Differentiation (AI SL)
Differentiation gives you the gradient function of a curve - a new function that tells you the rate of change at every point. This topic covers finding that gradient function with the power rule, using its sign to decide where a function is increasing or decreasing, locating stationary points where the gradient is zero, and applying all of this to real optimisation problems like maximising area or minimising cost.
What the syllabus says
This topic maps onto six points in the official IB Applications & Interpretation syllabus, all within the Calculus unit.
| Code | Syllabus content |
|---|---|
| SL5.1 | Introduction to the concept of a limit, estimated from a table or graph. The derivative interpreted as a gradient function and as a rate of change, using the notation \(\dfrac{dy}{dx}\), \(f'(x)\), \(\dfrac{dV}{dt}\), etc. |
| SL5.2 | Increasing and decreasing functions. Graphical interpretation of \(f'(x)>0\), \(f'(x)=0\), \(f'(x)<0\). |
| SL5.3 | The derivative of \(f(x)=ax^n\) is \(f'(x)=anx^{n-1}\), for functions of the form \(f(x)=ax^n+bx^{n-1}+\cdots\) where all exponents are integers. |
| SL5.4 | Tangents and normals at a given point, and their equations, using both analytic approaches and technology. |
| SL5.6 | Values of \(x\) where the gradient of a curve is zero. Solution of \(f'(x)=0\). Local maximum and minimum points, with awareness that a local max/min is not necessarily the greatest/least value over the whole domain. |
| SL5.7 | Optimisation problems in context, for example maximising profit, minimising cost, or maximising volume for a given surface area. |
SL5.1-SL5.5 are shared content with Analysis & Approaches - the calculus techniques here are identical either way.
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 the gradient function of a curve - a new function, written \(f'(x)\) or \(\dfrac{dy}{dx}\), that gives the gradient at any point \(x\). It's found using the power rule and tells you the instantaneous rate of change at that point.
e.g. If \(f(x)=x^2\), then \(f'(x)=2x\), so at \(x=3\) the gradient is \(2(3)=6\).
What does it mean for a function to be increasing or decreasing?
A function is increasing where its gradient is positive (\(f'(x)>0\)) and decreasing where its gradient is negative (\(f'(x)<0\)). You find these intervals by differentiating and checking the sign of the derivative.
e.g. For \(f(x)=x^2-4x\), \(f'(x)=2x-4\); at \(x=1\), \(f'(1)=-2<0\), so \(f\) is decreasing there.
What is a stationary point?
A stationary point is a point on a curve where the gradient is exactly zero - \(f'(x)=0\). It's found by differentiating, setting the result equal to zero, and solving for \(x\). Stationary points include local maxima and minima.
e.g. \(f(x)=x^3-3x^2\) gives \(f'(x)=3x^2-6x=3x(x-2)=0\), so \(x=0\) or \(x=2\).
What is the power rule for differentiation?
The power rule says that to differentiate \(ax^n\), multiply by the exponent and reduce the exponent by one: \(\dfrac{d}{dx}(ax^n)=anx^{n-1}\). Apply it to each term of a polynomial separately.
e.g. \(\dfrac{d}{dx}(4x^3)=4\times3x^2=12x^2\).
What is optimisation?
Optimisation uses differentiation to find the largest or smallest value a function can take in a real context - maximum area, minimum cost, and so on. Set up a model, differentiate it, solve \(f'(x)=0\), and check the result makes sense.
e.g. For \(A=120x-2x^2\), \(\dfrac{dA}{dx}=120-4x=0\) gives \(x=30\), the value that maximises \(A\).
Key formulas
Three ideas cover almost every question on this topic. The tables below summarise them at a glance - the explanations underneath go into more depth on each one.
Formula reference
Differentiation at AI SL is built entirely from the power rule - none of it is listed as a separate formula in the booklet, since it's treated as a technique you apply rather than a value you look up.
| Formula | Used for | Booklet? |
|---|---|---|
| \(f'(x)=anx^{n-1}\) | Power rule for differentiation | Not in booklet - prior knowledge |
| \(f'(x)=0\) | Locate stationary points | Not in booklet - prior knowledge |
| \(y-y_1=m(x-x_1)\) | Equation of a tangent at a known point | Not in booklet - prior knowledge |
Local maximum vs local minimum
Both are stationary points where \(f'(x)=0\) - what tells them apart is how the gradient behaves either side of them.
| Feature | Local maximum | Local minimum |
|---|---|---|
| Sign of \(f'(x)\) just before | Positive | Negative |
| Sign of \(f'(x)\) just after | Negative | Positive |
| Shape of the curve | Rises then falls | Falls then rises |
| Example (\(y=x^3-3x^2\)) | \((0,0)\) | \((2,-4)\) |
The power rule
Differentiate a polynomial one term at a time - each term follows the same rule independently of the others.
