What is the significance of the second derivative in Differentiation 2?

In Differentiation 2, the second derivative is crucial for understanding the concavity of a function and identifying points of inflection. It helps determine whether a function is concave up or concave down at a particular point, which is essential for analyzing the behavior of graphs in IB DP Mathematics.

How do you find the second derivative of a function?

To find the second derivative of a function, you first need to differentiate the function to obtain the first derivative. Then, differentiate the first derivative to get the second derivative. This process involves applying differentiation rules such as the power rule, product rule, or chain rule, depending on the function's complexity.

What role does the second derivative test play in optimization problems?

The second derivative test is used in optimization problems to determine the nature of critical points. If the second derivative at a critical point is positive, the function has a local minimum there. If it is negative, the function has a local maximum. This test is a powerful tool in IB DP Mathematics for solving real-world optimization problems.

How can you use the second derivative to find points of inflection?

Points of inflection occur where the concavity of a function changes. To find these points, set the second derivative equal to zero and solve for the variable. Verify that the concavity changes by checking the sign of the second derivative on either side of these points. This method is essential for graph analysis in Differentiation 2.

Why is understanding the second derivative important in IB DP Mathematics?

Understanding the second derivative is important in IB DP Mathematics because it provides deeper insights into the behavior of functions. It helps in analyzing the curvature and optimizing functions, which are key skills in solving complex mathematical problems and real-world applications covered in the curriculum.

Why is "$d/dx(y^3) = 3y^2$." a common mistake in Differentiation 2?

Why it happens: Treating y as independent. How to avoid it: Chain rule: d/dx(y^3) = 3y^2 (dy/dx).

Why is "Claiming inflection because $f''(x) = 0$." a common mistake in Differentiation 2?

Why it happens: Skipping the sign-change test. How to avoid it: f'' = 0 is necessary but not sufficient — also verify f'' CHANGES sign across the point.

Why is "Forgetting that 'sliding down' means $dy/dt < 0$." a common mistake in Differentiation 2?

Why it happens: Ignoring direction. How to avoid it: Always interpret the SIGN of the rate physically.

DifferentiationIB Maths AA HL

Differentiation 2

Q: Why is "Forgetting that 'sliding down' means $dy/dt < 0$." a common mistake in Differentiation 2?

Why it happens: Ignoring direction. How to avoid it: Always interpret the SIGN of the rate physically.

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Differentiation 2 — IB Maths AA HL: implicit, parametric, second derivative, optimisation, kinematics, related rates

AA HL extends differentiation with implicit and parametric derivatives, concavity analysis, optimisation, kinematics (displacement → velocity → acceleration) and related rates of change.

What you’ll learn

Mapped to the IB DP Maths AA HL subject guide (2021 onwards (applies to 2026 exams)).

AO1 — State and apply implicit and parametric derivatives.
AO1 — Interpret $f''(x)$ for concavity and inflection.
AO2 — Solve optimisation problems.
AO2 — Apply kinematics formulas.
AO2 — Set up and solve related-rates problems via chain rule.

Implicit and parametric differentiation

When $y$ is not isolated.

Implicit differentiation. When $y$ is given implicitly by an equation like $x^2 + y^2 = 25$ , differentiate both sides treating $y$ as a function of $x$ :

$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25) \Rightarrow 2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$ .

Worked example. Find $dy/dx$ for $x^2 y + y^3 = 4$ .

Differentiate: $2xy + x^2 \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 0$ .

$\frac{dy}{dx}(x^2 + 3y^2) = -2xy$ , so $dy/dx = -2xy/(x^2 + 3y^2)$ .

Parametric differentiation. When $x = f(t), y = g(t)$ :

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.$

Worked example. $x = t^2, y = t^3$ . Find $dy/dx$ at $t = 2$ .

$dx/dt = 2t = 4$ . $dy/dt = 3t^2 = 12$ . $dy/dx = 12/4 = 3$ .

