1.7 Differentiation
- Syllabus
- 9709–2028–2029
- Topic
- 1.7
- Level
- AS
Learning objectives
1.7.1Gradient as limit• understand the gradient of a curve at a point as the limit of the gradients of a suitable sequence of chords, and use the notations f ′(x), f ″(x), d d x y, and d d x y 2 2 for first and second derivatives Only an informal understanding of the idea of a limit is expected. e.g. includes consideration of the gradient of the chord joining the points with x coordinates 2 and (2 + h) on the curve y = x3. Formal use of the general method of differentiation from first principles is not required.1.7.2Differentiation rules• use the derivative of xn (for any rational n), together with constant multiples, sums and differences of functions, and of composite functions using the chain rule e.g. find x y d d, given yx 253= +.1.7.3Differentiation applications• apply differentiation to gradients, tangents and normals, increasing and decreasing functions and rates of change Including connected rates of change, e.g. given the rate of increase of the radius of a circle, find the rate of increase of the area for a specific value of one of the variables.1.7.4Stationary points• locate stationary points and determine their nature, and use information about stationary points in sketching graphs. Including use of the second derivative for identifying maxima and minima; alternatives may be used in questions where no method is specified. Knowledge of points of inflexion is not included.
Let chord gradients approach the tangent gradient
At x=a, join (a,f(a)) to (a+h,f(a+h)). Its chord gradient is [f(a+h)−f(a)]/h. As non-zero h approaches 0, the chord approaches the tangent and its gradient approaches the derivative f′(a).
For $f(x)=x^3$ at $x=2$: rac{(2+h)^3-8}{h}=12+6h+h^2\longrightarrow12\quad ext{as }h o0.Thus the tangent gradient is $12$.
| Function notation | Leibniz notation | Meaning |
|---|---|---|
| f′(x) | dy/dx | first derivative: gradient/rate |
| f′′(x) | d2y/dx2 | second derivative: rate of change of the gradient |
This is an informal limiting picture. Do not substitute h=0 into the original fraction, and a formal general first-principles differentiation method is not required for Paper 1.
Differentiate Paper 1 powers and composites layer by layer
For rational $n$: rac d{dx}(x^n)=nx^{n-1},\qquad rac d{dx}[af(x)+bg(x)]=af'(x)+bg'(x).Workonintervalswheretheoriginalpowersaredefined.
For y=[g(x)]n, differentiate the outside power while keeping g(x) inside, then multiply by g′(x): dy/dx=n[g(x)]n−1g′(x).
If $y=(x^2+1)^3$, then rac{dy}{dx}=3(x^2+1)^2\cdot2x=6x(x^2+1)^2.
Rewrite roots and reciprocals as rational powers before differentiating, for example x=x1/2 and 1/x2=x−2. Simplify only when doing so preserves the function's domain.
Do not omit the inner derivative. Product and quotient rules are introduced in Pure Mathematics 3, so Paper 1 expressions are handled with powers, constant multiples, sums/differences and the chain rule.
Translate a derivative into the application the question asks for
| Application | Derivative decision |
|---|---|
| Tangent at x=a | m=f′(a), then y−f(a)=m(x−a) |
| Normal at x=a | gradient −1/f′(a) when $f'(a) |
| e0$ | |
| Increasing/decreasing | determine intervals where f′(x)>0 or f′(x)<0 |
| Rate of change | include units and restrict to the physical domain |
If $A=\pi r^2$ and $r$ changes with time, rac{dA}{dt}= rac{dA}{dr} rac{dr}{dt}=2\pi r rac{dr}{dt}.For $r=3$ cm and $dr/dt=0.4$ cm s$^{-1}$, $dA/dt=2.4\pi$ cm$^2$ s$^{-1}$.
Name the dependent variables, write the relation between them, differentiate with respect to the required variable (often time), substitute the specified instant only after differentiating, and state the signed result with units.
A normal gradient is the negative reciprocal, not merely the negative tangent gradient. For connected rates, match the derivative direction to the rate given and the rate required.
Locate, classify and place each stationary point on the graph
Solve f′(x)=0, substitute each solution into f to obtain full coordinates, then classify. If f′′(a)>0 the gradient is increasing through zero and the point is a local minimum; if f′′(a)<0 it is a local maximum.
| Sign of f′ around a | Nature |
|---|---|
| positive then negative | local maximum |
| negative then positive | local minimum |
| no sign change | neither of those classifications |
For $f(x)=x^3-3x$, $f'(x)=3(x^2-1)$ gives $x=\pm1$. Since $f''(x)=6x$, $(-1,2)$ is a local maximum and $(1,-2)$ is a local minimum.
Place the stationary coordinates, use their nature to set the local turning direction, then combine them with intercepts, domain and end behaviour. A local classification alone does not determine the whole graph.
f′(a)=0 identifies a candidate, not automatically a maximum or minimum. If f′′(a)=0, the second-derivative test is inconclusive; use the sign of f′ without introducing point-of-inflexion theory, which is excluded from Paper 1.