Year 1 Pure · Chapter 12

Differentiation

Specification: Edexcel Pure 7 — Differentiation · AQA section G

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In this chapter

1. The gradient function
2. Differentiation from first principles
3. The power rule
4. Interpreting derivatives
Practice

Key facts — this chapter

The derivative $\dfrac{dy}{dx}$ is the gradient of the curve at each point — the slope of its tangent there.
From first principles: $\dfrac{dy}{dx} = \displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ .
Power rule: if $y = x^n$ then $\dfrac{dy}{dx} = nx^{\,n-1}$ — multiply by the power, then drop the power by one.
Differentiate a polynomial term by term; constants differentiate to $0$ .
For negative or fractional powers, rewrite in index form first, then apply the rule.
The second derivative $\dfrac{d^2y}{dx^2}$ is the derivative of the derivative — the rate at which the gradient is changing.

Differentiation measures how steep a curve is at each point. A straight line has one gradient everywhere; a curve’s gradient changes as you move along it, and the derivative is the function that gives it to you.

1. The gradient function

At each point of a curve, the tangent has a gradient. Collect those gradients for every $x$ and you get a new function — the gradient function (or derivative), written $\dfrac{dy}{dx}$ or $f'(x)$ .

Key result

$\dfrac{dy}{dx}$ is the gradient of the curve at each $x$ . So:

where the curve is rising, $\dfrac{dy}{dx} > 0$ ;
where it is falling, $\dfrac{dy}{dx} < 0$ ;
at a turning point (a flat tangent), $\dfrac{dy}{dx} = 0$ .

A curve with tangents at three points above, and its gradient function (a parabola) below, with matching points — Read the slope of the tangent at each point and you trace out the gradient function below.

This is the picture to keep in your head: the derivative is just the original curve’s slope, plotted as its own graph.

2. Differentiation from first principles

Before the shortcut, here’s where the derivative comes from. Take two points on the curve a small distance $h$ apart and find the gradient of the chord joining them; then shrink $h$ to zero.

Key result

\frac{dy}{dx} = \lim_{h\to 0}\frac{f(x+h) - f(x)}{h}

The fraction is the gradient of the chord from $x$ to $x+h$ . As $h \to 0$ the chord swings round until it becomes the tangent, and its gradient becomes the derivative.

A curve with a chord between x and x plus h, with the rise f(x+h) minus f(x) over the run h — The chord's gradient is rise over run; let h → 0 and it becomes the tangent's.

On y = x², the chord from x = 1 to x = 1 + h has gradient (f(1+h) − f(1))/h. Shrink h and watch the chord swing onto the tangent — its gradient heading for 2.

the dashed line is the tangent at x = 1 (gradient 2) — the chord approaches it

Worked example 12.1 · Differentiating x² from first principles

Use first principles to differentiate $y = x^2$ .

Form the chord gradient with $f(x) = x^2$ :
$\frac{f(x+h) - f(x)}{h} = \frac{(x+h)^2 - x^2}{h}$
Expand and simplify the top:
$= \frac{x^2 + 2xh + h^2 - x^2}{h} = \frac{2xh + h^2}{h} = 2x + h$
Let $h \to 0$ :
$\frac{dy}{dx} = \lim_{h\to 0}(2x + h) = 2x$

3. The power rule

Doing that for every function would be exhausting — happily, a single rule handles all powers of $x$ .

Key result

y = x^n \;\;\Longrightarrow\;\; \frac{dy}{dx} = n\,x^{\,n-1}

Multiply by the power, then knock one off the power. For a sum, differentiate each term separately; a constant multiple stays, and a lone constant goes to $0$ (its graph is flat).

Worked example 12.2 · Differentiating a polynomial

Differentiate $y = 3x^4 - 2x^2 + 5x - 7$ .

Apply the rule term by term:

\frac{dy}{dx} = 3(4x^3) - 2(2x) + 5(1) - 0 = 12x^3 - 4x + 5

Worked example 12.3 · Negative and fractional powers

Differentiate $y = \dfrac{4}{x^2} + \sqrt{x}$ .