Single term
\[\dfrac{d}{dx}(ax^n)=anx^{n-1}\]
Multiply by the exponent, then reduce the exponent by one.
Not in the formula booklet - prior knowledgeSum of terms
\[\dfrac{d}{dx}\big(ax^n+bx^m\big)=anx^{n-1}+bmx^{m-1}\]
Differentiate each term of a polynomial separately, then add the results.
Not in the formula booklet - prior knowledgeConstant term
\[\dfrac{d}{dx}(c)=0\]
A constant has zero gradient, so it disappears when you differentiate.
Not in the formula booklet - prior knowledgeIncreasing, decreasing, and stationary points
The sign of the derivative tells you everything about a curve's direction at a point, without needing to see the graph.
Increasing
\[f'(x)>0\]
The curve is rising - as \(x\) increases, so does \(y\).
Decreasing
\[f'(x)<0\]
The curve is falling - as \(x\) increases, \(y\) decreases.
Stationary
\[f'(x)=0\]
The curve is momentarily flat - a candidate for a local maximum or minimum.
Optimisation in context
Optimisation questions always follow the same three-step pattern, whatever the real-world dressing.
Set up the model
Write the quantity you're optimising (area, cost, volume) as a function of a single variable, using any given constraint to eliminate a second variable.
Differentiate and solve
Differentiate the model, set the derivative to zero, and solve for the variable.
Interpret the answer
Substitute back to find the optimised value, and check the result is sensible for the context (e.g. a length can't be negative).
Worked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
A tank drains so that the volume of water is \(V(t) = 200 - 5t^2\) litres, where \(t\) is in minutes.
(a) Find \(\dfrac{dV}{dt}.\)
(b) Find the rate of change of volume at \(t = 3\) and interpret its sign.
(c) Find the time at which the tank is empty.
Worked solution
(a) \(\dfrac{dV}{dt} = -10t.\) M1 A1
(b) At \(t = 3\): \(-10(3) = -30\) L/min. A1
The negative sign means the volume is decreasing - the tank is draining at 30 L/min. A1
(c) \(200-5t^2=0\Rightarrow t^2=40\Rightarrow t=\sqrt{40}\) M1
\(\approx6.32\) min. A1
For \(y = x^3 - 3x^2.\)
(a)(i) Find the \(x\)-coordinate of the stationary point with the smaller \(x\)-value.
(a)(ii) Find the \(x\)-coordinate of the stationary point with the larger \(x\)-value.
(b)(i) Find the coordinates of the stationary point with the smaller \(x\)-value.
(b)(ii) Find the coordinates of the stationary point with the larger \(x\)-value, and state which is the local maximum.
Worked solution
(a)(i) \(\dfrac{dy}{dx} = 3x^2 - 6x = 3x(x - 2) = 0.\) M1
\(x = 0.\) A1
(a)(ii) \(x = 2.\) A1
(b)(i) \(y(0) = 0\), giving \((0, 0)\). M1
\((0, 0).\) A1
(b)(ii) \(y(2) = 8 - 12 = -4\), giving \((2, -4)\). A1
The curve rises then falls at \(x=0\), so \((0, 0)\) is the local maximum (and \((2, -4)\) the local minimum). 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.
- Solving \(f(x)=0\) instead of \(f'(x)=0\). Finding roots gives you where the curve crosses the \(x\)-axis, not where it's stationary - always differentiate first when a question asks for a stationary point.
- Sign or exponent errors in the power rule. \(\dfrac{d}{dx}(2x^3)=6x^2\), not \(6x^3\) - the exponent must drop by one, not stay the same.
- Assuming any solution of \(f'(x)=0\) is the answer needed. A cubic can have two stationary points; check which one the context is actually asking about (maximum, minimum, or a specific \(x\)-value).
- Forgetting that a constant term differentiates to zero. \(\dfrac{d}{dx}(5)=0\), not \(5\) - constants have no gradient, so they vanish entirely.
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 does the derivative of a function tell you?
The derivative \(f'(x)\) gives the gradient of the curve \(y=f(x)\) at any point \(x\) - it's the instantaneous rate of change. A positive value means the function is increasing there; a negative value means it's decreasing.
How do I find a stationary point?
Differentiate the function, set \(f'(x)=0\), and solve for \(x\). Each solution is a stationary point - substitute it back into the original function to find its \(y\)-coordinate.
Is a GDC allowed for differentiation questions in AI SL?
Yes. Every AI SL paper permits a GDC, so you can graph a function and read off turning points directly, or use the numerical derivative feature to check your working.
What's the difference between a local maximum and the greatest value of a function?
A local maximum is only the highest point in its immediate neighbourhood - the function can be higher elsewhere in its domain. Always check the endpoints of a restricted domain as well as the stationary points.
Sub-topics
Differentiation broken down into its individual skills, each with its own focused page.
Related topics
More Calculus topics from the same AI SL syllabus unit, in case you want to keep going.