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Applications of the derivative

Level 7 · Ages 17–18

Quick recall· Let's wake up what you already know before we start.

What is the derivative of f(x) = x³?

The whole lesson

Read it again any time, including while you work through the exercises.

You already know how to compute a derivative: f'(x) measures the slope of the curve of f at the point with x-coordinate x. Time to turn this computing tool into a problem-solving machine. The idea fits in one sentence: the slope tells you everything about the motion of the curve.

If f'(x) > 0 on an interval, the slope is positive everywhere there: the curve rises, so f is increasing. If f'(x) < 0, the slope is negative: f is decreasing. The study plan is therefore always the same: compute f', study the sign of f', then deduce the variations of f in a table.

The points where f'(x) = 0 are the places where the tangent is horizontal. If, at such a point, f' goes from positive to negative, the curve rises then falls: that is a local maximum. If f' goes from negative to positive, that is a local minimum. Careful, one condition is essential: f' must change sign. For f(x) = x³, we have f'(0) = 0, but f' stays positive on both sides: the curve makes a brief plateau and keeps climbing, with no extremum at all.

This method becomes formidable in optimisation problems: finding the largest possible area, the minimal cost, the maximal volume. The approach: express the quantity to optimise as a function of a single variable, state the interval of study, differentiate, study the sign of the derivative, and conclude with the variation table. The best choice then shows up at the peak or the dip of the table.

The worked example, step by step

Study the variations of f(x) = x³ - 3x on its domain.

  1. f'(x) = 3x² - 3

    I differentiate term by term: the derivative of x³ is 3x², that of -3x is -3.

  2. f'(x) = 3(x² - 1) = 3(x - 1)(x + 1)

    I factor to study the sign. The derivative vanishes at x = -1 and x = 1.

  3. f' > 0 si x < -1 ou x > 1 ; f' < 0 entre -1 et 1

    It is a quadratic with a positive leading coefficient: negative between its roots, positive outside.

  4. f(-1) = -1 + 3 = 2 ; f(1) = 1 - 3 = -2

    I compute f at the points where f' vanishes and changes sign: these are the local extrema.

Result: f is increasing on ]-∞, -1], decreasing on [-1, 1], increasing on [1, +∞[. Local maximum 2 at x = -1, local minimum -2 at x = 1.

With 20 m of fencing, you want to enclose a rectangular pen against a wall (the wall replaces one of the long sides). Which dimensions give the maximal area?

  1. Largeur x, longueur 20 - 2x, avec 0 < x < 10

    I call the width x. The fencing covers two widths and one length: the length is therefore 20 - 2x. I state the interval of study.

  2. A(x) = x(20 - 2x) = 20x - 2x²

    I express the area to maximise as a function of the single variable x.

  3. A'(x) = 20 - 4x, qui s'annule en x = 5

    I differentiate, then look for where the derivative vanishes: 20 - 4x = 0 gives x = 5.

  4. A' > 0 si x < 5, A' < 0 si x > 5

    The area rises then falls: the maximum is reached at x = 5, where the derivative vanishes and changes sign.

  5. A(5) = 5 × 10 = 50

    I conclude: width 5 m, length 20 - 10 = 10 m, for a maximal area of 50 m².

Result: The area is maximal for a width of 5 m and a length of 10 m: it is then 50 m².