Rewrite in index form first:
$y = 4x^{-2} + x^{1/2}$
Apply the power rule to each:
$\frac{dy}{dx} = 4(-2)x^{-3} + \tfrac{1}{2}x^{-1/2} = -8x^{-3} + \tfrac{1}{2}x^{-1/2}$
(Optional) write back in surd/fraction form:
$\frac{dy}{dx} = -\frac{8}{x^3} + \frac{1}{2\sqrt{x}}$

Common mistake

You must rewrite roots and fractions as powers before differentiating. $\sqrt{x}$ is $x^{1/2}$ , and $\frac{1}{x^2}$ is $x^{-2}$ — you can’t differentiate them while they’re still dressed as a root or a fraction.

4. Interpreting derivatives

Once you have $\dfrac{dy}{dx}$ , substitute a value of $x$ to get the gradient at that point — and differentiate again for the second derivative.

Key result

$\dfrac{dy}{dx}$ evaluated at $x = a$ is the gradient of the curve at $x = a$ .
The second derivative $\dfrac{d^2y}{dx^2}$ (or $f''(x)$ ) is found by differentiating a second time. It measures how the gradient itself is changing.

A cubic with a maximum and a minimum, each marked with a flat tangent where the gradient is zero — Where the gradient is zero, the tangent is flat — a turning point.

Worked example 12.4 · Gradient at a point, and the second derivative

For $y = x^3 - 4x$ , find the gradient at $x = 2$ and the second derivative.

Differentiate:
$\frac{dy}{dx} = 3x^2 - 4$
The gradient at $x = 2$ :
$\left.\frac{dy}{dx}\right|_{x=2} = 3(2)^2 - 4 = 12 - 4 = 8$
Differentiate again for the second derivative:
$\frac{d^2y}{dx^2} = 6x$

Common mistake

A gradient is a number at a chosen point; the derivative is a function. “Find the gradient at $x = 2$ ” means differentiate first, then substitute $x = 2$ — never substitute into $y$ before differentiating.

Practice

Twelve questions in three tiers. Try each one properly before opening the solution — the solutions show full method, not just answers.

Warm-up

1 [3 marks]

Differentiate with respect to $x$ :

$y = x^5$
$y = 3x^2$
$y = 7$

Show solution

(a) $\dfrac{dy}{dx} = 5x^4$ .

(b) $\dfrac{dy}{dx} = 6x$ .

(c) A constant has zero gradient: $\dfrac{dy}{dx} = 0$ .

2 [3 marks]

Differentiate $y = 4x^3 - 5x^2 + 2x - 9$ .

Show solution

Term by term:

\frac{dy}{dx} = 12x^2 - 10x + 2

3 [2 marks]

Find the gradient of $y = x^2 - 3x$ at the point where $x = 4$ .

Show solution

Differentiate, then substitute:

\frac{dy}{dx} = 2x - 3 \;\;\Longrightarrow\;\; \left.\frac{dy}{dx}\right|_{x=4} = 8 - 3 = 5

4 [2 marks]

Given $y = 2x^4$ , find $\dfrac{d^2y}{dx^2}$ .

Show solution

\frac{dy}{dx} = 8x^3, \qquad \frac{d^2y}{dx^2} = 24x^2

Standard

5 [3 marks]

Differentiate $y = \dfrac{6}{x^3}$ .

Show solution

Rewrite as $y = 6x^{-3}$ , then apply the power rule:

\frac{dy}{dx} = 6(-3)x^{-4} = -18x^{-4} = -\frac{18}{x^4}

6 [3 marks]

Differentiate $y = 4\sqrt{x}$ .

Show solution

Rewrite as $y = 4x^{1/2}$ :

\frac{dy}{dx} = 4 \times \tfrac{1}{2}x^{-1/2} = 2x^{-1/2} = \frac{2}{\sqrt{x}}

7 [4 marks]

A curve has equation $y = x^3 - 6x^2 + 9x$ . Find the values of $x$ at which the gradient is zero.

