Year 12 Maths - Methods (Unit 3 & Unit 4) Differentiation

The graph of the derivative function

Q: How do you sketch the graph of f prime of x from the graph of f of x?

Mark every stationary point of f: each gives an x-intercept of the derivative graph. Between those points, put the derivative graph above the x-axis wherever f is increasing and below wherever f is decreasing, and make it further from the axis where f is steeper. The result is a curve one degree lower than f.

Q: What does the derivative graph do at a stationary point of f?

At a stationary point the tangent to f is horizontal, so f prime equals zero there and the graph of f prime crosses or touches the x-axis at that x value. A local maximum is where f prime changes from positive to negative, and a local minimum is where it changes from negative to positive.

Q: How can you tell where f is increasing or decreasing from the derivative graph?

Where the graph of f prime is above the x-axis, f prime is positive and f is increasing; where the graph of f prime is below the x-axis, f prime is negative and f is decreasing. The x-intercepts of f prime mark where f changes between increasing and decreasing.

Q: What shape is the graph of the derivative of a cubic?

The derivative of a cubic is a quadratic, so its graph is a parabola. In general the derivative graph is one degree lower than f: a quadratic gives a straight line, and a straight line gives a constant horizontal line.

Q: How do you read the features of f from a given graph of f prime of x?

The x-intercepts of the f prime graph give the x-coordinates of the stationary points of f. Where f prime is positive f is increasing, and where it is negative f is decreasing; a sign change from positive to negative is a local maximum of f and from negative to positive is a local minimum.

Q: What is the difference between the graph of f and the graph of f prime?

The graph of f shows the value of the function; the graph of f prime shows its gradient at each x. Heights on the f prime graph are the slopes of f, so the f prime graph is zero at the turning points of f and is one degree lower in shape.

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Theory

The graph of the derivative \(y=f'(x)\) can be read straight off the graph of \(y=f(x)\): it crosses the \(x\)-axis at every stationary point (where \(f'(x)=0\)), lies above the axis where \(f\) is increasing and below where \(f\) is decreasing, and its shape is one degree lower — a cubic gives a parabola. Running the same reasoning in reverse lets you read the features of \(f\) from a given graph of \(f'\).

The derivative \(f'(x)\) is the gradient of \(y=f(x)\) at each value of \(x\). So the height of the point on the graph of \(y=f'(x)\) is simply the slope of the original curve directly above or below it. This one idea — height on the \(f'\) graph = slope of \(f\) — is all you need to translate between the two pictures.

Where \(f\) is increasing the slope is positive, so the \(f'\) graph is above the \(x\)-axis; where \(f\) is decreasing the slope is negative, so the \(f'\) graph is below the axis. At a stationary point the tangent is horizontal, the slope is \(0\), and the \(f'\) graph crosses (or touches) the \(x\)-axis. A local maximum of \(f\) is where \(f'\) changes from \(+\) to \(-\); a local minimum is where \(f'\) changes from \(-\) to \(+\).

Because differentiating lowers a polynomial's degree by one, the graph of \(f'\) is one degree simpler than \(f\): the derivative of a cubic is a parabola, of a quadratic a straight line, and of a straight line a constant. The steeper \(f\) is, the further the \(f'\) graph sits from the axis. Reversing every step lets you recover \(f\)'s stationary points and its increasing/decreasing behaviour from a given graph of \(f'\).

Key idea. The height of \(y=f'(x)\) is the slope of \(y=f(x)\). So \(f'\) cuts the \(x\)-axis at the turning points of \(f\), is positive where \(f\) rises and negative where \(f\) falls, and is one degree lower in shape.

