = −4 < 0, so it is a maximumConcave down everywhere50 m by 100 m, area 5000 m²Answer✓ Check: x = 49 and x = 51 both give 4998 m², a little less than 5000.
Find the absolute maximum and minimum of f(x) = x³ − 3x on 0 ≤ x ≤ 3. Closed interval
Show answer and steps
f′(x) = 3x² − 3 = 0 at x = 1x = −1 is outside the intervalf(0) = 0, f(1) = −2, f(3) = 18Check the critical number and both endpointsmax 18 at x = 3, min −2 at x = 1Answer✓ On a closed interval the biggest and smallest values are always among these candidates (the Extreme Value Theorem).
A ripple’s radius grows at 3 cm/s. How fast is the circle’s area growing when the radius is 10 cm? Related rates
Show answer and steps
A = πr²Relate the quantitiesdA over dt = 2πr dr over dtDifferentiate both sides with respect to t= 2π(10)(3) = 60π ≈ 188.5 cm²/sPlug in r = 10 and dr/dt = 3 only now✓ Check: between r = 9.997 and r = 10.003 (0.001 s on each side) the area grows at 188.5 cm²/s.
A particle has position s(t) = t² − 8t + 12 metres at time t seconds. When is it at rest, where is it then, and which way is it moving before that? Motion
Show answer and steps
v(t) = 2t − 8, a(t) = 2Velocity and accelerationv = 0 at t = 4, where s(4) = −4At rest at 4 m left of the originv < 0 for 0 ≤ t < 4: moving leftThen right after t = 4✓ Check: s(3.9) = −3.99 and s(4.1) = −3.99, equal on both sides of the turning point.
Find the number c promised by the Mean Value Theorem for f(x) = x³ on 0 ≤ x ≤ 3. Mean Value Theorem
Show answer and steps
average slope = 27 − 0 over 3 − 0 = 9Slope of the secant linef′(c) = 3c² = 9Set the tangent slope equal to itc = √3 ≈ 1.732c must be inside 0 < c < 3✓ Check: f′(1.732) = 8.99947, very close to 9.
Section XIISelf-guided quiz: eight quick questions
Choose one answer for each question, then press “Check my answers.” Your best score is saved only in this browser.
Section XIIIFermat’s maxima and Newton’s fluxions
Finding the highest and lowest points came before calculus had a name. In the late 1620s, in Bordeaux, the French lawyer Pierre de Fermat worked out a method for maxima and minima. In 1636 he began writing to Marin Mersenne’s circle of mathematicians in Paris, and when they asked how he solved the problems he set them, he sent his Method for determining Maxima and Minima and Tangents to Curved Lines. His idea was to compare a quantity at a value with its value at a slightly changed one, which in modern terms amounts to setting the slope equal to zero.
René Descartes attacked the method, but Fermat was proved right, and Descartes eventually wrote that his way of finding tangents was “very good.” Fermat rarely polished his work for print; his method first appeared in a supplement that Pierre Hérigone added to his Cursus mathematicus.
Isaac Newton came at the same ideas through motion. In his method of fluxions, a changing quantity was a “fluent,” imagined as flowing in time, and its rate of change was its “fluxion,” what we now call a derivative with respect to time. Gottfried Wilhelm Leibniz, meanwhile, titled his first calculus paper of 1684 Nova Methodus pro Maximis et Minimis, “A new method for maxima and minima.”
NotesSources and credits
- J. J. O’Connor and E. F. Robertson, MacTutor History of Mathematics, University of St Andrews: Pierre de Fermat (biography), on his work on maxima and minima in Bordeaux, the 1636 letter to Mersenne, Method for determining Maxima and Minima and Tangents to Curved Lines, the dispute with Descartes and Hérigone’s supplement.
- MacTutor: Isaac Newton (biography), on the method of fluxions.
- MacTutor: A history of the calculus, on fluents and fluxions.
- MacTutor: Gottfried Wilhelm Leibniz (biography), on Nova Methodus pro Maximis et Minimis (1684).
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