Free Cheatsheet · SL 5.1–5.8 · AHL 5.12–5.15

IB Math AA HL Differentiation — Complete Cheatsheet

Every rule, derivative, trick, and trap you need for IB Mathematics Analysis & Approaches HL Differentiation. Hand-built by an IBO-certified Singapore tutor with 15+ years of IB experience.

Topic: Differentiation (Calculus) Syllabus: SL 5.1–5.8, AHL 5.12–5.15 Read time: ~14 minutes Last updated: Apr 2026

Differentiation is the single most heavily examined topic in IB Mathematics Analysis & Approaches HL, appearing on every Paper 1, Paper 2, and Paper 3. The HL syllabus extends the SL content (basic rules, optimisation, kinematics) with four high-leverage tools — L'Hôpital's rule, implicit differentiation, related rates, and the extended derivative table — and these account for the majority of the marks lost by HL students.

This cheatsheet condenses every formula, trick, and trap from SL 5.1–5.8 and AHL 5.12–5.15 into one page you can revise from. If you want the printable PDF version, the full set of notes, the worked tutorials, and the marked-up solutions, scroll to the bottom for the download links and the gated full library.

§1 — Limits & First Principles SL 5.1, AHL 5.12

Limit definition

Limit:$\displaystyle \lim_{x\to a} f(x) = L$ exists iff left-hand limit $=$ right-hand limit.
First principles:$\displaystyle f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$   (IB: polynomials only).

Key limits to memorise

$\displaystyle \lim_{x\to 0}\frac{\sin x}{x} = 1$
$\displaystyle \lim_{x\to 0}\frac{e^x - 1}{x} = 1$
$\displaystyle \lim_{x\to 0}\frac{1 - \cos x}{x} = 0$
$\displaystyle \lim_{x\to \infty}\frac{x^n}{e^x} = 0 \;\; \text{for any }n$
TrickFor first principles with $f(x) = x^n$: expand $(x+h)^n$ using the binomial theorem, cancel $h$ from the numerator, then let $h\to 0$. Every term except the first-order $h$ term vanishes.
TrapNever substitute $h = 0$ before cancelling $h$ from the numerator — you get $\frac{0}{0}$ and the method fails. Always simplify first.

§2 — Basic Differentiation Rules SL 5.3, 5.6

Core rules

Power:$\dfrac{d}{dx}[x^n] = n x^{n-1}, \; n\in\mathbb{Q}$
Constant:$\dfrac{d}{dx}[c] = 0$
Sum / diff:$(f \pm g)' = f' \pm g'$
Scalar:$(cf)' = c f'$
Chain:$[f(g(x))]' = f'(g(x))\cdot g'(x)$
Product:$(uv)' = u'v + uv'$
Quotient:$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$

Standard derivatives (SL)

$f(x)$$f'(x)$
$e^x$$e^x$
$\ln x$$\dfrac{1}{x}$
$\sin x$$\cos x$
$\cos x$$-\sin x$
$\sin(ax+b)$$a\cos(ax+b)$
$e^{ax+b}$$a e^{ax+b}$
$\ln(ax)$$\dfrac{1}{x}$   (the $a$ cancels!)
TrickLinear composite shortcut: $[f(ax+b)]' = a\cdot f'(ax+b)$. Just multiply by the inner derivative $a$.
Trap$\sin^2 x = (\sin x)^2 \neq \sin(x^2)$. These differentiate completely differently: $(\sin x)^2 \to 2\sin x\cos x$; $\sin(x^2) \to 2x\cos(x^2)$.
TrapQuotient rule order: it is $u'v - uv'$, NOT $uv' - u'v$. Subtraction is not commutative.

§3 — Extended HL Derivatives AHL 5.15

Reciprocal trig (NOT in the formula booklet — memorise!)

$f(x)$$f'(x)$
$\tan x$$\sec^2 x$
$\sec x$$\sec x \tan x$
$\csc x$$-\csc x \cot x$
$\cot x$$-\csc^2 x$

Memory aid: all the co-functions ($\cos$, $\csc$, $\cot$) carry a minus sign in their derivative.