Top: \(f(x)=\dfrac{1}{3}x^{3}-x\). Bottom: \(f'(x)=x^{2}-1\) crosses the axis exactly under the max and min of \(f\)

Top: \(f(x)=x^{2}-2\). Bottom: \(f'(x)=2x\) is a straight line, zero at the minimum and one degree lower than \(f\)

The sign of the derivative controls whether \(f\) rises or falls:

\[f'(x)>0 \iff f \text{ increasing} \qquad f'(x)<0 \iff f \text{ decreasing}\]

f^{'} (x) > 0 \Leftrightarrow f increasing

The derivative graph meets the \(x\)-axis exactly at the stationary points of \(f\):

\[f'(a)=0 \quad\Longleftrightarrow\quad x=a \text{ is a stationary point of } f\]

f^{'} (a) = 0

Differentiating lowers a polynomial's degree by one, so the derivative graph is one degree simpler:

\[\deg\!\big(f'\big)=\deg\!\big(f\big)-1 \qquad \dfrac{d}{dx}\!\left(ax^{n}\right)=nax^{n-1}\]

\deg (f^{'}) = \deg (f) - 1

Sign change = shape of turning point. If \(f'\) crosses the axis from \(+\) to \(-\), \(f\) has a local maximum; from \(-\) to \(+\), a local minimum. If \(f'\) only touches the axis without changing sign, \(f\) has a stationary point of inflection.

Sketching \(y=f'(x)\) from the graph of \(y=f(x)\)

Mark the stationary points. Every turning point or horizontal-tangent point of \(f\) gives an \(x\)-intercept of the \(f'\) graph, since \(f'=0\) there.
Fix the sign in each interval. Where \(f\) is increasing, draw \(f'\) above the \(x\)-axis; where \(f\) is decreasing, draw it below.
Classify the crossings. A max of \(f\) is where \(f'\) goes \(+\to-\); a min is where \(f'\) goes \(-\to+\); a stationary inflection is where \(f'\) just touches the axis.
Match the steepness and shape. Steeper parts of \(f\) push \(f'\) further from the axis; remember the graph is one degree lower (cubic \(\to\) parabola, quadratic \(\to\) line).
To go the other way, read a given \(f'\) graph the same way: its intercepts are \(f\)'s stationary points, and its sign tells you where \(f\) rises or falls.

Quick check. Count the turning points of \(f\): that is exactly the number of \(x\)-intercepts the \(f'\) graph must have. A cubic with two turning points gives a parabola cutting the axis twice.

Example 1 — derivative of a straight line

The graph of \(y=f(x)\) is a straight line with gradient \(3\). Describe the graph of \(y=f'(x)\).

Solution

The gradient is the same everywhere — that constant slope is the height of the \(f'\) graph:

\(\text{slope of } f\)	\(=\)	\(3 \text{ for every } x\)
\(f'(x)\)	\(=\)	\(3\)

A constant value plots as a horizontal line:

\(\therefore\) \(y=f'(x)\) is the horizontal line \(y=3\) (above the axis, since \(f\) is always increasing).

f^{'} (x) = 3

Example 2 — derivative of a parabola

A parabola \(y=f(x)\) has its minimum turning point at \(x=2\), falling to the left of it and rising to the right. Sketch \(y=f'(x)\), given \(f(x)=x^{2}-4x+1\).

Solution

Differentiate to get the gradient function:

\(f'(x)\)

\(=\)

\(2x-4\)

The stationary point of \(f\) becomes the \(x\)-intercept of \(f'\):

\(f'(x)=0\)

\(\Rightarrow\)

\(x=2\)

To the left \(f\) falls (\(f'<0\)); to the right it rises (\(f'>0\)). So \(f'\) is a line rising through \((2,0)\):

\(\therefore\) \(y=f'(x)=2x-4\): a straight line crossing the axis at \(x=2\), negative before it and positive after.

f^{'} (x) = 2 x - 4

Example 3 — derivative of a cubic

A cubic \(y=f(x)\) has a local maximum at \(x=-1\) and a local minimum at \(x=3\). What are the key features of the graph of \(y=f'(x)\)?

