IB Math AI HL Differentiation — Complete Cheatsheet
Every differentiation skill for IB Mathematics Applications & Interpretation HL — the power rule, tangents & normals, stationary points, the second derivative and the AHL rules — built into real optimisation problems like maximum volume and minimum cost. Hand-built by an IBO-certified Singapore tutor, with a print-ready PDF to download.
Differentiation is the engine of the IB Mathematics Applications & Interpretation HL calculus strand. In AI you rarely differentiate for its own sake — you differentiate to answer a modelling question: how fast is a population growing, what cutting size gives the largest volume, what order quantity gives the lowest cost. A GDC is always allowed, so the marks are not in grinding algebra but in choosing the right tool (power rule, chain rule, second-derivative test) and reading the answer back into the context with the right units.
This cheatsheet condenses the whole range — from the power rule and tangents in SL 5.1–5.4 to the second derivative and the AHL rules in AHL 5.6–5.8 — onto one page, and flags the traps that quietly cost method marks. The print-ready PDF is at the bottom, free to download.
§1 — The derivative: gradient & rate of change SL 5.1
The derivative $f'(x)$ measures the instantaneous rate of change of $y=f(x)$ — geometrically, the gradient of the curve (the slope of its tangent) at each point. Two notations mean the same thing:
Gradient function:$f'(x)=\dfrac{dy}{dx}$ — the gradient of $y=f(x)$ at each value of $x$
Rate of change:$\dfrac{dy}{dx}$ — how fast $y$ changes per unit increase in $x$
Kinematics link:$v=\dfrac{ds}{dt}$, $\;a=\dfrac{dv}{dt}=\dfrac{d^{2}s}{dt^{2}}$ — differentiate displacement for velocity, again for acceleration
A positive derivative means $y$ is rising, a negative derivative means it is falling, and a zero derivative marks a level (stationary) instant — the single idea behind every method on this sheet.
NoteRate of change is where AI meets the real world: $\dfrac{dP}{dt}$ is a growth rate, $\dfrac{dC}{dx}$ a marginal cost, $\dfrac{ds}{dt}$ a velocity. See Kinematics for the motion version in full.
§2 — The power rule & polynomials SL 5.3
Almost every AI derivative starts here. For any power term, multiply by the power and then lower it by one:
Power rule:$\dfrac{d}{dx}\left(ax^{n}\right)=an\,x^{\,n-1}$
Constant:$\dfrac{d}{dx}(c)=0$ — a constant has zero gradient
Sum rule:differentiate a polynomial term by term
For example $\dfrac{d}{dx}\left(5x^{3}-2x^{2}+7\right)=15x^{2}-4x$, and negative or fractional powers obey the same rule: $\dfrac{d}{dx}\left(\dfrac{3}{x}\right)=\dfrac{d}{dx}\left(3x^{-1}\right)=-3x^{-2}=-\dfrac{3}{x^{2}}$.
TrickBefore differentiating, rewrite every root and fraction as a power: $\sqrt{x}=x^{1/2}$ and $\dfrac{1}{x^{2}}=x^{-2}$. Then the power rule applies mechanically.
TrapBring the power down first, then subtract one from the exponent. Reducing the power before multiplying, or forgetting that a lone constant differentiates to $0$, are the classic slips.
§3 — Numerical derivative on the GDC SL 5.1
Because AI always allows a calculator, you can evaluate a gradient without differentiating by hand — invaluable for messy models and for checking your working.
Value at a point:$\left.\dfrac{d}{dx}\big(f(x)\big)\right|_{x=p}=f'(p)$ — the calculator's numerical-derivative template
From a graph:plot $y=f(x)$, then use the analyse/derivative ($\tfrac{dy}{dx}$) tool at $x=p$
TrickTo find where a graph is steepest, or to confirm a turning point, graph the model and read $\dfrac{dy}{dx}$ directly — the GDC returns $f'(p)$ to full accuracy in one step.
TrapThe numerical tool returns the gradient's value at one point, not the function $f'(x)$. When a question asks for $f'(x)$ in terms of $x$, or wants an exact Paper 1 answer, you must differentiate by hand.
§4 — Tangents & normals SL 5.4
At the point where $x=p$, first find the gradient $m=f'(p)$ and the point $(p,f(p))$ — then both lines drop straight out.
Tangent gradient:$m=f'(p)$
Tangent line:$y-f(p)=m(x-p)$ — its $y$-intercept is the constant $c$ in $y=mx+c$
Normal gradient:$-\dfrac{1}{f'(p)}$ — the negative reciprocal (perpendicular to the tangent)
Normal line:$y-f(p)=-\dfrac{1}{f'(p)}(x-p)$
TrapThe normal gradient is the negative reciprocal $-\dfrac{1}{f'(p)}$, not simply $-f'(p)$. If the tangent is horizontal ($f'(p)=0$) the normal is the vertical line $x=p$.
NoteRead the wording carefully: it may want just the gradient, the full line equation, or only the $y$-intercept. Find $m$ and the point once, then answer exactly what is asked.
