Advanced level

What the derivative is for

The sign of the derivative says whether a function rises or falls, and its zeros point at the candidates for extrema. Learn to write the equation of a tangent, read a sign table and find a largest value.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

All formulas

  • Equation of the tangent

    y=f′(x0)(x−x0)+f(x0)y = f'(x_0)(x - x_0) + f(x_0)

    a line of slope f′(x₀) through the point of tangency

  • Increasing function

    f′(x)>0 on an interval  ⟹  f increasingf'(x) > 0 \ \text{on an interval} \implies f \ \text{increasing}

    a positive derivative is a tangent leaning upwards

  • Decreasing function

    f′(x)<0 on an interval  ⟹  f decreasingf'(x) < 0 \ \text{on an interval} \implies f \ \text{decreasing}

    a negative derivative is a tangent leaning downwards

  • Necessary condition for an extremum

    f′(x0)=0f'(x_0) = 0

    a critical point — a candidate, not yet an extremum

  • Sufficient condition (maximum)

    f′:+→− at x0  ⟹  maximum at x0f' : + \to - \ \text{at} \ x_0 \implies \text{maximum at} \ x_0

    the derivative must CHANGE sign from plus to minus

  • Sufficient condition (minimum)

    f′:−→+ at x0  ⟹  minimum at x0f' : - \to + \ \text{at} \ x_0 \implies \text{minimum at} \ x_0

    minus to plus — a trough instead of a crest

The derivative is not an end in itself. Its value tells you how steeply a graph runs, and its sign tells you which way. Those two pieces of information are enough to read off the shape of a function without drawing it.

The equation of a tangent

A tangent at x0x_0 is a straight line, so a slope and one point describe it fully. The slope is the derivative and the point is (x0, f(x0))(x_0,\ f(x_0)):

y=f′(x0)(x−x0)+f(x0)y = f'(x_0)(x - x_0) + f(x_0)
012345−20246810xyy = 2x − 4P(3, 2)f(x) = x² − 4x + 5
A tangent touches the parabola at exactly one point and shares its slope there. Here f′(3) = 2, so the tangent climbs by 2 for every unit to the right.
Write the equation of the tangent to f(x) = x² − 4x + 5 at the point with x₀ = 3.

The sign of the derivative and monotonicity

Since the derivative is the slope of the tangent, its sign says outright which way the graph runs:

f′(x)>0 on an interval  ⟹  f increasing theref'(x) > 0 \ \text{on an interval} \implies f \ \text{increasing there} f′(x)<0 on an interval  ⟹  f decreasing theref'(x) < 0 \ \text{on an interval} \implies f \ \text{decreasing there}

This turns a question about monotonicity — which we used to read off a drawing — into an inequality. Instead of inspecting the graph, we solve f′(x)>0f'(x) > 0.

Mind the phrase "on an interval". Monotonicity is a property of an interval, never of a point, so the answer is always an interval and never a number.

Critical points

Between an interval where a function rises and one where it falls there has to be a place where it stops doing the first and starts doing the second. The derivative passes through zero there:

f′(x0)=0f'(x_0) = 0

Such an argument is called a critical point, and the condition is necessary for an extremum. Necessary, but not sufficient: it only produces a shortlist of candidates.

A counterexample worth remembering: for f(x)=x3f(x) = x^3 the derivative 3x23x^2 vanishes at zero, yet it is positive on both sides. The function keeps increasing and merely flattens for an instant — there is no extremum there.

The sufficient condition

An extremum requires the derivative to change sign:

f′:+→− at x0  ⟹  local maximum at x0f' : + \to - \ \text{at} \ x_0 \implies \text{local maximum at} \ x_0 f′:−→+ at x0  ⟹  local minimum at x0f' : - \to + \ \text{at} \ x_0 \implies \text{local minimum at} \ x_0
−3−2−10123−3−2−10123xyf′(−1) = 0f′(1) = 0maximumminimumf(x) = x³ − 3x
At both critical points the tangent is horizontal. Left of x = −1 the function rises, between −1 and 1 it falls, and beyond it rises again — hence the crest and the trough.
Study the monotonicity and the extrema of f(x) = x³ − 3x.

Optimisation

The most practical use of a derivative: finding a largest or a smallest value. The pattern never changes — write the quantity as a function of one variable, set the derivative to zero, then check the sign to confirm which kind of extremum you found.

A ball thrown straight up has height h(t) = 20t − 5t² (metres, seconds). When does it reach its highest point, and how high is it?

Exercises

The three questions match the three steps of this lesson. For the tangent, type the right-hand side of y=…y = \ldots, that is an expression such as 2x - 4. For the zeros of the derivative, give both critical points in the separate x1x_1 and x2x_2 fields. For the extremum, the answer is its value, the height of the crest or the depth of the trough — not the argument at which it is reached.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0
Tangent to the graph: f(x) = 2x² + x + 6, x₀ = 2 → y

Common mistakes

  • A tangent without its point of tangency — f′(x0)f'(x_0) alone gives the slope; the line must still pass through (x0, f(x0))(x_0,\ f(x_0)).
  • An extremum from f′(x₀) = 0 alone — that condition is necessary; with no change of sign there is no extremum (witness x3x^3 at zero).
  • Giving the argument instead of the value of the extremum — an extremum is f(x0)f(x_0), not x0x_0.
  • Monotonicity at a point — a function increases on an interval, so the answer is an interval.
  • Confusing a local extremum with a largest value — "local" concerns a neighbourhood only.
  • Reading the sign on one side only — settling the question needs both intervals around the critical point.

Formula card

Topic: What the derivative is for

  • Equation of the tangent

    y=f′(x0)(x−x0)+f(x0)y = f'(x_0)(x - x_0) + f(x_0)

    a line of slope f′(x₀) through the point of tangency

  • Increasing function

    f′(x)>0 on an interval  ⟹  f increasingf'(x) > 0 \ \text{on an interval} \implies f \ \text{increasing}

    a positive derivative is a tangent leaning upwards

  • Decreasing function

    f′(x)<0 on an interval  ⟹  f decreasingf'(x) < 0 \ \text{on an interval} \implies f \ \text{decreasing}

    a negative derivative is a tangent leaning downwards

  • Necessary condition for an extremum

    f′(x0)=0f'(x_0) = 0

    a critical point — a candidate, not yet an extremum

  • Sufficient condition (maximum)

    f′:+→− at x0  ⟹  maximum at x0f' : + \to - \ \text{at} \ x_0 \implies \text{maximum at} \ x_0

    the derivative must CHANGE sign from plus to minus

  • Sufficient condition (minimum)

    f′:−→+ at x0  ⟹  minimum at x0f' : - \to + \ \text{at} \ x_0 \implies \text{minimum at} \ x_0

    minus to plus — a trough instead of a crest

012345−20246810xyy = 2x − 4P(3, 2)f(x) = x² − 4x + 5
The tangent to f(x) = x² − 4x + 5 at P(3, 2). Its equation y = 2x − 4 comes from the slope f′(3) = 2 and the coordinates of the point of tangency.
−3−2−10123−3−2−10123xyf′(−1) = 0f′(1) = 0maximumminimumf(x) = x³ − 3x
The graph of f(x) = x³ − 3x with two horizontal tangents. At x = −1 and x = 1 the derivative vanishes: the first is a local maximum of value 2, the second a local minimum of value −2.

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