Show solution

Differentiate:
$\frac{dy}{dx} = 3x^2 - 12x + 9$
Set equal to zero and solve:
$\begin{aligned} 3x^2 - 12x + 9 &= 0 \\[0.3em] x^2 - 4x + 3 &= 0 \\[0.3em] (x-1)(x-3) &= 0 \end{aligned}$

So the gradient is zero at $x = 1$ and $x = 3$ .

8 [4 marks]

Differentiate $y = x^2(2x - 3)$ .

Show solution

Expand first — the power rule needs separate powers of $x$ :

y = 2x^3 - 3x^2 \;\;\Longrightarrow\;\; \frac{dy}{dx} = 6x^2 - 6x

9 [5 marks]

Use differentiation from first principles to find the derivative of $y = x^2 + 3x$ .

Show solution

Form the chord gradient with $f(x) = x^2 + 3x$ :
$\frac{f(x+h) - f(x)}{h} = \frac{\big[(x+h)^2 + 3(x+h)\big] - \big[x^2 + 3x\big]}{h}$
Expand the top: $(x^2 + 2xh + h^2 + 3x + 3h) - (x^2 + 3x) = 2xh + h^2 + 3h$ .
Divide by $h$ and let $h \to 0$ :
$\frac{2xh + h^2 + 3h}{h} = 2x + h + 3 \;\;\xrightarrow{\;h\to 0\;}\;\; 2x + 3$

So $\dfrac{dy}{dx} = 2x + 3$ .

Exam style

10 [5 marks]

The curve $C$ has equation $y = \dfrac{x^3 + 2x}{\sqrt{x}}$ , for $x > 0$ .

Show that $y$ can be written as $x^{5/2} + 2x^{1/2}$ . [2]
Hence find $\dfrac{dy}{dx}$ . [3]

Show solution

(a) Divide each term by $\sqrt{x} = x^{1/2}$ :

y = \frac{x^3}{x^{1/2}} + \frac{2x}{x^{1/2}} = x^{3 - 1/2} + 2x^{1 - 1/2} = x^{5/2} + 2x^{1/2}

(b) Apply the power rule:

\frac{dy}{dx} = \tfrac{5}{2}x^{3/2} + 2 \times \tfrac{1}{2}x^{-1/2} = \tfrac{5}{2}x^{3/2} + x^{-1/2}

11 [5 marks]

The curve $y = x^3 - 3x^2 + 4$ passes through the point $P(1, 2)$ .

Find the gradient of the curve at $P$ . [3]
Find the value of $x$ , other than at $P$ , where the curve has the same gradient as at $P$ . [2]

Show solution

(a) $\dfrac{dy}{dx} = 3x^2 - 6x$ . At $x = 1$ : $3 - 6 = -3$ .

(b) Set the gradient equal to $-3$ :

3x^2 - 6x = -3 \;\;\Longrightarrow\;\; 3x^2 - 6x + 3 = 0 \;\;\Longrightarrow\;\; (x-1)^2 = 0

The only solution is $x = 1$ (a repeated root), so there is no other point with gradient $-3$ — the curve has that gradient only at $P$ .

12 [6 marks]

For the curve $y = 2x^3 - 9x^2 + 12x$ :

Find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ . [3]
Find the $x$ -values where $\dfrac{dy}{dx} = 0$ . [2]
Evaluate $\dfrac{d^2y}{dx^2}$ at each of those $x$ -values. [1]

Show solution

(a)

\frac{dy}{dx} = 6x^2 - 18x + 12, \qquad \frac{d^2y}{dx^2} = 12x - 18

(b) Solve $6x^2 - 18x + 12 = 0$ :

x^2 - 3x + 2 = 0 \;\;\Longrightarrow\;\; (x-1)(x-2) = 0 \;\;\Longrightarrow\;\; x = 1 \text{ or } x = 2

(c) At $x = 1$ : $\dfrac{d^2y}{dx^2} = 12 - 18 = -6$ . At $x = 2$ : $24 - 18 = 6$ .

(The negative value flags a maximum at $x = 1$ , the positive a minimum at $x = 2$ — a Chapter 13 idea.)