Differentiation and the gradient of a curve

A straight line has one gradient everywhere, but a curve gets steeper and shallower as you move along it. Differentiation is the tool that gives the gradient of a curve at every point at once, as a formula called the derivative. Everything else in the Paper 1 calculus questions (tangents and normals, stationary points, rates of change, optimisation, and integration as the reverse process) is built on the rules in this note, so they have to be fast and error-free.

The gradient of a curve

Gradient at a point means gradient of the tangent

The tangent to a curve at a point $P$ is the straight line that touches the curve at $P$ and has the same direction as the curve there. The gradient of the curve at $P$ is defined to be the gradient of that tangent.

The difficulty is that a tangent passes through only one known point, and you need two points to calculate a gradient. The way round this is to approximate the tangent by chords.

Chords that close in on the tangent

Take the curve $y = x^3$ and the point $P(2, 8)$. Pick a second point $Q$ on the curve with $x$-coordinate $2 + h$, so $Q$ is $\big(2 + h,\ (2 + h)^3\big)$. The gradient of the chord $PQ$ is

$$\frac{(2 + h)^3 - 8}{(2 + h) - 2} = \frac{8 + 12h + 6h^2 + h^3 - 8}{h} = 12 + 6h + h^2$$

Now slide $Q$ towards $P$ by making $h$ smaller:

$h$	Gradient of chord $PQ$
$1$	$19$
$0.1$	$12.61$
$0.01$	$12.0601$
$0.001$	$12.006001$

As $h$ gets closer to $0$, the chord gradients get closer and closer to $12$. The chord turns into the tangent, so the gradient of the curve at $(2, 8)$ is $12$. In the language of limits, $12 + 6h + h^2 \to 12$ as $h \to 0$.

The graph shows $y = x^3$, the chord from $(2, 8)$ to $(3, 27)$ (the case $h = 1$, gradient $19$) and the tangent $y = 12x - 16$ at $(2, 8)$, gradient $12$. Smaller values of $h$ give chords that lie closer and closer to the tangent.

Only an informal idea of a limit is expected: you should be able to set up the chord gradient for a specific curve, simplify it, and say what it tends to as $h \to 0$. You are not asked to differentiate general functions from first principles.

The derivative and its notation

Repeat the chord argument at a general point $(x, x^2)$ on $y = x^2$, with $Q$ at $x + h$:

$$\frac{(x + h)^2 - x^2}{h} = \frac{2xh + h^2}{h} = 2x + h \;\to\; 2x \text{ as } h \to 0$$

So the gradient of $y = x^2$ at any point is $2x$. At $x = 3$ the gradient is $6$; at $x = -1$ it is $-2$. A formula like this, which gives the gradient at every point, is the gradient function or derivative.

$\dfrac{dy}{dx}$ is a single symbol, read "dee $y$ by dee $x$". It is not a fraction you can cancel, although in the chain rule it often behaves like one. The notation $\dfrac{d}{dx}(\ldots)$ means "the derivative of $(\ldots)$ with respect to $x$", so $\dfrac{d}{dx}(x^2) = 2x$.

The second derivative measures how fast the gradient itself is changing. You will use it to decide whether a stationary point is a maximum or a minimum.

The power rule

The same chord argument for $x^3$, $x^4$, and so on reveals a pattern:

$y$	$x$	$x^2$	$x^3$	$x^4$
$\dfrac{dy}{dx}$	$1$	$2x$	$3x^2$	$4x^3$

The power comes down in front, and the power drops by one. This works for every rational power, including negative powers and fractions.

For example, if $y = 4x^3 - 5x^2 + 7x - 2$ then $\dfrac{dy}{dx} = 12x^2 - 10x + 7$, and $\dfrac{d^2y}{dx^2} = 24x - 10$.

Rewrite as powers first

The power rule only works on terms of the form $kx^n$. Roots, reciprocals, products and quotients must be rewritten before you differentiate.

