5.7 The second derivative (IB Maths: Analysis and Approaches SL): Revision notes

Section 1

What is the second derivative?

Differentiating $f'(x)$ again gives the second derivative, written $f''(x)$ or $\frac{d^{2}y}{dx^{2}}$ (read "d two y by d x squared").

Example: $f(x)=x^{4}-2x^{3}+5x$ gives $f'(x)=4x^{3}-6x^{2}+5$ and $f''(x)=12x^{2}-12x$ .

All the rules from 5.6 still apply, so a second derivative may need the chain, product or quotient rule twice: $y=e^{3x}-6x\Rightarrow\frac{dy}{dx}=3e^{3x}-6\Rightarrow\frac{d^{2}y}{dx^{2}}=9e^{3x}$ .

Key termssecond derivative

Common mistake

$\frac{d^{2}y}{dx^{2}}$ is not $\left(\frac{dy}{dx}\right)^{2}$ . Squaring the first derivative is a completely different quantity.

Exam tip

Simplify $f'(x)$ fully before differentiating again; it saves errors in the second step.

Section 2

What does the second derivative tell us?

$f''(x)$ is the rate of change of the gradient.

$f''(x)>0$ : the gradient $f'$ is increasing.
$f''(x)<0$ : the gradient $f'$ is decreasing.
$f''(x)=0$ : the gradient is momentarily not changing.

This says nothing directly about whether $f$ itself is increasing: that is decided by the sign of $f'$ . For $y=e^{3x}-6x$ , the gradient is always increasing ( $9e^{3x}>0$ ) even though the curve is decreasing for $x<\frac13\ln2$ .

Key termsrate of change of the gradient

Common mistake

Reading $f''<0$ as " $f$ is decreasing". $f''<0$ means the gradient is decreasing; the function may still be increasing, just more slowly.

Section 3

Linking the graphs of f, f′ and f″ in words

Each function describes the gradient of the one before:

where $f$ is increasing, $f'$ is positive; where $f$ is decreasing, $f'$ is negative;
a stationary point of $f$ is a zero of $f'$ ;
where $f'$ is increasing, $f''$ is positive; a maximum or minimum of $f'$ (steepest point of $f$ ) is a zero of $f''$ where it changes sign.

Example: $f'(x)=(x+1)(x-2)^{2}$ touches zero at $x=2$ without changing sign, so $f$ keeps increasing through $x=2$ . There $f''$ changes from negative to positive, so $f'$ has a minimum value of 0 and the graph of $f$ flattens momentarily before getting steeper again.

Key termsgradient functionsteepest point

Exam tip

When a question describes a graph in words, translate each phrase: "rising" is $f'>0$ , "levelling off" is $f'\to0$ , "getting steeper" is $|f'|$ increasing.

Section 4

Second derivatives in context

If $P(t)$ is a population, $P'(t)$ is its growth rate and $P''(t)$ says whether that growth is speeding up or slowing down. For $P(t)=400+60t^{2}-2t^{3}$ : $P''(5)=60>0$ (growth accelerating), $P''(15)=-60<0$ (growth slowing), but $P'(15)=450>0$ , so the population is still rising.

The growth rate is greatest where $P''=0$ and changes from positive to negative, here $t=10$ with $P'(10)=600$ fish per month. Units of $P''$ are (units of $P$ ) per (unit of time) squared, e.g. fish per month per month.

Key termsgrowth rate

Common mistake

Saying the population is falling because $P''<0$ . Always check the sign of $P'$ before commenting on $P$ .

Example

Distance, velocity and acceleration are the classic trio: acceleration is the second derivative of displacement.

Must know

$f''(x)=\frac{d}{dx}f'(x)$ ; notation $f''(x)$ or $\frac{d^{2}y}{dx^{2}}$ .
$f''>0$ : gradient increasing. $f''<0$ : gradient decreasing.
Sign of $f'$ decides whether $f$ increases; sign of $f''$ decides whether $f'$ increases.
The maximum rate of change of $f$ occurs where $f''=0$ and changes sign from $+$ to $-$ .
In context, interpret $f''$ as the rate of change of a rate, with squared time units.

That's the notes covered.

Carry on to the next subtopic.