For second derivative: $d^2y/dx^2 = \frac{d/dt(dy/dx)}{dx/dt}$ .

Implicit: every $y$ term gets a $dy/dx$ factor.
Parametric: divide $t$ -derivatives.
Second derivative — differentiate again w.r.t. $t$ , then divide by $dx/dt$ once more.

Optimisation, kinematics and related rates

Applications of derivatives.

Optimisation steps.

Define variables and write the quantity to optimise as a function of one variable.
Set derivative $= 0$ .
Confirm max/min via $f''$ or sign change.
Check boundary values for global extrema.
State the answer with units.

Kinematics. For displacement $s(t)$ :

Velocity: $v = ds/dt$ .
Acceleration: $a = dv/dt = d^2s/dt^2$ .

Object is at rest when $v = 0$ ; speeding up when $v$ and $a$ have the same sign; slowing down when opposite.

Related rates. Two quantities related by an equation both change with $t$ . Differentiate w.r.t. $t$ and use chain rule.

Worked example (related rates). A spherical balloon's volume increases at $20$ cm³/s. Find the rate at which the radius increases when $r = 5$ .

$V = (4/3)\pi r^3 \Rightarrow dV/dt = 4\pi r^2 (dr/dt)$ .

$20 = 4\pi (25)(dr/dt) \Rightarrow dr/dt = 20/(100\pi) = 1/(5\pi) \approx 0.0637$ cm/s.

Worked example (optimisation). Open rectangular box, square base. Volume $= 32$ cm³. Minimise surface area.

Base side $x$ , height $h$ . $x^2 h = 32 \Rightarrow h = 32/x^2$ .

Surface: $S = x^2 + 4xh = x^2 + 128/x$ . $S' = 2x - 128/x^2 = 0 \Rightarrow x^3 = 64 \Rightarrow x = 4$ .

$S'' = 2 + 256/x^3 > 0$ , so minimum. $h = 32/16 = 2$ . Min surface $= 16 + 32 = 48$ cm².

Optimisation: derivative = 0 then confirm with $f''$ .
Kinematics: derivative chain $s \to v \to a$ .
Related rates: differentiate the link equation w.r.t. $t$ .

How it’s examined

Paper 1: implicit/parametric derivatives, basic kinematics. Paper 2: optimisation, complex related rates with GDC. Paper 3: extended modelling problems.

Sources: IB Diploma Programme Mathematics: Analysis and Approaches subject guide (IBO, official). Last reviewed 2026-05-31.

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Step-by-step worked examples — Differentiation 2

Step-by-step solutions to past-paper-style questions on differentiation 2, written exactly the way a tutor would explain them at the board.

