Practice Maths

The Derivative as a Function

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Key Terms

f′(x) as a gradient function
: f′(x) gives the gradient (slope) of y = f(x) at every point x.
f′(x) > 0
: f is increasing — the curve slopes upward.
f′(x) < 0
: f is decreasing — the curve slopes downward.
f′(x) = 0
: f has a stationary point — the tangent is horizontal.
To sketch f′(x) from f(x): where f has a local max or min, f′ crosses the x-axis. Where f is steepest, |f′(x)| is largest.
The degree of f′ is always one less than the degree of f.
📚 Key Facts
f′(x) > 0 on interval ⇒ f is increasing on that interval
f′(x) < 0 on interval ⇒ f is decreasing on that interval
f′(x) = 0 at x = a ⇒ stationary point at x = a
Steepest point of f ⇒ largest |f′(x)|
x y f(x)=x³−3x −2 −1 0 1 2 max min x y f′(x)=3x²−3 −2 −1 0 1 2 f′=0 f′=0 x=−1 x=1
Hot Tip The x-intercepts of f′(x) always line up with the turning points of f(x). Above the x-axis (f′ > 0) means f is rising; below (f′ < 0) means f is falling. Read the sign of f′ to determine increasing/decreasing intervals without drawing f at all.

Worked Example 1 — Reading sign of f′(x)

Question: Given f′(x) = (x − 2)(x + 1), state the intervals where f is increasing and decreasing.

Find zeros of f′: x = −1 and x = 2

Sign test:

  • x < −1: test x = −2 → f′(−2) = (−4)(−1) = 4 > 0 → increasing
  • −1 < x < 2: test x = 0 → f′(0) = (−2)(1) = −2 < 0 → decreasing
  • x > 2: test x = 3 → f′(3) = (1)(4) = 4 > 0 → increasing

Answer: Increasing on (−∞, −1) and (2, ∞); decreasing on (−1, 2).

Worked Example 2 — Sketching f′(x) from f(x)

Question: f(x) is a cubic that increases on (−∞, −2), decreases on (−2, 1), and increases on (1, ∞). Sketch f′(x).

Step 1: f′(x) = 0 at x = −2 and x = 1 (turning points of f).

Step 2: f′(x) > 0 for x < −2 and x > 1; f′(x) < 0 for −2 < x < 1.

Step 3: Since f is cubic (degree 3), f′ is quadratic (degree 2) — a parabola opening upward, crossing x-axis at x = −2 and x = 1.

Sketch: f′(x) is an upward parabola with x-intercepts at −2 and 1.

Introduction

When we differentiate f(x) using the power rule or first principles, we get f′(x) — not just a number at one point, but a whole new function defined at every x. This derivative function (also called the gradient function) tells us the slope of f at any x-value. Understanding how to read and sketch f′(x) is one of the most important skills in calculus.

What f′(x) Tells Us

The sign of f′(x) reveals everything about how f is behaving:

  • f′(x) > 0 on an interval ⇒ f is increasing there (curve going uphill as x increases).
  • f′(x) < 0 on an interval ⇒ f is decreasing there (curve going downhill).
  • f′(x) = 0 at a point ⇒ stationary point (tangent is horizontal). Could be a local max, local min, or stationary inflection.

The magnitude of f′(x) tells us how steep the curve is: large |f′(x)| means the curve is steep; small |f′(x)| means nearly flat.

Sketching f′(x) from f(x)

Use these steps to go from a graph of f to a sketch of f′:

  1. Mark stationary points of f — these give x-intercepts of f′.
  2. Identify increasing regions — f′ is positive (above x-axis) here.
  3. Identify decreasing regions — f′ is negative (below x-axis) here.
  4. Identify steepest points — f′ has a local max or min here (in terms of |f′|).
  5. Match degrees — if f is degree n, f′ is degree n−1. A cubic gives a parabola, a quadratic gives a line, a line gives a constant.
Worked Example — Sketching f′ from f

f(x) = x² − 4x + 3. Sketch f′(x).

Step 1: Find f′: f′(x) = 2x − 4 (a line with gradient 2, x-intercept at x = 2).

Step 2: f′(2) = 0 corresponds to the minimum of f at x = 2.

Step 3: f′ < 0 for x < 2 (f is decreasing); f′ > 0 for x > 2 (f is increasing).