Exponential, log & inverse trig (in the formula booklet)

$f(x)$$f'(x)$
$a^x$$a^x \ln a$
$\log_a x$$\dfrac{1}{x \ln a}$
$\arcsin x$$\dfrac{1}{\sqrt{1 - x^2}}$
$\arccos x$$-\dfrac{1}{\sqrt{1 - x^2}}$
$\arctan x$$\dfrac{1}{1 + x^2}$
Trick$a^x = e^{x\ln a}$, so by chain rule $\frac{d}{dx}[a^x] = e^{x\ln a}\cdot \ln a = a^x \ln a$. Derive it on the spot if you forget.
Trap$\arcsin'$ and $\arccos'$ are negatives of each other. Don't mix them up under exam pressure.
Note$\arctan x$ has no square root in its derivative because it is defined for all $x\in\mathbb{R}$ — the denominator $1 + x^2$ is always positive.

§4 — Higher Derivatives AHL 5.12

$f''(x) = \dfrac{d^2 y}{dx^2}$ — second derivative. $f^{(n)}(x) = \dfrac{d^n y}{dx^n}$ — $n$-th derivative. Differentiate $f'(x)$ to get $f''(x)$; repeat for higher orders.

Concavity from $f''$: $f''(x) > 0$ → concave up (smile); $f''(x) < 0$ → concave down (frown); $f''(x) = 0$ AND sign change → point of inflexion.

Trap$f''(a) = 0$ does NOT automatically mean a point of inflexion. You must verify $f''$ changes sign at $x = a$. Classic counter-example: $f(x) = x^4$ has $f''(0) = 0$ but $(0,0)$ is a minimum, not an inflexion.

§5 — Implicit Differentiation AHL 5.14

Key rule: $\dfrac{d}{dx}[f(y)] = f'(y) \cdot \dfrac{dy}{dx}$.

4-step method

  1. Differentiate every term with respect to $x$.
  2. Apply the product rule to any $xy$-terms.
  3. Collect every $\dfrac{dy}{dx}$ on one side.
  4. Factorise and divide.

Essential examples

$\dfrac{d}{dx}[y^3] = 3y^2 \dfrac{dy}{dx}$
$\dfrac{d}{dx}[xy] = y + x\dfrac{dy}{dx}$
$\dfrac{d}{dx}[x^2 y] = 2xy + x^2 \dfrac{dy}{dx}$
$\dfrac{d}{dx}[\sin y] = \cos y \cdot \dfrac{dy}{dx}$
$\dfrac{d}{dx}[e^{xy}] = e^{xy}\!\left(y + x\dfrac{dy}{dx}\right)$
TrickThe gradient at a point on an implicit curve still requires substituting both $x$ AND $y$ coordinates — the formula for $\dfrac{dy}{dx}$ will contain both variables.
TrapThe most common mark lost: forgetting to multiply by $\dfrac{dy}{dx}$ when differentiating a $y$-term. $\dfrac{d}{dx}[y^2] = 2y\dfrac{dy}{dx}$, NOT just $2y$.

§6 — L'Hôpital's Rule AHL 5.13

Only for $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ indeterminate forms:

$$\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}$$

Differentiate numerator and denominator separately. Repeat if still indeterminate.

IB mark scheme requirements

  1. State the indeterminate form (R1).
  2. Write $\lim$ notation at every step.
  3. Show derivatives of numerator AND denominator.
  4. Final evaluated value (A1).

Omitting $\lim$ notation loses the final A1.

TrickIf the limit is still $\frac{0}{0}$ after one application, apply L'Hôpital again. In IB exams the rule is almost always needed twice.
TrapL'Hôpital cannot be used if the form is NOT indeterminate — e.g. $\frac{\sin x}{x+1}$ at $x=0$ gives $\frac{0}{1}=0$ directly. Applying L'Hôpital to a non-indeterminate form gives the wrong answer.

§7 — Stationary Points & Optimisation SL 5.8

Classification

$f'(a) = 0$$f''(a)$Type
$> 0$Local minimum
$< 0$Local maximum
$= 0$Use sign of $f'$ test

Inflexion: $f'' = 0$ and sign changes. Increasing: $f' > 0$. Decreasing: $f' < 0$.

