The derivative function graphically

What is it about?

The derivative $f'(x)$ is the slope of the tangent to the graph of $f$ at $x$ . It is the limit of the slopes of secants (difference quotient):

f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

Example $f(x) = x^2$ at $1$ with $h = 0.1$ : from the right $\frac{1.21 - 1}{0.1} = 2.1$ , from the left $\frac{1 - 0.81}{0.1} = 1.9$ , symmetric $\frac{1.21 - 0.81}{0.2} = 2$ . As $h \to 0$ all of them approach $f'(1) = 2$ . If the left and the right limit are not equal, the graph has a kink and $f$ is not differentiable there (e.g. $|x|$ at $0$ : left $-1$ , right $1$ ).

Assigning the slope $f'(x)$ to every $x$ gives the derivative function $f'$ . You can sketch its graph without calculating:

Where $f$ increases, $f'(x) > 0$ ; where $f$ decreases, $f'(x) < 0$ . The steeper $f$ , the further $f'$ is from the $x$ -axis.
A horizontal tangent of $f$ is a zero of $f'$ . With a sign change ( $+ \to -$ maximum, $- \to +$ minimum) it is an extremum, without a sign change a saddle point (stationary point of inflection).
Where $f$ increases or decreases most steeply (point of inflection), $f'$ has an extremum. There $f''$ changes sign: $f'' > 0$ means concave up, $f'' < 0$ concave down.

The rules of differentiation can be justified with pictures: constant multiple rule $(a \cdot f)' = a \cdot f'$ (all heights and so all slopes become $a$ times as large), sum rule $(g + h)' = g' + h'$ (the increases add up), power rule $(x^n)' = n \cdot x^{n-1}$ , product rule $(u \cdot v)' = u' \cdot v + u \cdot v'$ (increase of a rectangle’s area) and chain rule $(\sin(kx))' = k \cdot \cos(kx)$ .

In everyday life $f'$ is a rate of change: distance → velocity → acceleration, amount of water → inflow, cost → marginal cost, elevation profile → gradient ( $100\,\tfrac{\text{m}}{\text{km}} = 10\,\%$ ).

Try it

Running tangent: press “Let P run along”. Compare the vertical side of the slope triangle (with $\Delta x = 1$ ) with the vertical bar in the picture of $f'$ : it is exactly $f'(x)$ high.
With the second derivative: the small tangent to $f'$ has the slope $f''(x)$ . Where is $f''$ zero?
Parabola, sine, exponential and ln x: which well-known function appears below each time?
Secant → tangent: press “Let h tend to 0”. How does the point in the chart of the difference quotient move? Also choose “left” and “symmetric”.
Kink of |x|: why do the two branches in the lower picture not meet at $h = 0$ ?
Features: press “Features step by step” and follow the guide lines through all three pictures. Saddle point of x³: what is different?
Rules: constant multiple rule, sum rule, power rule, product rule, chain rule. Make $\Delta x$ smaller for the product rule: what happens to the small corner piece $\Delta u \cdot \Delta v$ ?
Everyday life: car, 100 m sprint, lift, vertical throw, temperature, epidemic, cost, hike, bathtub. What do a maximum, a zero of $f'$ and a point of inflection mean in each case?

Tasks

Task 1: Features of x³ − 3x

Load task

For $f(x) = x^3 - 3x$ , find $f'(2)$ , the extrema and the points of inflection, first graphically, then by calculation.

Show solution

$f'(x) = 3x^2 - 3$ , so $f'(2) = 12 - 3 = 9$ . $f'(x) = 0$ for $x = \pm 1$ : at $-1$ the sign of $f'$ changes from $+$ to $-$ (maximum $(-1, 2)$ ), at $1$ from $-$ to $+$ (minimum $(1, -2)$ ). $f''(x) = 6x$ is zero at $0$ and changes sign: point of inflection $(0, 0)$ . Show solution

Task 2: Difference quotient

Load task

For $f(x) = x^2$ at $x = 1$ with $h = 0.1$ , calculate the right, the left and the symmetric difference quotient. Which one is closest to $f'(1)$ ?

