Differentiation (AA HL)
Differentiation gives you the gradient function of a curve - a formula that tells you the slope at any point. This topic covers the rules for differentiating standard and composite functions, the second derivative and what it says about concavity, finding and classifying stationary points, and the HL extensions of implicit differentiation, related rates of change and optimisation problems.
What the syllabus says
This topic maps onto four points in the official IB Analysis & Approaches syllabus, spanning SL foundations and the HL extension.
| Code | Syllabus content |
|---|---|
| SL5.6 | Derivative of \(x^n\), \(\sin x\), \(\cos x\), \(e^x\) and \(\ln x\). Differentiation of a sum and a multiple of these functions. The chain rule for composite functions. The product and quotient rules. |
| SL5.7 | The second derivative, using both \(\dfrac{d^2y}{dx^2}\) and \(f''(x)\) notation. Graphical behaviour of functions, including the relationship between the graphs of \(f\), \(f'\) and \(f''\). |
| SL5.8 | Local maximum and minimum points, tested using the change of sign of the first derivative or the sign of the second derivative. Optimisation problems. Points of inflexion with zero and non-zero gradients. |
| AHL5.14 | Implicit differentiation. Related rates of change. Optimisation problems, including cases where the optimum solution is at an endpoint. |
SL5.6-5.8 are core AA content examinable at both levels; AHL5.14 is HL-only.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is a derivative?
A derivative is a function that gives the gradient of the original curve at any point. It's found using differentiation rules and written \(f'(x)\) or \(\dfrac{dy}{dx}\). Where the derivative is positive the curve is increasing, and where it's negative the curve is decreasing.
e.g. If \(f(x) = x^3\), then \(f'(x) = 3x^2\), so \(f'(2) = 3(2)^2 = 12\).
What is the chain rule?
The chain rule differentiates a composite function - a function inside another function - by multiplying the derivative of the outer function by the derivative of the inner one. It's essential whenever you can't expand the expression first.
e.g. For \(f(x) = (2x+1)^3\), \(f'(x) = 3(2x+1)^2 \cdot 2 = 6(2x+1)^2\), so \(f'(1) = 6(3)^2 = 54\).
What is implicit differentiation?
Implicit differentiation is used when \(y\) isn't isolated on one side of an equation. You differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\), so every term containing \(y\) picks up a \(\dfrac{dy}{dx}\) factor via the chain rule.
e.g. For \(xy = 6\): \(y + x\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{y}{x}\). At \((2,3)\), \(\dfrac{dy}{dx} = -\dfrac{3}{2}\).
What is a stationary point?
A stationary point is a point on a curve where the gradient is zero, found by solving \(f'(x) = 0\). It can be a local maximum, a local minimum, or a point of inflexion with a zero gradient - the second derivative tells you which.
e.g. For \(f(x) = x^2 - 4x + 5\): \(f'(x) = 2x - 4 = 0 \Rightarrow x = 2\), and \(f(2) = 1\), giving the stationary point \((2, 1)\).
What is a point of inflexion?
A point of inflexion is where a curve changes concavity - from concave up to concave down, or vice versa. It occurs where \(f''(x) = 0\) and the sign of \(f''(x)\) actually changes either side of that point.
e.g. For \(f(x) = x^3 - 3x^2\): \(f''(x) = 6x - 6 = 0 \Rightarrow x = 1\), and \(f(1) = -2\), giving the inflexion point \((1, -2)\).
Key formulas
A small set of rules covers almost every differentiation question at this level. The tables below summarise them at a glance - the explanations underneath go into more depth on each one.
