You can sketch $f'$ without knowing the rule for $f$. Every feature of the derivative graph is read off the shape of the original.
Given only a picture of $f$, you can draw $f'$. Where $f$ is flat, $f'$ crosses zero. Where $f$ rises, $f'$ is positive. The whole sketch follows from reading steepness.
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
Sketch $y = x^2$ roughly. Where is it falling, where is it flat, and where is it rising? Now guess what a graph of its gradient would look like.
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
Work through the core explanation before applying it.
Where $f$ is increasing, $f'$ is above the axis. Where $f$ is decreasing, $f'$ is below it. Where $f$ has a stationary point, $f'$ crosses or touches zero.
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
Work through the core explanation before applying it.
Walk along the graph of $f$ from left to right and ask one question at each point: am I going up, down, or level?
Going up means $f' > 0$, so the derivative graph sits above the axis there. Going down means $f' < 0$, below the axis. Level means $f' = 0$, on the axis.
For $f(x) = x^2$: falling for $x < 0$, so $f'$ is negative there; stationary at $x = 0$, so $f'$ is zero there; rising for $x > 0$, so $f'$ is positive. That is exactly the line $y = 2x$.
Where $f$ rises, $f'$ is above the axis; where $f$ falls, $f'$ is below it; where $f$ is stationary, $f'$ is on it. For $f(x) = x^2$ that reasoning reproduces the line $y = 2x$.
Pause, copy the three-case rule and the worked $x^2$ giving $2x$, into your book.
We just saw how to get the sign of $f'$. That raises a question: where exactly does the derivative graph cross the axis? This card answers it → at every stationary point of $f$.
Start your sketch by marking every stationary point of $f$ and dropping a zero of $f'$ directly below or above each one. Those intercepts are the skeleton.
Then fill in the sign between them, using whether $f$ rises or falls on each interval.
Finally, judge the height. Where $f$ is steepest, $f'$ reaches its largest magnitude; where $f$ flattens out, $f'$ approaches the axis.
Sketch $f'$ by marking a zero under every stationary point of $f$, filling in the sign from whether $f$ rises or falls, then setting the height from how steep $f$ is. The degree of $f'$ is one less than that of $f$.
Pause, copy the three-step sketching order and the degree check, into your book.
We just saw how to sketch $f'$ for a smooth curve. That raises a question: what if $f$ has a sharp corner? This card answers it → $f'$ breaks there, because no single gradient exists.
At a sharp corner the gradient approaching from the left differs from the gradient approaching from the right, so there is no single value of $f'$.
On the sketch, $f'$ has a break at that $x$-value: an open circle on each side, jumping from one level to another.
For $f(x) = |x|$: the gradient is $-1$ for $x < 0$ and $+1$ for $x > 0$, with nothing defined at $x = 0$. The derivative graph is two horizontal rays with a hole between them.
At a sharp corner the left and right gradients differ, so $f'$ is undefined and its graph breaks there. $y = |x|$ is continuous at $x = 0$ but not differentiable: continuity is necessary for differentiability, not sufficient.
Pause, copy the corner case, the two rays with a hole for $y = |x|$, and the continuity note, into your book.
Worked examples · 3 in a row, reveal as you go
Follow the reasoning through complete worked solutions.
The graph of $f$ is a parabola with a minimum at $x = 3$, falling to the left of it and rising to the right. Sketch $f'$.
A cubic $f$ has a maximum at $x = -1$ and a minimum at $x = 2$. Describe the graph of $f'$.
Sketch the derivative of $f(x) = |x|$.
Work through the core explanation before applying it.
Run the quick drill and copy the summary into your book.
At the start you described where $y = x^2$ falls, is flat, and rises, then guessed the shape of its gradient graph. Compare your guess with $y = 2x$, and explain which feature of the parabola fixed the position of the $x$-intercept.
Answer the drill bank and rate your confidence.
Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Write full responses, then check them against the model answers.
Q1. A cubic function $f$ has a maximum at $x = -2$ and a minimum at $x = 1$, and is increasing for large positive $x$. Sketch the graph of $f'$, marking its $x$-intercepts and stating where it is positive and negative. (4 marks)
Q2. Explain why the graph of $f'$ for a quadratic $f$ must be a straight line, referring to the stationary point of the parabola. (3 marks)
Q3. A student claims that because $f(x) = |x - 3|$ is continuous everywhere, $f'$ exists everywhere. Explain the error, and describe the graph of $f'$. (3 marks)
Practice 1: positive. Practice 2: degree $3$, a cubic. Practice 3: three. Practice 4: it breaks there, because the left and right gradients differ.
Q1 (4 marks): $x$-intercepts of $f'$ at $x = -2$ and $x = 1$ [1]. $f' > 0$ for $x < -2$, since $f$ rises into the maximum [1]. $f' < 0$ for $-2 < x < 1$, between the maximum and the minimum [1]. $f' > 0$ for $x > 1$, so $f'$ is an upward parabola through those two intercepts [1].
Q2 (3 marks): Differentiating a polynomial reduces its degree by one, so a degree-two $f$ has a degree-one $f'$, which is a straight line [1]. A parabola has exactly one stationary point, so $f'$ has exactly one $x$-intercept [1]. A straight line with one $x$-intercept and constant gradient is consistent with the parabola falling on one side of the vertex and rising on the other [1].
Q3 (3 marks): Continuity is necessary for differentiability but not sufficient [1]. At $x = 3$ the graph has a sharp corner: the gradient approaching from the left is $-1$ and from the right is $1$, so no single gradient exists and $f'(3)$ is undefined [1]. The graph of $f'$ is therefore $y = -1$ for $x < 3$ and $y = 1$ for $x > 3$, with an open circle at $x = 3$ [1].
Take the module quiz if you are ready, then mark the lesson complete.
Read a graph of f and produce the matching graph of f prime. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaTick when you've finished the practice and review.
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