This topic is DfE content area G · Differentiation. Differentiation measures rate of change: how fast $y$ changes as $x$ changes. On a graph that is the gradient of the curve; in context it is a speed, a growth rate or a marginal cost.
Tangents and normals · increasing and decreasing functions
Stationary points, the second derivative, and optimisation
| Word | Meaning | Example |
|---|---|---|
| Derivative / gradient function | The function giving the gradient at each point. | $\mathrm{f}'(x) = 2x$ for $\mathrm{f}(x) = x^2$ |
| $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ | The rate of change of $y$ with respect to $x$. | $y = x^3 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2$ |
| Second derivative | The derivative of the derivative. | $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ or $\mathrm{f}''(x)$ |
| Chord | A straight line joining two points on a curve. | Used in first principles |
| Limit | The value approached as $h$ gets closer to $0$. | $\displaystyle\lim_{h \to 0}(2x + h) = 2x$ |
| Tangent | Line touching the curve with the same gradient. | Gradient $\mathrm{f}'(a)$ |
| Normal | Line perpendicular to the tangent at the point. | Gradient $-1/\mathrm{f}'(a)$ |
| Stationary point | Where the gradient is zero. | Maximum, minimum or inflection |
| Increasing | $y$ rises as $x$ rises: $\mathrm{f}'(x) > 0$. | $x^3 + x$ for all $x$ |
The curve $C$ has equation $y = x^3 - 3x^2 - 9x + 5$. Find the stationary points of $C$ and their nature, and the equation of the tangent where $C$ crosses the $y$-axis.
Content area G for Year 1, split into four pages.
- GDifferentiation from First PrinciplesThe gradient of a chord, the limit, and proofs for $x^2$, $x^3$ and other simple functions.
- GDifferentiating Powers of x$x^n$ for rational $n$, rewriting before differentiating, and the second derivative.
- GTangents and NormalsEquations of tangents and normals, and working backwards from a gradient.
- GStationary Points and OptimisationIncreasing and decreasing functions, maxima and minima, gradient graphs and optimisation.
Definition
$\displaystyle\lim_{h\to0}\dfrac{\mathrm{f}(x+h)-\mathrm{f}(x)}{h}$
Power rule
$kx^n \to knx^{n-1}$
Constants
$c \to 0$, $kx \to k$
Rewrite
$\sqrt x = x^{1/2}$, $\dfrac1{x^n} = x^{-n}$
Tangent
Gradient $\mathrm{f}'(a)$ through $(a, \mathrm{f}(a))$.
Normal
Gradient $-1/\mathrm{f}'(a)$.
Increasing
$\mathrm{f}'(x) > 0$; decreasing $\mathrm{f}'(x) < 0$.
Nature
$\mathrm{f}'' < 0$ max; $\mathrm{f}'' > 0$ min.
Optimise
One variable, differentiate, $= 0$, check.
Ten questions drawing on the whole of content area G for Year 1. Work each one through before opening the solution.
Differentiate $y = 3x^4 - 5x^2 + \dfrac{2}{x}$.
▶ Show solution
$y = 3x^4 - 5x^2 + 2x^{-1}$
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 12x^3 - 10x - 2x^{-2} = 12x^3 - 10x - \dfrac{2}{x^2}$
Prove from first principles that the derivative of $4x^2$ is $8x$.
▶ Show solution
$\dfrac{4(x+h)^2 - 4x^2}{h} = \dfrac{8xh + 4h^2}{h} = 8x + 4h$
$\mathrm{f}'(x) = \displaystyle\lim_{h\to0}(8x + 4h) = 8x$ $\blacksquare$
Find the gradient of $y = x\sqrt x$ at the point where $x = 4$.
▶ Show solution
$y = x^{3/2}$, so $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \tfrac32 x^{1/2}$.
At $x = 4$: $\tfrac32 \times 2 = 3$.
Given $\mathrm{f}(x) = (x - 2)\left(x^2 + 1\right)$, find $\mathrm{f}'(x)$ and $\mathrm{f}''(x)$.
▶ Show solution
Expand: $\mathrm{f}(x) = x^3 - 2x^2 + x - 2$.
$\mathrm{f}'(x) = 3x^2 - 4x + 1$; $\mathrm{f}''(x) = 6x - 4$.
Find the equation of the tangent to $y = x^2 - 2x$ at the point where $x = 3$.
▶ Show solution
Point $(3, 3)$. $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x - 2 = 4$.
$y - 3 = 4(x - 3) \Rightarrow y = 4x - 9$.
Find the equation of the normal to $y = \dfrac{8}{x}$ at $(2, 4)$.
▶ Show solution
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{8}{x^2} = -2$ at $x = 2$, so the normal gradient is $\dfrac12$.
$y - 4 = \dfrac12(x - 2) \Rightarrow y = \dfrac12 x + 3$.
Find and classify the stationary points of $y = x^3 - 3x + 2$.
▶ Show solution
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 3 = 0 \Rightarrow x = \pm 1$.
$x = 1$: $y = 0$. $x = -1$: $y = -1 + 3 + 2 = 4$.
$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6x$: at $x = 1$, $6 > 0$, minimum $(1, 0)$; at $x = -1$, $-6 < 0$, maximum $(-1, 4)$.
Show that $\mathrm{f}(x) = 2x^3 + x - 4$ is increasing for all values of $x$.
▶ Show solution
$\mathrm{f}'(x) = 6x^2 + 1$. Since $x^2 \geqslant 0$, $\mathrm{f}'(x) \geqslant 1 > 0$ for all $x$.
So $\mathrm{f}$ is increasing for all $x$.
Show that there is exactly one point on $y = x^3 - 6x^2$ where the gradient is $-12$, and find the equation of the tangent there.
▶ Show solution
$3x^2 - 12x = -12 \Rightarrow x^2 - 4x + 4 = 0 \Rightarrow (x - 2)^2 = 0$.
A repeated root, so exactly one point: $x = 2$, $y = 8 - 24 = -16$.
Tangent: $y + 16 = -12(x - 2) \Rightarrow y = -12x + 8$.
A closed cuboid box has a base measuring $x$ cm by $2x$ cm, and a volume of $36$ cm³.
(a) Show that the height is $\dfrac{18}{x^2}$ cm and that the surface area is $A = 4x^2 + \dfrac{108}{x}$. (b) Find the value of $x$ that minimises $A$, giving it exactly and to 3 s.f. (c) Show that this gives a minimum. (d) Find the minimum surface area.
▶ Show solution
(a) $2x^2h = 36 \Rightarrow h = \dfrac{18}{x^2}$.
$A = 2(x \cdot 2x) + 2(x \cdot h) + 2(2x \cdot h) = 4x^2 + 6xh = 4x^2 + 6x \cdot \dfrac{18}{x^2} = 4x^2 + \dfrac{108}{x}$ ✓
(b) $\dfrac{\mathrm{d}A}{\mathrm{d}x} = 8x - 108x^{-2} = 0 \Rightarrow x^3 = 13.5 \Rightarrow x = \sqrt[3]{13.5} \approx 2.38$.
(c) $\dfrac{\mathrm{d}^2A}{\mathrm{d}x^2} = 8 + 216x^{-3} = 8 + \dfrac{216}{13.5} = 24 > 0$: a minimum.
(d) $A = 4(2.381)^2 + \dfrac{108}{2.381} \approx 22.7 + 45.4 = 68.0$ cm² (3 s.f.)