Formula reference
The standard derivatives and the three combination rules are on the official formula booklet; deciding what a stationary point is once you've found it is a reasoning skill, not a formula.
| Formula | Used for | Booklet? |
|---|---|---|
| \(f(x)=x^n \Rightarrow f'(x)=nx^{n-1}\) | Power rule | ✓ Yes |
| \(\sin x \to \cos x,\ \cos x \to -\sin x,\ e^x \to e^x,\ \ln x \to \tfrac1x\) | Standard derivatives | ✓ Yes |
| \(\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}\) | Chain rule | ✓ Yes |
| \(y=uv \Rightarrow y'=u'v+uv'\) | Product rule | ✓ Yes |
| \(y=\dfrac{u}{v} \Rightarrow y'=\dfrac{u'v-uv'}{v^2}\) | Quotient rule | ✓ Yes |
| \(f''(x)>0 \Rightarrow\) minimum; \(f''(x)<0 \Rightarrow\) maximum | Second derivative test | Not in booklet - reasoning skill |
First derivative test vs second derivative test
Both methods classify a stationary point, but they check different things - and one of them also works when the second derivative is zero.
| Feature | First derivative test | Second derivative test |
|---|---|---|
| What you check | Sign of \(f'(x)\) either side of the point | Sign of \(f''(x)\) at the point |
| Maximum | \(f'\) changes \(+ \to -\) | \(f''(x) < 0\) |
| Minimum | \(f'\) changes \(- \to +\) | \(f''(x) > 0\) |
| When it's inconclusive | Never - always works | When \(f''(x) = 0\) |
Differentiation rules
These four rules cover every function you'll be asked to differentiate at this level, alone or in combination.
Power rule
\[f(x)=x^n \Rightarrow f'(x)=nx^{n-1}\]
Bring the exponent down as a multiplier, then reduce the exponent by one.
✓ In the formula bookletChain rule
\[\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}\]
Differentiate the outer function, leave the inner function alone, then multiply by the derivative of the inner function.
✓ In the formula bookletProduct rule
\[y=uv \Rightarrow y'=u'v+uv'\]
Differentiate each factor in turn, keeping the other factor unchanged, and add the two results.
✓ In the formula bookletQuotient rule
\[y=\dfrac{u}{v} \Rightarrow y'=\dfrac{u'v-uv'}{v^2}\]
"Low d-high minus high d-low, over low squared" - order matters because subtraction isn't commutative.
✓ In the formula bookletStationary points and concavity
Once you have \(f'(x)\) and \(f''(x)\), classifying any point on the curve is a matter of checking signs.
Local maximum / minimum
Solve \(f'(x)=0\), then use the sign of \(f''(x)\) or the sign change of \(f'(x)\) to classify each stationary point.
Not in the formula booklet - reasoning skillConcavity
\(f''(x)>0\) means the curve is concave up; \(f''(x)<0\) means concave down. This is exactly what the second derivative test relies on.
Not in the formula booklet - reasoning skillPoint of inflexion
Requires \(f''(x)=0\) and a genuine sign change in \(f''(x)\) either side - a zero second derivative alone isn't sufficient (e.g. \(y=x^4\) at \(x=0\)).
Not in the formula booklet - reasoning skillHL extensions
At HL you also apply differentiation to equations that aren't solved for \(y\), and to situations where a quantity changes over time.
Implicit differentiation
Differentiate both sides of an equation in \(x\) and \(y\) with respect to \(x\), applying the chain rule so every \(y\)-term gains a \(\dfrac{dy}{dx}\) factor, then rearrange to isolate \(\dfrac{dy}{dx}\).
Not in the formula booklet - techniqueRelated rates of change
Connect two changing quantities with an equation, differentiate both sides with respect to time \(t\), then substitute known values to find the unknown rate.
Not in the formula booklet - techniqueOptimisation
Write the quantity to optimise as a function of one variable using a constraint, differentiate, set equal to zero, and check the result is a genuine maximum or minimum (or that an endpoint isn't better).
Not in the formula booklet - techniqueWorked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
For \(f(x) = x^3 - 6x^2 + 9x.\)
(a) Find \(f''(x).\)
(b) Find the \(x\)-coordinate of the point of inflexion.
(c) State the intervals where the graph is concave up.
