Q: What does H2 Maths Notes (JC 1-2): 5.1) Differentiation cover? A: Differentiation rules, implicit methods, and optimisation for H2 Maths Topic 5.1.
Before you revise Memorise derivative rules with the full statement (chain, product, quotient). Practise switching between coordinate, parametric, and implicit forms so you can differentiate anything the paper throws at you.
Core Derivative Rules
Power rule: dxdxn=nxn−1
for rational
n
.
Product rule: dxd[uv]=u′v+uv′.
Quotient rule: dxd(vu)=v2u′v−uv′.
Chain rule: dxdf(g(x))=f′(g(x))g′(x).
Exponential and logarithmic derivatives: dxdekx=kekx, dxdlnx=x1.
Trigonometric and Hyperbolic Functions
dxdsinx=cosx, dxdcosx=−sinx, dxdtanx=sec2x.
Inverses: dxdsin−1x=1−x21
Hyperbolic: dxdsinhx=coshx, dxdcoshx=sinhx
Implicit Differentiation
Differentiate both sides treating y as function of x and solve for dxdy.
Example -- Implicit derivative
Given x2+xy+y2=7.
Differentiate: 2x+y+xdxdy+2ydxdy=0.
Factor: (x+2y)dxdy=−(2x+y).
Hence dxdy=−x+2y2x+y
Parametric Differentiation
For x=f(t), y=g(t): dxdy=dtdxdtdy.
Second derivative: dx2d2y=dtdx1dtd(dxdy)
Example -- Parametric slope
Given x=t2+1, y=ln(1+t).
dtdx=2tdtdy=1+t1.
dxdy=2t(1+t)1.
Tangents and Normals
Tangent gradient mt=dxdy at point; normal gradient mn=−mt1.
Tangent line: y−y0=mt(x−x0)
Normal line: y−y0=mn(x−x0)
Example -- Tangent to curve
Find tangent at x=1 for y=xe−x.
dxdy=e−x−xe−x=e−x(1−x).
At x=1, gradient zero → tangent horizontal: y=e1.
Optimisation and Stationary Points
Stationary points satisfy dxdy=0; classify with second derivative test.
Second derivative: dx2d2y>0 (minimum), <0 (maximum). If dx2d2y(x0)=0, differentiate repeatedly until the first non-zero dxmdmy(x0) appears - an even-order derivative reveals a minimum (positive) or maximum (negative), while an odd-order derivative signals a point of inflection. Alternatively, chart the sign of dxdy on either side of x0 to confirm behaviour.
For optimisation, interpret result in context (length, area, cost).
Example -- Optimisation
Minimise surface area of open-top box with base square of side x and volume 108cm3.
Height h=x2108.
Surface area S=x2+4xh=x2+x432.
Differentiate: dxdS=2x−x2432=0⇒2x3=432
x=3216=6, h=3
Related Rates
Differentiating implicit relationships with respect to time: dtdz=dxdzdtdx.
Example -- Expanding circle
A circle radius r grows at dtdr=0.2cm⋅s−1. Find rate of change of area when r=5cm.
Area A=πr2.
dtdA=2πrdtdr=2π×5×0.2=2πcm2⋅s−1.
Calculator Workflow
The graphing calculator (GC) differentiation function evaluates derivatives numerically; record inputs (e.g. d/dx at specific points).
Use TABLE to evaluate derivative sign around stationary points for classification.
Store intermediate expressions to avoid algebra slips when differentiating complex functions.
Exam Watch Points
Present exact derivatives before substituting numerical values.
State the method used (implicit, parametric) and justify each step clearly.
Include units in related-rates answers.
For optimisation, verify solutions satisfy constraints (positive dimensions, etc.).
Practice Quiz
Test differentiation fluency, stationary point analysis, and related-rate workflows under exam-style pressure.
Quick Revision Checklist
Apply product, quotient, and chain rules without hesitation.
Differentiate implicit and parametric relations, reporting dxdy clearly.
Evaluate tangents, normals, and stationary points with proper classification.
Tackle optimisation and related-rate problems with a structured plan and unit-aware answers.
Appendix: Higher-Order Derivative Test (Proof Sketch)
Suppose f is m-times differentiable in a neighbourhood of x0 and that f′(x0)=f′′(x0)=⋯=f(m−1)(x0)=0 while f(m)(x0)=0. Taylor's theorem with remainder gives
f(x)=f(x0)+m!f(m)(x0)(x−x0)m+Rm(x),
where the remainder satisfies ∣Rm(x)∣≤K∣x−x0∣m+1 for some constant K when x is close to x0. For sufficiently small ∣x−x0∣, the dominant term is therefore m!f(m)(x0)(x−x0)m.
If m is even, (x−x0)m≥0 on both sides of x0. A positive coefficient f(m)(x0) forces f(x)≥f(x0) nearby (local minimum), while a negative coefficient gives f(x)≤f(x0) (local maximum).
If m is odd, the sign of (x−x0)m flips across x0
This argument justifies the higher-order derivative test and matches the sign-chart alternative described earlier.
