Differentiation (AA SL)
Differentiation gives you the gradient function of a curve - a formula that tells you the slope at any point, without having to draw the graph. This topic covers the power, chain, product and quotient rules, how to build the equation of a tangent or normal at a point, how to find and classify stationary points, and how derivatives model rates of change in kinematics.
What the syllabus says
This topic maps onto five points in the official IB Analysis & Approaches syllabus.
| Code | Syllabus content |
|---|---|
| SL5.2 | Increasing and decreasing functions. Graphical interpretation of \(f'(x)>0\), \(f'(x)=0\), \(f'(x)<0\). |
| SL5.3 | Derivative of \(f(x)=ax^n\) is \(f'(x)=anx^{n-1}\), \(n\in\mathbb{Z}\). The derivative of functions of the form \(f(x)=ax^n+bx^{n-1}+\dots\), where all exponents are integers. |
| SL5.4 | Tangents and normals at a given point, and their equations. Use of both analytic approaches and technology. |
| SL5.6 | Derivative of \(x^n\) (\(n\in\mathbb{Q}\)), \(\sin x\), \(\cos x\), \(e^x\) and \(\ln x\). Differentiation of a sum and a multiple of these functions. The chain rule for composite functions. The product and quotient rules. |
| SL5.8 | Local maximum and minimum points, tested using change of sign of \(f'(x)\) or the sign of \(f''(x)\). Optimisation. Points of inflexion with zero and non-zero gradients. |
Kinematics questions on this topic (velocity and acceleration as derivatives of displacement) draw on the related SL5.9 syllabus point.
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?
The derivative of a function is another function that gives the gradient of the curve at any point. It's written \(f'(x)\) or \(\dfrac{dy}{dx}\), and it comes from differentiating the original function using one of a small set of rules.
e.g. If \(f(x)=x^2\), then \(f'(x)=2x\), so the gradient at \(x=3\) is \(f'(3)=6\).
What is the power rule?
The power rule is the basic method for differentiating a term of the form \(ax^n\): multiply by the exponent, then reduce the exponent by 1. It's the starting point for almost every derivative you'll calculate.
e.g. \(\dfrac{d}{dx}(4x^3) = 3\times4x^{3-1} = 12x^2\).
What is a tangent line?
A tangent is the straight line that touches a curve at one point and has the same gradient as the curve there. You find it by differentiating to get the gradient, then using \(y-y_1=m(x-x_1)\) with the point of contact.
e.g. For \(y=x^2\) at \(x=2\): gradient \(=4\), point \((2,4)\), so the tangent is \(y=4x-4\).
What is a normal line?
A normal is the straight line perpendicular to the tangent at the same point on a curve. Its gradient is the negative reciprocal of the tangent's gradient, since perpendicular lines satisfy \(m_1 m_2 = -1\).
e.g. For \(y=x^2\) at \(x=2\), the tangent gradient is 4, so the normal gradient is \(-\tfrac14\), giving \(y=-\tfrac14x+4.5\).
What is a stationary point?
A stationary point is any point on a curve where the gradient is zero, so \(f'(x)=0\). It could be a local maximum, a local minimum, or a point of inflexion with zero gradient - the second derivative or a sign table tells you which.
e.g. For \(f(x)=x^2-4x+3\), \(f'(x)=2x-4=0\Rightarrow x=2\), and \(f(2)=-1\), a minimum at \((2,-1)\).
Key formulas
A handful of rules cover every differentiation question on this topic. The tables below summarise them at a glance - the explanations underneath go into more depth on each one.
Formula reference
The chain, product and quotient rules are on the official formula booklet; the power rule and the equation of a straight line are assumed prior knowledge and aren't listed separately.
| Formula | Used for | Booklet? |
|---|---|---|
| \(\dfrac{d}{dx}(ax^n) = anx^{n-1}\) | Power rule | Not in booklet |
| \(y=g(u), u=h(x) \Rightarrow \dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}\) | Chain rule | ✓ Yes |
| \(y=uv \Rightarrow \dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\) | Product rule | ✓ Yes |
| \(y=\dfrac{u}{v} \Rightarrow \dfrac{dy}{dx}=\dfrac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\) | Quotient rule | ✓ Yes |
| \(y-y_1=m(x-x_1)\) | Equation of tangent/normal | Not in booklet - prior knowledge |
Tangent vs normal
Both lines pass through the same point on the curve, but their gradients are related, not equal.
| Feature | Tangent | Normal |
|---|---|---|
| Direction relative to curve | Touches the curve, same slope | Perpendicular to the tangent |
| Gradient | \(m = f'(x_1)\) | \(m = -\dfrac{1}{f'(x_1)}\) |
| Undefined when | Never (if \(f'\) exists) | \(f'(x_1)=0\) (normal is vertical) |
| Example at \(x=2\) on \(y=x^2\) | \(y=4x-4\) | \(y=-\tfrac14x+4.5\) |
Rules for differentiating
Every derivative on this topic reduces to one (or a combination) of these four rules.