Worked solution
(a) \(f'(x) = 3x^2 - 12x + 9\), M1
\(f''(x) = 6x - 12.\) A1
(b) \(f''(x) = 0 \Rightarrow x = 2.\) A1
(c) \(f''(x) > 0\) for \(x > 2\): concave up on \((2, \infty).\) M1 A1
For the curve \(x^2 + y^2 = 25\):
(a) Find \(\dfrac{dy}{dx}\) by implicit differentiation.
(b)(i) Find the gradient of the tangent at \((3, 4).\)
(b)(ii) Find the value of \(c\), giving the equation of the tangent in the form \(y=mx+c.\)
Worked solution
(a) \(2x + 2y\tfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx}\) M1
\(= -\dfrac{x}{y}.\) A1
(b)(i) At \((3,4)\): gradient \(= -\tfrac34.\) M1 A1
(b)(ii) Tangent: \(y - 4 = -\tfrac34(x - 3) \Rightarrow y = -\tfrac34 x + \tfrac{25}{4},\) so \(c=\tfrac{25}{4}\) M1
\(=6.25.\) A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Forgetting the chain rule on composite functions. Differentiating \(\sin(3x)\) as \(\cos(3x)\) instead of \(3\cos(3x)\) drops the "multiply by the derivative of the inner function" step - always check for a function inside a function.
- Confusing \(\dfrac{d^2y}{dx^2}\) with \(\left(\dfrac{dy}{dx}\right)^2\). The second derivative means differentiate twice, not square the first derivative - these give completely different results.
- Assuming \(f'(x)=0\) always means a maximum or minimum. It could be a point of inflexion with a zero gradient, like \(y=x^3\) at \(x=0\) - always check the second derivative or the sign change either side.
- Dropping \(\dfrac{dy}{dx}\) in implicit differentiation. Every term containing \(y\) needs a \(\dfrac{dy}{dx}\) factor from the chain rule - forgetting it on even one term gives a completely wrong equation to rearrange.
Using your GDC
Every step below is a real button sequence, not a vague "use your calculator" hint - covering the TI-84 Plus, TI-Nspire, and Casio fx-9860/fx-CG50. Pick your model to filter down to just the steps that apply to you.
Gives a gradient instantly to check your differentiation or when a function is awkward.
- MATH → 8:nDeriv(, then enter nDeriv(f(x), x, a). Or graph and use 2nd → CALC → 6:dy/dx.TI-84
- menu → Calculus → Numerical Derivative at a Point.Nspire
- Run-Matrix → MATH (F4) → d/dx, then enter the function and the x-value.Casio
- Read off the gradient - useful for tangent slopes without algebra.
Tip: Handy for checking the gradient at a point or finding a tangent's slope.
See the full GDC guide for more calculator models and topics.
Ready to practise properly?
Differentiation questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
What's the difference between the first and second derivative?
The first derivative \(f'(x)\) gives the gradient of the curve - how \(y\) changes as \(x\) changes. The second derivative \(f''(x)\) gives the rate of change of the gradient itself, which tells you about concavity: whether the curve bends upward or downward.
How do I know if a stationary point is a maximum, minimum or inflexion?
Find where \(f'(x) = 0\), then check \(f''(x)\) at that point. If \(f''(x) > 0\) it's a local minimum, if \(f''(x) < 0\) it's a local maximum. If \(f''(x) = 0\), check whether the sign of \(f'(x)\) actually changes either side - if it does, it's a point of inflexion, not a turning point.
When do I need implicit differentiation?
Use implicit differentiation whenever \(y\) isn't isolated on one side of the equation, such as \(x^2 + y^2 = 25.\) Differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\), and remember every \(y\) term picks up a \(\dfrac{dy}{dx}\) factor from the chain rule.
Is differentiation from first principles examined?
Yes, but only for polynomials at HL, using the limit definition \(f'(x) = \lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}.\) In practice you'll almost always use the standard rules instead, but you should understand where they come from.
Sub-topics
Differentiation broken down into its individual skills, each with its own focused page.
Related topics
More Calculus topics from the same AA HL syllabus unit, in case you want to keep going.