The sign of the derivative describes the shape of the graph, and the places where it vanishes are the turning points a model cares about most.
Increasing:$f'(x)>0$ — the curve rises
Decreasing:$f'(x)<0$ — the curve falls
Stationary:$f'(x)=0$ — solve this for the turning points; a quadratic vertex sits at $x=-\dfrac{b}{2a}$
At a turning point the tangent is horizontal, so $f'(x)=0$: a local maximum at $x_1$ and a local minimum at $x_2$. The curve rises where $f'(x)>0$ and falls where $f'(x)<0$.
TrickA local maximum is where $f'$ changes from $+$ to $-$; a local minimum is where it changes from $-$ to $+$. For a positive cubic the smaller root of $f'(x)=0$ is the maximum, the larger the minimum.
TrapOnce you have the stationary $x$, substitute it back into $f(x)$ — not $f'(x)$ — to get the height $y$ of the turning point.
§6 — The second derivative: nature & inflexion AHL 5.6
Differentiating twice gives $f''(x)$, which measures how the gradient itself is changing — the fastest way to classify a stationary point.
Maximum:$f'(x)=0$ and $f''(x)<0$ — concave down (a peak)
Minimum:$f'(x)=0$ and $f''(x)>0$ — concave up (a trough)
Inflexion:$f''(x)=0$ with a change of concavity — solve $f''(x)=0$, then use $f(x)$ for the height
TrapIf $f''(x)=0$ at a stationary point the test is inconclusive — fall back on the sign of $f'(x)$ on either side. And $f''(x)=0$ on its own is not a point of inflexion unless the concavity actually changes.
NoteConcave up where $f''(x)>0$, concave down where $f''(x)<0$. Running differentiation backwards is anti-differentiation — integrate $\dfrac{dy}{dx}$ and add $+C$, which a boundary condition pins down (see Integration).
Beyond polynomials, AI HL expects the derivatives of the key modelling functions. Learn these four cold:
Function $f(x)$
Derivative $f'(x)$
$e^{x}$
$e^{x}$ — unchanged
$\ln x$
$\dfrac{1}{x}$
$\sin x$
$\cos x$
$\cos x$
$-\sin x$
NoteCalculus of $\sin$ and $\cos$ assumes radians — set the GDC to radian mode before differentiating or graphing a trig model. The one sign to remember is that $\cos x$ differentiates to $-\sin x$.
TrickCombined with the chain rule in §8, these give exponential rates ($\dfrac{d}{dx}e^{kx}=k\,e^{kx}$) and the slopes of periodic models — exactly the derivatives AI uses for growth, decay and oscillation.
§8 — Chain, product & quotient rules AHL 5.8
These three rules differentiate the combinations AI models are built from. Spot the structure first, then apply the matching rule.
Chain rule:$\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$; e.g. $\dfrac{d}{dx}(ax+b)^{n}=n(ax+b)^{n-1}\cdot a$
TrapThe chain rule's most-dropped piece is the derivative of the inside: $\dfrac{d}{dx}(ax+b)^{n}$ carries a factor $a$. Multiply by the derivative of the inside every single time.
NoteIn the quotient rule the denominator is squared and the numerator is $u'v-uv'$ (order matters — keep the $u'v$ term first). The sign of the result tells you whether the model is increasing or decreasing at that point.
§9 — Optimisation in context AHL 5.7
This is where AI HL differentiation pays off: finding the input that makes a real quantity as large or as small as possible. The method is always the same five steps.
1 · Model:write the quantity as a function of one variable, using the constraint to eliminate the other
3 · Solve:set the derivative to $0$ and solve for the critical value
4 · Justify:confirm a maximum or minimum with the second derivative (or a sign check)
5 · Answer:substitute back to report the required area, volume or cost
Two AI staples: a wall on one side of a rectangular field means only three sides are fenced, so the constraint is $2x+y=F$ and you maximise the area $A=xy$; a cost of the form $C(x)=ax+\dfrac{b}{x}$ is minimised by solving $C'(x)=a-\dfrac{b}{x^{2}}=0$ (keeping $x>0$).
TrickGraph the model on the GDC and use the maximum/minimum feature to confirm your answer instantly — then present the calculus so you still collect the method marks.
TrapAnswer the actual question — the area, volume or cost — not just the value of $x$. Reject any root that makes a length zero or negative, and keep only values inside the model's domain.
§10 — Exam attack plan All sections
Question cue
What to do
Watch for
"Find the gradient at $x=p$"
Differentiate, then substitute $x=p$ (or use the GDC $\tfrac{dy}{dx}$ tool)
Differentiate first, substitute second
"Equation of the tangent"
$m=f'(p)$, point $(p,f(p))$, then $y-y_1=m(x-x_1)$
You need the point too; $y$-intercept is $c$
"Equation of the normal"
Gradient $-\dfrac{1}{f'(p)}$ through $(p,f(p))$
Negative reciprocal, not $-f'(p)$
"Increasing or decreasing?"