Power rule
\[\dfrac{d}{dx}(ax^n) = anx^{n-1}\]
Multiply by the exponent, then subtract 1 from it.
Not in the formula booklet - prior knowledgeChain rule
\[\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}\]
Differentiate the outer function, then multiply by the derivative of the inner function.
✓ In the formula bookletProduct rule
\[\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\]
For a product of two functions, differentiate one at a time and add the results.
✓ In the formula bookletQuotient rule
\[\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right)=\dfrac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\]
For a fraction of two functions - order matters in the numerator.
✓ In the formula bookletApplications of the derivative
Once you can differentiate, the derivative answers several different exam questions.
Tangents and normals
Evaluate \(f'(x)\) at the given point to get the gradient, then substitute into \(y-y_1=m(x-x_1)\). Use \(-1/m\) for the normal's gradient.
Stationary points
Solve \(f'(x)=0\), then use \(f''(x)\) (or a sign table for \(f'(x)\)) to classify each solution as a maximum, minimum or point of inflexion.
Kinematics
Velocity is the derivative of displacement, \(v=\dfrac{ds}{dt}\), and acceleration is the derivative of velocity, \(a=\dfrac{dv}{dt}\). "At rest" means \(v=0\).
Worked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
The function \(f\) is defined by \(f(x)=x^3-4x+1\).
(a) Find \(f'(x)\).
(b) Find the gradient of the graph at \(x=2\).
Worked solution
(a) \(f'(x)=3x^2-4.\) A1
(b) \(f'(2)=3(4)-4\) M1
\(=8.\) A1
A particle's displacement is \(s(t) = t^3 - 6t^2 + 9t\) metres, \(t \ge 0\).
(a) Find the velocity \(v(t)\).
(b)(i) Find the earlier time at which the particle is at rest.
(b)(ii) Find the later time.
(c) Find the acceleration when \(t = 4\).
Worked solution
(a) Velocity is the rate of change of displacement: \(v(t)=\dfrac{ds}{dt}=3t^2-12t+9.\) M1
\(v(t)=3t^2-12t+9.\) A1
(b) "At rest" means \(v=0\). \(3(t-1)(t-3)=0\Rightarrow t=1\) s and \(t=3\) s. M1
\(t=1\) s. A1 A1
(c) Acceleration is the rate of change of velocity: \(a(t)=\dfrac{dv}{dt}=6t-12.\) M1
At \(t=4\), \(a=12\) m/s². 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 inner derivative in the chain rule. Differentiating \(\sin(3x-1)\) as \(\cos(3x-1)\) instead of \(3\cos(3x-1)\) drops the multiplier from the inner function \(3x-1\).
- Confusing the normal's gradient with the tangent's. The normal uses the negative reciprocal \(-1/m\), not \(-m\) - a very easy sign-and-flip slip under exam pressure.
- Sign errors with negative or fractional exponents in the power rule. \(\dfrac{d}{dx}(x^{-2})=-2x^{-3}\), not \(2x^{-3}\) - the exponent's own sign carries through to the new coefficient.
- Writing a tangent or normal equation without finding the \(y\)-coordinate first. \(y-y_1=m(x-x_1)\) needs both coordinates of the point, not just the gradient - substitute into the original function to get \(y_1\).
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 a derivative and a tangent line?
The derivative is a function - it gives the gradient at any point on the curve. A tangent line is one specific straight line, built by evaluating the derivative at a single point and using that gradient with the point's coordinates.
How do I know whether a stationary point is a maximum or a minimum?
Check the sign of the second derivative: if \(f''(x)>0\) the point is a minimum, if \(f''(x)<0\) it's a maximum. If \(f''(x)=0\), use a sign table for \(f'(x)\) either side of the point instead.
Is differentiation examined without a calculator?
Yes - large parts of Paper 1 test differentiation by hand, including the power, chain, product and quotient rules. Paper 2 lets you use a GDC to check gradients and stationary points numerically.
Can my GDC find a derivative for me?
Yes - most GDCs have a numerical derivative function that gives the gradient of a function at a specific point instantly. It's useful for checking your algebra, but you still need the rules for general working. See the GDC guide for model-specific instructions.
Sub-topics
Differentiation broken down into its individual skills, each with its own focused page.
Related topics
More Calculus topics from the same AA SL syllabus unit, in case you want to keep going.