Optimisation 7-step method

  1. Define your variables.
  2. Write the objective function.
  3. Apply the constraint to eliminate one variable.
  4. Differentiate; set $= 0$.
  5. Classify (max or min).
  6. Check the domain endpoints.
  7. Answer what was actually asked.
TrickFor optimisation, if the domain is bounded (e.g. $0 < x < 4$), always evaluate at the endpoints too — the global maximum may be there, not at the stationary point.

§8 — Related Rates of Change AHL 5.14

Chain rule for rates: $\dfrac{dy}{dt} = \dfrac{dy}{dx} \cdot \dfrac{dx}{dt}$.

Common formulae

  • Sphere: $V = \tfrac{4}{3}\pi r^3 \Rightarrow \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}$
  • Circle: $A = \pi r^2 \Rightarrow \dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}$
  • Equilateral triangle: $A = \tfrac{\sqrt{3}}{4} x^2 \Rightarrow \dfrac{dA}{dt} = \tfrac{\sqrt{3}}{2} x \dfrac{dx}{dt}$

5-step method

  1. Draw a diagram, label variables.
  2. Write the geometric formula.
  3. Differentiate both sides w.r.t. $t$.
  4. Find any unknown variable values first.
  5. Substitute rates and solve. Always state units.
TrapSubstituting numerical values BEFORE differentiating is the most common related-rates error. Differentiate the formula first, THEN substitute the specific values.

§9 — Kinematics SL 5.8

$v(t) = s'(t)$   and   $a(t) = v'(t) = s''(t)$.

$v > 0$Moving in positive direction
$v < 0$Moving in negative direction
$v = 0$Stationary (check sign change)
$a > 0, v > 0$  or  $a < 0, v < 0$Speeding up
$a > 0, v < 0$  or  $a < 0, v > 0$Slowing down

Displacement vs total distance

Displacement: $s(b) - s(a)$ (signed). Total distance: split at every $t$ where $v = 0$, then add $|\Delta s|$ for each segment. On the GDC (P2): nInt(abs(v(t)),t,a,b). Speed $= |v(t)|$.

Trap$|s(b) - s(a)|$ gives net displacement, NOT total distance. If the particle reverses direction, these are different. This is the single most dropped mark in HL kinematics.

§10 — Tangents & Normals SL 5.4

Tangent at $(a, f(a))$: gradient $= f'(a)$  →  $y - f(a) = f'(a)(x - a)$.

Normal at $(a, f(a))$: gradient $= -\dfrac{1}{f'(a)}$  →  $y - f(a) = -\dfrac{1}{f'(a)}(x - a)$.

Tangent to a curve — discriminant method: if line $y = mx + c$ is tangent to $y = f(x)$, set them equal and demand $\Delta = 0$ (exactly one solution).

TrickFor "find $m$ such that the line is tangent to the curve": substitute the line into the curve equation, collect into a quadratic, then set $b^2 - 4ac = 0$. This gives the gradient(s) directly without finding the contact point first.
TrapThe normal gradient is $-\dfrac{1}{f'(a)}$, NOT $-f'(a)$. Take the negative reciprocal — don't just negate.

§11 — GDC Skills (TI-Nspire CX II) Paper 2

TaskSyntax
$f'(a)$d(f(x),x)|x=a
$f''(x)$d(f(x),x,2)
Graph of $f'$f2(x)=d(f1(x),x)
Solve $f'(x) = k$solve(d(f(x),x)=k,x)
Total distancenInt(abs(v(t)),t,a,b)
Max / minMenu → Analyze Graph

Common syntax traps: sinxsin(x); e^xe^(x); 3x3*x; sin^2(x)(sin(x))^2; arctan(x)atan(x).