Test the sign of $f'(x)$
$f'(x)>0$ rises, $f'(x)<0$ falls
"Turning / stationary points"
Solve $f'(x)=0$, then $f''$ for the nature
Sub $x$ into $f$, not $f'$, for $y$
"Maximum / minimum in context"
One variable via the constraint, $f'=0$, justify, substitute back
Give area/volume/cost; reject bad roots
"Velocity / acceleration"
Differentiate $s(t)$ once for $v$, twice for $a$
Keep units; $a=0$ gives the time, then read $v$
"Composite / product / fraction"
Chain / product / quotient rule
Multiply by the inside's derivative; square the denominator
Worked Example — Maximising a Box Volume (AI HL Optimisation)
Question (AI HL Paper 2 style — 9 marks)
An open-topped box is made from a square sheet of card measuring $24\text{ cm}$ by $24\text{ cm}$. A square of side $x\text{ cm}$ is cut from each corner and the sides are folded up. (a) Show that the volume is $V(x)=4x^{3}-96x^{2}+576x$. (b) Find $\dfrac{dV}{dx}$. (c) Find the value of $x$ that maximises the volume, and justify that it is a maximum. (d) Hence find the maximum volume.
Solution
(a) The base is a square of side $24-2x$ and the height is $x$, so $V=x(24-2x)^{2}=x\left(576-96x+4x^{2}\right)=4x^{3}-96x^{2}+576x$. (M1)(A1)
(b) Differentiating term by term, $\dfrac{dV}{dx}=12x^{2}-192x+576$. (A1)
(c) Set $\dfrac{dV}{dx}=0$: $\;12x^{2}-192x+576=0\Rightarrow x^{2}-16x+48=0\Rightarrow (x-4)(x-12)=0$, so $x=4$ or $x=12$. (M1)(A1) A valid cut needs $0 < x < 12$ (at $x=12$ the base vanishes), so reject $x=12$ and take $x=4$. (R1) The second derivative is $\dfrac{d^{2}V}{dx^{2}}=24x-192$; at $x=4$ it is $24(4)-192=-96<0$, confirming a maximum. (M1)(A1)
Examiner's note: in AI you may instead graph $V(x)$ on the GDC and use the maximum feature to read $x=4,\ V=1024$ directly — but the working above earns the method marks, so show the calculus and quote the GDC value as the check. Always give the volume the question asks for (not just $x$), keep the units ($\text{cm}^{3}$), and reject any root that makes a length zero or negative.
Common Student Questions
Do I still have to differentiate by hand if my GDC can do it?
Usually yes. On Paper 2 the GDC's numerical derivative and graph-maximum tools are ideal for checking and for evaluating $f'(x)$ at a point, but the marks are still awarded for showing $f'(x)$ and setting it to zero. On Paper 1 (no GDC) you differentiate by hand throughout. Best practice: show the calculus, then quote the GDC value as confirmation.
What is the difference between a tangent and a normal?
Both pass through the same point $(p,f(p))$. The tangent has gradient $m=f'(p)$ and just grazes the curve; the normal is perpendicular to it, so its gradient is the negative reciprocal $-\dfrac{1}{f'(p)}$. A frequent slip is to negate the gradient ($-f'(p)$) instead of taking the negative reciprocal.
How do I tell whether a stationary point is a maximum or a minimum?
Find the second derivative and substitute the stationary $x$. If $f''(x)<0$ the curve is concave down — a maximum; if $f''(x)>0$ it is concave up — a minimum. If $f''(x)=0$ the test is inconclusive, so check the sign of $f'(x)$ just before and just after the point instead.
When do I use the chain rule versus the product or quotient rule?
Look at how the function is built. A composite (a function inside another, like $(3x+1)^{5}$ or $e^{2x}$) needs the chain rule — multiply by the derivative of the inside. Two functions multiplied together use the product rule $u'v+uv'$; a quotient (one function divided by another) uses $\dfrac{u'v-uv'}{v^{2}}$. Many exam expressions need two rules at once.
In an optimisation problem, how do I know which solution to keep?
Solving $f'(x)=0$ often gives more than one root. Reject any root that is physically impossible — a negative length, or a value that makes a dimension zero — and keep those inside the model's domain. Then confirm it is the maximum or minimum you want with the second-derivative test, and substitute back to report the area, volume or cost, not the value of $x$ itself.
What's NOT in this cheatsheet
This page gives you the formulas and traps for free. The full Photon Academy IB Math AI HL library (enrolled students, or the lifetime resource library) adds:
Notes PDF — every rule, from the power rule to the chain, product and quotient rules, with fully worked derivations and edge cases.
Tutorial booklet — 25+ AI HL-style questions from power-rule fluency to full optimisation and modelling applications.
Tutorial Solutions — mark-scheme-style solutions with M1/A1 annotations and GDC screenshots.
Predicted-paper questions — the exact differentiation and optimisation question types most likely in the next session.