Show solution

Right $\frac{1.21 - 1}{0.1} = 2.1$ , left $\frac{1 - 0.81}{0.1} = 1.9$ , symmetric $\frac{1.21 - 0.81}{0.2} = 2$ . The symmetric quotient even hits $f'(1) = 2$ exactly here (always for parabolas); the one-sided ones are off by exactly $h$ . Show solution

Task 3: Sketch f′

Load task

Sketch the graph of the derivative of $f(x) = \frac{1}{4}(x^3 - 6x)$ : drag the seven points in the lower picture, then press “Reveal the solution”.

Show solution

$f$ increases, decreases between the extrema at $x = \pm\sqrt{2} \approx \pm 1.41$ and increases again. So $f'$ is an upward-opening parabola with zeros at $\pm\sqrt{2}$ and its minimum at $x = 0$ (point of inflection of $f$ ): $f'(x) = \frac{3}{4}x^2 - 1.5$ , e.g. $f'(0.5) = -1.3125$ . Show solution

Task 4: Car

Load task

In $t$ seconds a car covers the distance $s(t) = -0.05t^3 + 1.5t^2$ (in m) ( $0 \le t \le 20$ ). How fast is it after $4$ s? When is it fastest, and what is the acceleration at the start?

Show solution

$v(t) = s'(t) = -0.15t^2 + 3t$ , so $v(4) = -2.4 + 12 = 9.6\,\tfrac{\text{m}}{\text{s}} \approx 34.6\,\tfrac{\text{km}}{\text{h}}$ . $a(t) = v'(t) = -0.3t + 3$ is zero at $t = 10$ : there $v$ is greatest, $v(10) = 15\,\tfrac{\text{m}}{\text{s}} = 54\,\tfrac{\text{km}}{\text{h}}$ , and $s$ has a point of inflection. At the start $a(0) = 3\,\tfrac{\text{m}}{\text{s}^2}$ . Show solution

Task 5: Applying the rules

Load task

a) Differentiate $f(x) = 2\sin x + x^3$ and calculate $f'(0)$ . b) Use the product rule to calculate $(u \cdot v)'(2)$ for $u(x) = 1 + 0.5x$ and $v(x) = 1 + 0.25x^2$ .

Show solution

a) Constant multiple, sum and power rule: $f'(x) = 2\cos x + 3x^2$ , $f'(0) = 2$ . b) $u(2) = 2$ , $u' = 0.5$ , $v(2) = 2$ , $v'(x) = 0.5x$ , so $v'(2) = 1$ : $(u \cdot v)'(2) = 0.5 \cdot 2 + 2 \cdot 1 = 3$ . Show

Task 6: True or false?

“If $f'$ has a maximum at $x_0$ , then $f$ also has a maximum there.”

Show solution

False. A maximum of $f'$ means that $f$ increases most steeply there – $f$ has a point of inflection at $x_0$ . Example epidemic: the most new cases per day ( $N'$ greatest, after about $13.2$ days) occur at the point of inflection of the number of cases $N$ . Example

Ideas for the classroom

Use: the views work well as a sequence: secant → tangent, running tangent, features, your own sketch, rules. The simulation fits upper secondary calculus (local and global differentiation, monotonicity and extrema) and later the derivatives of sine and cosine and the product and chain rules.
Predictions: with “Show terms and values” switched off, students first sketch $f'$ on paper, then check with the trace or in the task mode “sketch f′”, which tells them for each point whether it fits.
Differentiability: the chart “difference quotient against h” turns limits into a picture: for smooth functions the left and right branches run towards the same point, for $|x|$ they do not.
Real-world context: nine examples with units (m/s, m/s², L/min, MU/unit, m/km and percent) – good for interpreting maxima, zeros of $f'$ and points of inflection in context.
Common stumbling blocks:
- A maximum of $f$ is confused with a maximum of $f'$ .
- The function value $f(x)$ is confused with the slope $f'(x)$ (“ $f$ is negative, so $f$ decreases”).
- A zero of $f'$ without a sign change is taken for an extremum.
- Points of inflection are overlooked because “nothing special” can be seen on the graph there.