Solution

Each stationary point of \(f\) is an \(x\)-intercept of \(f'\):

\(f'(x)=0\)

\(\text{at}\)

\(x=-1 \text{ and } x=3\)

Read the sign of \(f'\) from the shape of \(f\):

\(x<-1\)	\(:\)	\(f\) rising, so \(f'>0\)
\(-1	\(:\)	\(f\) falling, so \(f'<0\)
\(x>3\)	\(:\)	\(f\) rising, so \(f'>0\)

Positive, then negative, then positive means an upward parabola:

\(\therefore\) \(y=f'(x)\) is a parabola opening upward, with \(x\)-intercepts at \(x=-1\) and \(x=3\).

f^{'} (x) = 0 at x = - 1, x = 3

Example 4 — reading \(f\) from a given \(f'\)

The graph shown is \(y=f'(x)\), a parabola with \(x\)-intercepts at \(x=0\) and \(x=4\). Find the \(x\)-coordinates of the stationary points of \(f\), classify each, and state where \(f\) is increasing.

Solution

Stationary points of \(f\) are the \(x\)-intercepts of the given \(f'\) graph:

\(f'(x)=0\)

\(\text{at}\)

\(x=0 \text{ and } x=4\)

Classify by the sign change of \(f'\) at each intercept:

at \(x=0\)	\(:\)	\(f'\) goes \(+\to-\), a local maximum
at \(x=4\)	\(:\)	\(f'\) goes \(-\to+\), a local minimum

\(f\) increases wherever the \(f'\) graph is above the axis:

\(f'>0\)

\(\text{for}\)

\(x<0 \text{ or } x>4\)

\(\therefore\) max at \(x=0\), min at \(x=4\); \(f\) is increasing for \(x<0\) and \(x>4\), and decreasing for \(0

max at x = 0, min at x = 4

Common pitfalls

The \(f'\) graph shows slope, not height. A point where \(f\) is high but flat has a large \(y\)-value on the \(f\) graph yet \(f'=0\). Always plot the gradient, not the value of \(f\).

A maximum of \(f\) is not a maximum of \(f'\). At a local max the \(f'\) graph is crossing the \(x\)-axis (from \(+\) to \(-\)), so its height there is \(0\), not a peak.

Do not forget the degree drops. The derivative of a cubic is a parabola, so it can cut the axis twice — matching two turning points. Sketching another cubic for \(f'\) is a common slip.

Frequently asked questions

How do you sketch the graph of f prime of x from the graph of f of x?

Mark every stationary point of \(f\): each gives an \(x\)-intercept of the derivative graph. Between those points, put \(f'\) above the axis where \(f\) is increasing and below where \(f\) is decreasing, further from the axis where \(f\) is steeper. The result is one degree lower than \(f\).

What does the derivative graph do at a stationary point of f?

The tangent to \(f\) is horizontal, so \(f'=0\) and the \(f'\) graph crosses or touches the \(x\)-axis there. A local maximum is where \(f'\) changes from positive to negative; a local minimum is where it changes from negative to positive.

How can you tell where f is increasing or decreasing from the derivative graph?

Where \(f'\) is above the \(x\)-axis it is positive and \(f\) is increasing; where \(f'\) is below the axis it is negative and \(f\) is decreasing. The \(x\)-intercepts of \(f'\) mark the changeovers.

What shape is the graph of the derivative of a cubic?

A parabola. Differentiating drops the degree by one, so a cubic gives a quadratic, a quadratic gives a straight line, and a straight line gives a constant horizontal line.

How do you read the features of f from a given graph of f prime of x?

The \(x\)-intercepts of \(f'\) give the stationary points of \(f\). Where \(f'>0\) the function increases and where \(f'<0\) it decreases; \(+\to-\) is a local maximum and \(-\to+\) is a local minimum.

What is the difference between the graph of f and the graph of f prime?

The graph of \(f\) shows the function's value; the graph of \(f'\) shows its gradient at each \(x\). Heights on the \(f'\) graph are the slopes of \(f\), so \(f'\) is zero at the turning points of \(f\) and one degree lower in shape.

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The chain rule

The graph of the derivative function

📖 Theory

Sketching \(y=f'(x)\) from the graph of \(y=f(x)\)

Common pitfalls

Frequently asked questions

Theory