The Derivative as a Limit

See the gradient of a curve as the limit of chord gradients, read dy/dx as a rate of change, and find second derivatives.

In the exam for Edexcel International AS Maths · 12 real past-paper questions below

From chord to tangent

You already find the gradient at a point with $\frac{d y}{d x}$ . But why does the power rule give the gradient of the tangent? You only know one point on the tangent, P, so you cannot do up ÷ across.

The trick: use a second point Q close to P. The straight line PQ is a chord. Then slide Q towards P and watch its gradient.

Example

On $y = x^{2}$ , take P(3, 9). Let Q have x coordinate $3 + h$ , where $h$ is a small step.

$h$	Q	Gradient of PQ
1	(4, 16)	$\frac{16 - 9}{1} = 7$
0.1	(3.1, 9.61)	$\frac{0.61}{0.1} = 6.1$
0.01	(3.01, 9.0601)	$\frac{0.0601}{0.01} = 6.01$
0.001	(3.001, 9.006001)	$\frac{0.006001}{0.001} = 6.001$

The chord gradients close in on 6. And $\frac{d y}{d x} = 2 x$ gives $2 \times 3 = 6$ at P. The tangent is where the chords are heading.

Key idea

In the 1630s the French lawyer Pierre de Fermat found tangents and maximum points with a similar trick: a tiny step he called E. A few decades later, Newton and Leibniz turned ideas like this into calculus.

Key idea

The gradient of the tangent is the number the chord gradients get closer and closer to, as Q slides into P.

Chord gradient in terms of h

The exam question: P is on a curve, and Q has x coordinate $a + h$ . Find the gradient of PQ in terms of $h$ . Keep $h$ as a letter, so one answer covers every step size.

Worked example

1
The curve $y = 3 x^{2} - 2$ passes through P(2, 10). Q on the curve has x coordinate $2 + h$ . Find the gradient of PQ in terms of $h$ .
2
$y_{Q} = 3 (2 + h)^{2} - 2$ — Put $2 + h$ into the curve. Brackets round all of it.
3
$= 3 (4 + 4 h + h^{2}) - 2 = 12 + 12 h + 3 h^{2} - 2$ — Square the bracket first, then multiply by 3.
4
$y_{Q} = 10 + 12 h + 3 h^{2}$ — Tidy. The 10 is P's y value, so you are on track.
5
$\frac{( 10 + 12 h + 3 h ^{2} ) - 10}{( 2 + h ) - 2}$ — Gradient is up ÷ across: the change in y over the change in x.
6
$= \frac{12 h + 3 h ^{2}}{h}$ — The 10s cancel on top. The 2s cancel underneath.
7
$= 12 + 3 h$ — Divide each term on top by $h$ . This is the simplest form.

Check: put $h = 0.1$ . The gradient is 12.3, just above the tangent gradient $\frac{d y}{d x} = 6 x = 12$ at P.

Rule

Marks: one for $y_{Q}$ , one for up ÷ across with your $y_{Q}$ , one for the simplified answer. That last mark needs this algebra. If $12 + 3 h$ comes from calculus instead, you lose it.

Common mistake

Writing $(2 + h)^{2} = 4 + h^{2}$ . Square the whole bracket: $(2 + h)^{2} = 4 + 4 h + h^{2}$ .

Common mistake

Stopping at $\frac{12 h + 3 h ^{2}}{h}$ . “Simplest form” means divide by $h$ .

Key idea

Find $y_{Q}$ , do up ÷ across, and divide by $h$ . In your answer, the number with no $h$ is the tangent gradient.

Let h shrink to 0

As Q slides into P, the step $h$ shrinks towards 0. So $12 + 3 h$ gets closer and closer to 12, the gradient of the tangent.

We write $h \to 0$ , said “ $h$ tends to 0”. We never put $h = 0$ : then up ÷ across would be $\frac{0}{0}$ , which means nothing.

