Graphs of Derived Functions

Video lesson · from 1:43:583 worked examples

One lesson video covers all of Differentiation, so it opens at 1:43:58 for this topic — not from the beginning.

We can sketch $f'(x)$ by examining the features of $f(x)$ and its gradients.

Original Graph: $f(x)$	Derived Graph: $f'(x)$
Increasing	Positive (above x-axis)
Decreasing	Negative (below x-axis)
Turning Point	Root (cuts x-axis)
Point of Inflection	Root and Turning Point (touches x-axis)

For example, if $f(x)$ is a cubic with two turning points, $f'(x)$ will be a parabola with two roots.

This is one of the worst-answered questions in the course. Most candidates gain no marks on it at all. It is very winnable if you follow the rules below.
Get the degree right: many candidates draw a curve of the wrong type — a quartic differentiates to a cubic, a cubic to a parabola. Decide the shape before you draw anything.
Turning points become roots: every stationary point of $f(x)$ is a point where $f'(x)$ crosses the x-axis. Mark those first — they fix the sketch.
Read the sign of the slope: where $f(x)$ rises, $f'(x)$ is above the axis; where it falls, below. Check each region.
Use the conditions given: the stated conditions are often ignored, and some candidates mistakenly treat the question as a differential equation. Read what you are told about the original graph.

Example 1

Shown is the graph of $f(x)$ . Sketch $f'(x)$ .

The original graph is a parabola with a minimum turning point at $(-1, -11)$ .

When $x < -1$ , the function is decreasing, so the gradient is negative ( $f'(x) < 0$ ).
At $x = -1$ , there is a turning point, so the gradient is zero ( $f'(-1) = 0$ ).
When $x > -1$ , the function is increasing, so the gradient is positive ( $f'(x) > 0$ ).

The derivative of a quadratic is a linear function (straight line), crossing the x-axis at $-1$ .

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Example 2

Shown is the graph of $f(x)$ . Sketch $f'(x)$ .

The original graph is a cubic with a maximum turning point at $(0, -6)$ and a minimum turning point at $(3, -18)$ .

At $x = 0$ and $x = 3$ , there are turning points, so $f'(0) = 0$ and $f'(3) = 0$ (roots of the derived graph).
Between $0$ and $3$ , the function is decreasing, so $f'(x)$ is negative (below the x-axis).
Outside of this interval ( $x < 0$ and $x > 3$ ), the function is increasing, so $f'(x)$ is positive (above the x-axis).

The derivative of a cubic is an upward-opening parabola crossing at $0$ and $3$ .

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Example 3

Shown is the graph of $f(x)$ . Sketch $f'(x)$ .

The original graph has a rising point of inflection at $(0, 5)$ and a maximum turning point at $(1.5, 8.375)$ .

At $x = 0$ and $x = 1.5$ , the gradient is zero, so $f'(0) = 0$ and $f'(1.5) = 0$ .
Since $(0, 5)$ is a rising point of inflection, the graph is increasing both before and immediately after $x = 0$ . So $f'(x)$ is positive, touches $0$ , and stays positive.
After the maximum at $x = 1.5$ , the graph decreases, so the gradient becomes negative.

The derived graph touches the x-axis at $0$ and crosses it at $1.5$ .

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