Differentiation
1. Differentiation rules you must own
Differentiate algebraically first; only substitute a point after you have a simplified derivative.
Power rule
\[\frac{d}{dx}x^n=nx^{n-1}\]Exponential & trig
\[\frac d{dx}e^{ax}=ae^{ax},\quad \frac d{dx}\sin ax=a\cos ax,\quad \frac d{dx}\cos ax=-a\sin ax\]Product rule
\[(uv)\prime=u\prime v+uv\prime\]Quotient rule
\[\left(\frac uv\right)\prime=\frac{u\prime v-uv\prime}{v^2}\]Chain rule
\[\frac d{dx}f(g(x))=f\prime(g(x))g\prime(x)\]Chain + product
Differentiate \(y=x^2\sin 3x\): \[\frac{dy}{dx}=2x\sin3x+3x^2\cos3x.\]Exam tip. On product/quotient questions, define \(u\) and \(v\) in your working. It makes the rule visible and reduces sign mistakes.
2. Tangents, normals and stationary points
Tangent/normal equations are coordinate-geometry questions after the gradient is found.
Tangent gradient
\[m_t=f\prime(a)\]Normal gradient
\[m_n=-\frac1{f\prime(a)}\]Stationary condition
\[f\prime(x)=0\]Second derivative test
\[f\prime\prime(x)>0\Rightarrow\min,\qquad f\prime\prime(x)<0\Rightarrow\max\]Stationary point
For \(f(x)=x^3-6x^2+9x+2\), \[f\prime(x)=3x^2-12x+9=3(x-1)(x-3).\] Stationary x-values are 1 and 3. Then \[f\prime\prime(x)=6x-12,\] so \(x=1\) gives a maximum and \(x=3\) a minimum.Common mistake. If the question says “show that it is a maximum/minimum”, \(f\prime(x)=0\) is not enough. Use the second derivative or a sign-change argument.
3. Optimisation and proof
Optimisation is modelling + differentiation + interpretation.
- Draw/interpret the geometry and define the variable.
- Use the constraint to express the target quantity in one variable.
- Differentiate and set the derivative equal to zero.
- Solve for admissible stationary values.
- Justify max/min and state the requested quantity with units.
Generic box pattern
If a volume is \(V(x)=x(20-2x)(12-2x)\), first restrict the physical domain \(0<x<6\), then solve \(V\prime(x)=0\) and justify that the relevant stationary point lies inside the domain.Exam tip. State the physical domain before optimising. An algebraic stationary point outside the geometry is not a valid answer.
4. Rates of change
Related rates are chain-rule questions with a physical interpretation.
Chain of rates
\[\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}\]Inverse relation
\[\frac{dx}{dt}=\frac{dx}{dy}\frac{dy}{dt}\]Related rates
If \(A=\pi r^2\) and \(dr/dt=0.4\), then \[\frac{dA}{dt}=2\pi r\frac{dr}{dt}.\] At \(r=5\), \(dA/dt=4\pi\) square units per unit time.Common mistake. Carry units through rate questions. They help detect whether you have differentiated the right quantity.
5. Small changes, approximations and percentage change
Small-change questions are tangent-line approximations in disguise.
Linear approximation
\[\delta y\approx\frac{dy}{dx}\,\delta x\]Percentage change
\[\%\Delta y\approx\frac{\delta y}{y}\times100\%\]Approximate change
For \(y=x^{1/2}\) at \(x=25\), \(dy/dx=1/(2\sqrt x)=1/10\). If \(x\) rises by \(0.2\), then \[\delta y\approx\frac1{10}(0.2)=0.02.\]Exam tip. This is a local approximation; use the derivative at the stated base value, not at the changed value.