Sketch: f′(x) = 2x − 4 is a straight line passing through (2, 0) with positive slope.

Worked Example — Intervals from f′

Given f′(x) = x² − 5x + 6 = (x − 2)(x − 3), find where f increases and decreases.

Step 1: Zeros of f′: x = 2 and x = 3.

Step 2: Test x = 0: f′(0) = 6 > 0 → f increasing on (−∞, 2).

Step 3: Test x = 2.5: f′(2.5) = (0.5)(−0.5) = −0.25 < 0 → f decreasing on (2, 3).

Step 4: Test x = 4: f′(4) = (2)(1) = 2 > 0 → f increasing on (3, ∞).

💡 Key Reminder: Always find the zeros of f′ first — these are the boundaries between increasing and decreasing intervals. A sign chart (number line with +/− marked) is the clearest way to organise your answer.

Summary

f′(x) is a function in its own right. Its sign tells you whether f is increasing (+) or decreasing (−). Its zeros correspond to stationary points of f. To sketch f′ from f: locate turning points (x-intercepts of f′), determine sign of f′ in each interval, and match the degree of f′ to be one less than f.

Mastery Practice

See Answers ➔
  1. Fluency

    State the sign of f′(x) and whether f is increasing or decreasing at each point.

    1. (a) f′(3) = 5
    2. (b) f′(−1) = −2
    3. (c) f′(0) = 0
    4. (d) f′(4) = −7
  2. Fluency

    For each derivative function, find the x-values where the original function has stationary points.

    1. (a) f′(x) = 2x − 6
    2. (b) f′(x) = 3x² − 12
    3. (c) f′(x) = x² − x − 6
    4. (d) f′(x) = 4x(x − 3)
  3. Fluency

    Given f(x) = x² − 4x + 1:

    1. (a) Find f′(x).
    2. (b) Find where f′(x) = 0.
    3. (c) State the interval(s) where f is increasing.
    4. (d) State the interval(s) where f is decreasing.
  4. Fluency

    Given f′(x) = (x − 1)(x + 3), complete the sign chart and state where f increases and decreases.

    Method: Identify zeros, test a value in each interval, record sign of f′, then state conclusion about f.
  5. Understanding

    Given f(x) = x³ − 3x² − 9x + 2:

    1. (a) Find f′(x).
    2. (b) Solve f′(x) = 0.
    3. (c) State the intervals where f is increasing and where it is decreasing.
  6. Understanding

    The graph of f(x) has a local maximum at x = −2, a local minimum at x = 3, and is a degree 3 polynomial.

    1. (a) State the x-intercepts of f′(x).
    2. (b) Describe the shape of f′(x) (what type of function, and in which direction does it open?).
    3. (c) Sketch a possible graph of f′(x).
  7. Understanding

    Match each function with its derivative graph description.

    1. (a) f(x) = 3x + 1
    2. (b) f(x) = x² − 2x
    3. (c) f(x) = x³ − 3x

    Descriptions:   (i) A horizontal line at y = 3   (ii) An upward parabola crossing x-axis at −1 and 1   (iii) A line with x-intercept at x = 1

  8. Understanding

    The derivative of a function is f′(x) = −2x + 4.

    1. (a) Where is f′(x) = 0?
    2. (b) Where is f increasing? Where is f decreasing?
    3. (c) What is the gradient of f at x = 0 and at x = 5?
    4. (d) Describe the shape of f (what type of function is it, and does it have a max or min?).
  9. Problem Solving

    The graph of y = f(x) is shown below (described): it is a cubic that has a local maximum at (−1, 4) and a local minimum at (2, −3). It is increasing on (−∞, −1) and (2, ∞) and decreasing on (−1, 2).

    1. (a) Sketch f′(x), clearly showing x-intercepts and indicating where f′ is positive and negative.
    2. (b) Using your sketch, estimate the value of x at which |f′(x)| is greatest (i.e., where f is steepest).
    3. (c) For what values of x is f′(x) > 0?
  10. Problem Solving

    Water level application.

    Challenge. The water level in a tank is modelled by W(t) = −t³ + 6t² + 1 for 0 ≤ t ≤ 6, where W is in litres and t is in minutes.
    1. (a) Find W′(t).
    2. (b) Find when the water level is rising fastest (maximum of W′).
    3. (c) Find the intervals during which the water level is rising and falling.
    4. (d) What does it mean in context when W′(t) = 0?