1Implicit derivative with mixed terms
Stretch• implicit
Question
Find $dy/dx$ for $x^3 + y^3 = 6xy$ at $(3, 3)$ .
Step-by-step solution
1. Step 1
  Differentiate: $3x^2 + 3y^2 dy/dx = 6y + 6x dy/dx$ .
2. Step 2
  $(3y^2 - 6x)(dy/dx) = 6y - 3x^2 \Rightarrow dy/dx = (6y - 3x^2)/(3y^2 - 6x)$ .
3. Step 3
  At $(3, 3)$ : $dy/dx = (18 - 27)/(27 - 18) = -9/9 = -1$ .
Answer
$dy/dx = -1$ at $(3, 3)$ .
2Parametric curve gradient
Building confidence• parametric
Question
Curve: $x = \cos t, y = \sin 2t$ . Find $dy/dx$ at $t = \pi/4$ .
Step-by-step solution
1. Step 1
  $dx/dt = -\sin t$ . $dy/dt = 2\cos 2t$ .
2. Step 2
  $dy/dx = (2\cos 2t)/(-\sin t)$ .
3. Step 3
  At $t = \pi/4$ : $\cos(\pi/2) = 0$ , so $dy/dx = 0$ .
Answer
$dy/dx = 0$ at $t = \pi/4$ .
3Kinematics with sign analysis
Building confidence• kinematics
Question
$s(t) = t^3 - 6t^2 + 9t$ for $t \in [0, 4]$ . Find the times when the particle is at rest, and when it is slowing down.
Step-by-step solution
1. Step 1
  $v = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)$ .
2. Step 2
  At rest: $v = 0 \Rightarrow t = 1, 3$ .
3. Step 3
  $a = 6t - 12$ . $a = 0$ at $t = 2$ .
4. Step 4
  Slowing: $v$ and $a$ opposite signs. On $[1, 2]$ : $v < 0$ , $a < 0$ — speeding up. On $[2, 3]$ : $v < 0$ , $a > 0$ — slowing.
5. Step 5
  Also on $[0, 1]$ : $v > 0$ , $a < 0$ — slowing.
Answer
At rest at $t = 1, 3$ . Slowing on $[0, 1] \cup [2, 3]$ .
4Related rates — sliding ladder
Stretch• related rates
Question
A 5 m ladder slides down a wall. The bottom slides at $0.2$ m/s. Find the rate at which the top slides when the bottom is 3 m from the wall.
Step-by-step solution
1. Step 1
  Let $x$ = bottom distance, $y$ = top height. $x^2 + y^2 = 25$ .
2. Step 2
  Differentiate: $2x dx/dt + 2y dy/dt = 0 \Rightarrow dy/dt = -(x/y)dx/dt$ .
3. Step 3
  At $x = 3$ : $y = 4$ . $dy/dt = -(3/4)(0.2) = -0.15$ m/s.
Answer
Top slides DOWN at $0.15$ m/s.

Key Definitions and Keywords — Differentiation 2

Definitions to memorise and the exact keywords mark schemes credit for differentiation 2 answers — sharpened from recent examiner reports for the 2026 IB DP Maths AA HL sitting.

Implicit differentiation
Examiner keyword
Differentiating an equation in $x$ and $y$ treating $y = y(x)$ ; uses chain rule on $y$ -terms.
Inflection point
Examiner keyword
Point where concavity changes; $f''$ changes sign (not merely $f''(x) = 0$ ).
Related rates
Examiner keyword
Quantities linked by an equation whose rates of change w.r.t. time are connected via the chain rule.

Common Mistakes and Misconceptions — Differentiation 2

The traps other students keep falling into on differentiation 2 questions — taken from recent IB DP Maths AA HL examiner reports and mark schemes — and how to avoid them.

✕ $d/dx(y^3) = 3y^2$ .
Why it happens
Treating $y$ as independent.
How to avoid it
Chain rule: $d/dx(y^3) = 3y^2 (dy/dx)$ .
✕Claiming inflection because $f''(x) = 0$ .
Why it happens
Skipping the sign-change test.
How to avoid it
$f'' = 0$ is necessary but not sufficient — also verify $f''$ CHANGES sign across the point.
✕Forgetting that 'sliding down' means $dy/dt < 0$ .
Why it happens
Ignoring direction.
How to avoid it
Always interpret the SIGN of the rate physically.

Past paper style quiz

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Differentiation 2 — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Differentiation 2

Short Study Notes in the form of Slides

Short Study Notes — Differentiation 2

Detailed Notes

What you’ll learn

How it’s examined

1Implicit derivative with mixed terms

2Parametric curve gradient

3Kinematics with sign analysis

4Related rates — sliding ladder

Implicit differentiation

Inflection point

Related rates

✕d/dx(y3)=3y2d/dx(y^3) = 3y^2d/dx(y3)=3y2.

✕Claiming inflection because f′′(x)=0f''(x) = 0f′′(x)=0.

✕Forgetting that 'sliding down' means dy/dt<0dy/dt < 0dy/dt<0.

Past paper style quiz

Differentiation 2 — frequently asked questions

✕ $d/dx(y^3) = 3y^2$ .

✕Claiming inflection because $f''(x) = 0$ .

✕Forgetting that 'sliding down' means $dy/dt < 0$ .