§12 — Exam Attack Plan All sections

When you see this in the question — reach for that:

Question triggerReach for
Polynomial, power, or rationalPower rule; rewrite roots/fractions as powers first
$f(g(x))$ — composite functionChain rule; identify outer and inner
Two functions multipliedProduct rule: $u'v + uv'$
$F(x, y) = c$, mixed equationImplicit differentiation
$\frac{0}{0}$ or $\frac{\infty}{\infty}$ limitState form (R1), then L'Hôpital
"Maximise / minimise"Write objective, eliminate variable, differentiate, classify
"Rate of change / increasing at"Related rates: chain rule for $t$
"Total distance"Split at $v = 0$; sum $|\Delta s|$ per segment
"Show that $\dfrac{dy}{dx} = \ldots$"AG question — every step must be shown explicitly
First principlesExpand $f(x+h)$, cancel $h$, take $\lim_{h\to 0}$

Worked Example — IB-Style Implicit Differentiation

Question (HL Paper 1 style — 6 marks)

The curve $C$ is defined by $x^3 + y^3 = 6xy$ (the Folium of Descartes). Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$, and hence find the equation of the tangent to $C$ at the point $(3, 3)$.

Solution

  1. Differentiate both sides with respect to $x$ — apply the product rule to $6xy$:
    $3x^2 + 3y^2 \dfrac{dy}{dx} = 6y + 6x\dfrac{dy}{dx}$  (M1)(A1)
  2. Collect $\dfrac{dy}{dx}$ on one side: $3y^2 \dfrac{dy}{dx} - 6x\dfrac{dy}{dx} = 6y - 3x^2$  (M1)
  3. Factorise and divide: $\dfrac{dy}{dx} = \dfrac{6y - 3x^2}{3y^2 - 6x} = \dfrac{2y - x^2}{y^2 - 2x}$  (A1)
  4. Substitute $(3, 3)$: gradient $= \dfrac{2(3) - 9}{9 - 6} = \dfrac{-3}{3} = -1$  (A1)
  5. Tangent line: $y - 3 = -1(x - 3) \Rightarrow y = -x + 6$  (A1)

Examiner's note: Forgetting to apply the product rule on $6xy$ (writing just $6y$) is the single most common error here. The product rule gives $6y + 6x\dfrac{dy}{dx}$ — both terms are required.

Common Student Questions

What is the difference between $f''(x) = 0$ and a point of inflexion?
$f''(a) = 0$ is necessary but not sufficient for an inflexion. You must also check that $f''$ changes sign across $x = a$. The classic counter-example is $f(x) = x^4$ at $x = 0$: $f''(0) = 0$, but the point is a minimum, not an inflexion. In every IB inflexion question you must show the sign-change test, not just compute $f'' = 0$.
When can I use L'Hôpital's rule in IB Math AA HL?
Only when the limit gives an indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$. State the form explicitly to score the R1 mark, then differentiate the numerator and denominator separately. If still indeterminate, apply L'Hôpital again — IB exam questions almost always require two applications. Do not apply L'Hôpital to a non-indeterminate form like $\frac{\sin x}{x + 1}$ at $x = 0$ — you will get the wrong answer and lose all the marks.
Why do I keep losing marks on related rates questions?
The single most common error is substituting numerical values before differentiating. The geometric formula must first be differentiated with respect to $t$, then the values are substituted. Substituting first turns variables into constants, killing the derivative. Always follow the order: diagram → formula → differentiate w.r.t. $t$ → substitute → state units.
How do I find total distance vs displacement in kinematics?
Displacement is signed: $s(b) - s(a)$. Total distance is always positive: split the interval at every $t$ where $v(t) = 0$, compute $|\Delta s|$ on each segment, and add them up. On Paper 2, use the GDC: nInt(abs(v(t)),t,a,b). Confusing the two is the single most dropped mark in HL kinematics — Photon Academy students drill this in the first month.
Do I need to memorise the reciprocal trig derivatives ($\sec$, $\csc$, $\cot$)?
Yes — they are not in the IB formula booklet. You must memorise all four: $\dfrac{d}{dx}[\tan x] = \sec^2 x$, $\dfrac{d}{dx}[\sec x] = \sec x \tan x$, $\dfrac{d}{dx}[\csc x] = -\csc x \cot x$, $\dfrac{d}{dx}[\cot x] = -\csc^2 x$. The memory aid: every co-function ($\cos$, $\csc$, $\cot$) carries a minus sign in its derivative.

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