Formula

f^{'} (x) = h \to 0 l i m \frac{f ( x + h ) - f ( x )}{h}

$x$ — the x coordinate of P, where you want the gradient.
$h$ — the step across to Q. It shrinks towards 0 but never equals 0.
$lim$ — “limit”: the value the chord gradient gets closer and closer to.

Worked example

1
Use the limit to find the gradient of $y = x^{2}$ at any point.
2
$\frac{( x + h ) ^{2} - x ^{2}}{h}$ — The chord gradient from $x$ to $x + h$ .
3
$= \frac{x ^{2} + 2 x h + h ^{2} - x ^{2}}{h}$ — Expand the bracket.
4
$= \frac{2 x h + h ^{2}}{h} = 2 x + h$ — The $x^{2}$ cancels. Divide by $h$ .
5
$As h \to 0, 2 x + h \to 2 x$ — The $h$ shrinks away. So the gradient is $2 x$ , just as the power rule says.

Rule

“Explain the relationship” is worth 1 mark. Say both parts: as $h \to 0$ , the gradient of PQ tends to 12, which is the tangent gradient (usually found in part (a)).

Memory trick

Zoom in on a smooth curve and it looks straight. Shrink the chord, and it becomes the tangent.

Common mistake

Writing only “they are the same”. Mention $h \to 0$ (or “limit”), and link it to the tangent gradient.

Key idea

The derivative is the limit of chord gradients. That is why the power rule gives the gradient of the tangent.

dy/dx as a rate of change

A gradient is up ÷ across: how much y changes for each 1 that x moves. So $\frac{d y}{d x}$ is the rate of change of y with respect to x.

Formula

\frac{d y}{d x}

$y$ — the quantity that changes.
$x$ — what it changes with, often time. The units are “y units per x unit”.

Worked example

1
The depth of water in a tank is $D = 40 + 6 t - \frac{1}{2} t^{2}$ cm, $t$ hours after a valve opens. How fast is the depth changing at $t = 2$ and at $t = 8$ ?
2
$\frac{d D}{d t} = 6 - t$ — Differentiate: $6 t \to 6$ , $- \frac{1}{2} t^{2} \to - t$ , and $40 \to 0$ .
3
$t = 2 : 6 - 2 = 4$ — Positive, so the water is rising, at 4 cm per hour.
4
$t = 8 : 6 - 8 = - 2$ — Negative, so the water is falling, at 2 cm per hour.

Common mistake

Putting $t = 2$ into D. That gives the depth (50 cm), not how fast it is changing.

Key idea

Positive rate: growing. Negative rate: shrinking. Give the units as “per”, like cm per hour.

The second derivative

The gradient of a curve can change too. On $y = x^{3}$ the gradient is $3 x^{2}$ :

$x$	0	1	2	3
Gradient $3 x^{2}$	0	3	12	27

The curve gets steeper and steeper. How fast the gradient changes is found by differentiating again: $6 x$ .

Formula

f^{''} (x) = \frac{d ^{2} y}{d x ^{2}}

$f^{''} (x)$ — “f double dash of x”: differentiate $f (x)$ , then differentiate the answer.
$\frac{d ^{2} y}{d x ^{2}}$ — the same thing when the curve is written $y = \dots$

In the tank, $\frac{d D}{d t} = 6 - t$ , so $\frac{d ^{2} D}{d t ^{2}} = - 1$ . The rate of change drops by 1 cm per hour, every hour.

Worked example

1
$y = 2 x^{3} + 8 x - \frac{3}{x}$ . Find $\frac{d ^{2} y}{d x ^{2}}$ and its value when $x = 1$ .
2
$y = 2 x^{3} + 8 x^{\frac{1}{2}} - 3 x^{- 1}$ — Write every term as a power of $x$ first.
3
$\frac{d y}{d x} = 6 x^{2} + 4 x^{- \frac{1}{2}} + 3 x^{- 2}$ — Power rule. $\frac{1}{2} \times 8 = 4$ and $(- 1) \times (- 3) = + 3$ .
4
$\frac{d ^{2} y}{d x ^{2}} = 12 x - 2 x^{- \frac{3}{2}} - 6 x^{- 3}$ — Differentiate again. $- \frac{1}{2} \times 4 = - 2$ and $- 2 \times 3 = - 6$ .
5
$x = 1 : 12 - 2 - 6 = 4$ — Every power of 1 is 1.

