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Year 12 Maths Advanced (2027) Applications of calculus

Sketching Derivative & Second-Derivative Graphs

20 practice questions 0 video lessons Theory + worked examples
NSW · Year 12 Mathematics Advanced · Applications of calculus

Sketching derivative graphs reads the gradient of \(y=f(x)\) to draw \(y=f'(x)\) and \(y=f''(x)\), linking turning points to zeros of the derivative.

Part of the NSW Year 12 Mathematics Advanced course, in the Calculus area of study (Applications of calculus focus area) of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.

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Theory

From the graph of \(f\) you can sketch \(f'\) by reading gradients. This Year 12 Mathematics Advanced topic (MAV-12-06) builds derivative and second-derivative graphs.

A stationary point of \(f\) becomes an \(x\)-intercept of \(f'\); where \(f\) rises, \(f'\) is above the axis; where \(f\) falls, \(f'\) is below. A point of inflection of \(f\) is a turning point of \(f'\).

f and its derivativeThe parabola f = x squared and its derivative f prime = 2x. xy f=x^2
\(f=x^{2}\) (navy) has its minimum where \(f'=2x\) (red) crosses the axis.
\[\text{stationary point of }f\ \leftrightarrow\ x\text{-intercept of }f'\]
a stationary point of f corresponds to an x intercept of f prime

Method

  1. Mark the stationary points of \(f\) as zeros of \(f'\).
  2. Sign: \(f'\) is positive where \(f\) rises, negative where it falls.
  3. Repeat for \(f''\) from \(f'\).
Example 1 — Derivative of a parabola
Sketch \(f'\) for \(f(x)=x^{2}\).
Solution
\(f'(x)\)\(=\)\(2x\)

A line through the origin.

Example 2 — Zero of f'
\(f\) has a maximum at \(x=3\). What does \(f'\) do there?
Solution

\(f'(3)=0\) — it crosses the axis.

Example 3 — Sign of f'
\(f\) is decreasing on an interval. Where is \(f'\)?
Solution

Below the axis (\(f'<0\)).

Example 4 — Second derivative
Sketch \(f''\) for \(f(x)=x^{3}\).
Solution
\(f'(x)\)\(=\)\(3x^{2}\)
\(f''(x)\)\(=\)\(6x\)

A line through the origin.

Common pitfalls

Peaks/troughs of \(f\) become zeros of \(f'\).
Rising \(f\) means \(f'>0\).
An inflection of \(f\) is a turning point of \(f'\).

Frequently asked questions

How do you sketch the derivative graph from f?

Mark zeros of f prime at the stationary points of f, put f prime above the axis where f increases and below where f decreases.

What happens to f prime at a maximum of f?

It is zero and crosses from positive to negative.

What does a point of inflection of f give on the f prime graph?

A turning point (maximum or minimum) of f prime.

How do you get the second-derivative graph?

Apply the same gradient reading to the graph of f prime.