Written as	Rewrite as	Derivative
$\sqrt{x}$	$x^{\frac{1}{2}}$	$\tfrac{1}{2}x^{-\frac{1}{2}} = \dfrac{1}{2\sqrt{x}}$
$\dfrac{1}{x}$	$x^{-1}$	$-x^{-2} = -\dfrac{1}{x^2}$
$\dfrac{5}{x^3}$	$5x^{-3}$	$-15x^{-4}$
$\dfrac{3}{\sqrt{x}}$	$3x^{-\frac{1}{2}}$	$-\tfrac{3}{2}x^{-\frac{3}{2}}$
$\sqrt[3]{x^2}$	$x^{\frac{2}{3}}$	$\tfrac{2}{3}x^{-\frac{1}{3}}$
$\dfrac{1}{2x}$	$\tfrac{1}{2}x^{-1}$	$-\tfrac{1}{2}x^{-2}$
$(x + 2)(x - 5)$	$x^2 - 3x - 10$	$2x - 3$
$\dfrac{x^2 + 3}{x}$	$x + 3x^{-1}$	$1 - 3x^{-2}$

Two of these are traps worth naming. $\dfrac{1}{2x}$ is $\tfrac{1}{2}x^{-1}$, not $2x^{-1}$: only the $x$ is in the denominator's power. And a fraction with several terms on top is split into separate terms, each divided by the denominator. There is no product rule or quotient rule in Paper 1; you never need one, because every P1 product or quotient can be expanded or split.

Using the derivative

Once you have $\dfrac{dy}{dx}$ there are three things you can do with it.

• Find the gradient at a given point: substitute the $x$-coordinate into $\dfrac{dy}{dx}$.
• Find where the gradient has a given value: set $\dfrac{dy}{dx}$ equal to that value and solve for $x$, then substitute into the curve's equation (not the derivative) for $y$.
• Find unknown constants: if the curve contains letters such as $a$ and $b$, each piece of information (a point on the curve, a gradient at a point) gives one equation.

The curve $y = x^3 - 6x^2 + 5$ has gradient $-9$ at two points, $(1, 0)$ and $(3, -22)$, so it has two parallel tangents of gradient $-9$. A derivative that is a quadratic can take the same value twice, which is why "find the points where the gradient is $m$" usually has two answers. This is Example 4 below.

Practice questions

Answers

1. $\dfrac{dy}{dx} = 20x^3 - 6x + 8$.

2. $y = 3x^{-2} - 4x^{\frac{1}{2}} + 2$, so $\dfrac{dy}{dx} = -6x^{-3} - 2x^{-\frac{1}{2}} = -\dfrac{6}{x^3} - \dfrac{2}{\sqrt{x}}$.

3. Expand: $y = 2x^3 + 3x^2 - 2x - 3$, so $\dfrac{dy}{dx} = 6x^2 + 6x - 2$.

4. $f(x) = x^{\frac{3}{2}} + 4x^{-\frac{1}{2}}$, so $f'(x) = \tfrac{3}{2}x^{\frac{1}{2}} - 2x^{-\frac{3}{2}}$. Then $f'(4) = \tfrac{3}{2}(2) - \dfrac{2}{8} = 3 - \tfrac{1}{4} = \tfrac{11}{4}$.

5. $f(x) = x^3 - 6x^{-1}$, $f'(x) = 3x^2 + 6x^{-2}$, $f''(x) = 6x - 12x^{-3}$. So $f''(1) = 6 - 12 = -6$.

6. Gradient of $PQ = \dfrac{(1 + h)^3 - 1}{h} = \dfrac{3h + 3h^2 + h^3}{h} = 3 + 3h + h^2$. As $h \to 0$ this tends to $3$, so the gradient at $P$ is $3$ (and $\dfrac{dy}{dx} = 3x^2 = 3$ at $x = 1$ agrees).

7. $\dfrac{dy}{dx} = 6x^2 - 6x - 10 = 2$, so $6x^2 - 6x - 12 = 0$, $x^2 - x - 2 = 0$, $(x - 2)(x + 1) = 0$. At $x = 2$, $y = 16 - 12 - 20 + 1 = -15$; at $x = -1$, $y = -2 - 3 + 10 + 1 = 6$. Points $(2, -15)$ and $(-1, 6)$.

8. $\dfrac{dy}{dx} = 3ax^2 + b$. At $x = 1$: $3a + b = 1$. At $x = 2$: $12a + b = 19$. Subtracting, $9a = 18$, so $a = 2$ and $b = -5$. Then $\dfrac{d^2y}{dx^2} = 6ax = 12x$, which is $24$ at $x = 2$.

9. $\dfrac{dy}{dx} = 3x^2 + 2kx + 3$. This quadratic has positive $x^2$ coefficient, so it is positive for all $x$ exactly when it has no real roots: discriminant $(2k)^2 - 4(3)(3) < 0$, so $4k^2 < 36$, $k^2 < 9$, giving $-3 < k < 3$.