Common mistake

Thinking $\frac{d ^{2} y}{d x ^{2}}$ means $(\frac{d y}{d x})^{2}$ . It means differentiate twice, not square.

Common mistake

Losing a minus sign on negative powers. $x^{- \frac{1}{2}}$ goes to $x^{- \frac{3}{2}}$ : taking 1 off makes it more negative.

Key idea

The second derivative is the rate of change of the gradient.

Solving with f″(x)

Two common questions: solve $f^{''} (x) =$ a number, or use $f^{''} (x) = 0$ at a point to find a missing constant. Both start the same way: differentiate twice.

Worked example

1
$f (x) = 4 x^{\frac{5}{2}} - 30 x + 1, x > 0$ . Solve $f^{''} (x) = 6$ .
2
$f^{'} (x) = 10 x^{\frac{3}{2}} - 30$ — $\frac{5}{2} \times 4 = 10$ . The $- 30 x$ gives $- 30$ .
3
$f^{''} (x) = 15 x^{\frac{1}{2}}$ — $\frac{3}{2} \times 10 = 15$ . The $- 30$ gives 0.
4
$15 x = 6$ — Set it equal to 6.
5
$x = \frac{2}{5}$ — Divide by 15 and simplify.
6
$x = \frac{4}{25}$ — Square both sides.

Worked example

1
$f^{'} (x) = a x + \frac{32}{x}$ , and $f^{''} (x) = 0$ when $x = 4$ . Find $a$ .
2
$f^{'} (x) = a x^{\frac{1}{2}} + 32 x^{- 1}$ — Powers of $x$ first. You are given $f^{'}$ , so differentiate only once more.
3
$f^{''} (x) = \frac{1}{2} a x^{- \frac{1}{2}} - 32 x^{- 2}$ — $a$ is just a number, so it stays in front.
4
$\frac{1}{2} a \times \frac{1}{2} - \frac{32}{16} = 0$ — Put in $x = 4$ : $4^{- \frac{1}{2}} = \frac{1}{2}$ and $4^{- 2} = \frac{1}{16}$ .
5
$\frac{a}{4} = 2, a = 8$ — Solve for $a$ .

Rule

These questions often say “show all stages of your working” or “solutions relying on calculator technology are not acceptable”. Then an answer with no differentiating shown scores 0. Show $f^{'}$ , then $f^{''}$ , then the equation.

Common mistake

Differentiating $f (x)$ twice when the question already gives $f^{'} (x)$ . Count from what you are given.

Key idea

Differentiate twice, set it equal to what you are told, and solve. A constant like $a$ just rides along.

Which method? Read the question

The wording tells you what to do.

You see	You do
Q has x coordinate $a + h$	Find $y_{Q}$ , then up ÷ across, then divide by $h$
“Explain the relationship” with part (a)	As $h \to 0$ , the chord gradient tends to the tangent gradient
“Rate of change”	Differentiate and put in the value; give units “per”
$f^{''} (x)$ or $\frac{d ^{2} y}{d x ^{2}}$	Differentiate twice
“Solve $f^{''} (x) = 6$ ”	Differentiate twice, set equal to 6, solve
$f^{''} (x) = 0$ at a point, constant unknown	Put the point into $f^{''} (x)$ , solve for the constant

Common mistake

Using calculus for the $h$ part. The question wants the chord gradient from the algebra.

Key idea

Chord, limit, tangent: that is what $\frac{d y}{d x}$ means. Differentiate again for how fast the gradient itself changes.

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