Mathematics 9709 · Pure Mathematics 1

Derivatives and the gradient function

Differentiation gives you a formula called the derivative. It tells you the gradient of a curve at any point.
Use it to measure how steep a curve is. A curve bends, so its steepness keeps changing.
Almost everything in this chapter is the same two steps: differentiate, then put a number in.

The power rule: $\dfrac{d}{dx}\big(x^n\big)=nx^{n-1}$ . Multiply by the power, then take one off it. A number in front stays in front. A plain number on its own differentiates to $0$ .

The rule only works on a power of x. So first turn roots and fractions into powers: $\sqrt{x}=x^{1/2}$ , $\tfrac{1}{x^2}=x^{-2}$ , $\tfrac{1}{\sqrt{x}}=x^{-1/2}$ . Doing this first stops most lost marks in this topic.

Where the rule comes from

Pick a point on the curve. Pick a second point a tiny step $h$ away. The line joining them is a chord. Its gradient is rise over run.
Now move the second point toward the first. The chord turns. Its gradient gets closer and closer to one value.
That value is the limit. It is the gradient of the curve at the point.

See it on $y=2x^2-3$ at $x=2$ . The chord to the point at $x=2+h$ has gradient

$\dfrac{\big[2(2+h)^2-3\big]-\big[2(2)^2-3\big]}{h}=\dfrac{8h+2h^2}{h}=8+2h$

As $h\to 0$ the $2h$ goes to zero. The gradient is exactly 8. This is differentiation from first principles. The power rule below is the short way to get the same answer.
The exam will not ask you to do this. But it does want you to know that the gradient at a point is this limit of chord gradients. (9709/12 Oct/Nov 2024 Q3)

The gradient at P is the limit of chord gradients: as Q moves toward P, the chord lines up with the gold tangent (slope 8).

Why the h goes away

After you expand the top, every term left on top has an $h$ in it. The parts with no $h$ cancel, because they are the same point twice. Now divide by $h$ . You are left with $8+2h$ . That is a fixed number plus a small leftover you can shrink to nothing. The derivative is what is left. The power rule does all this for you, so you never redo the algebra.

The power rule, term by term

That limit always gives the same rule:

$\dfrac{d}{dx}\big(x^n\big)=nx^{n-1}\qquad \dfrac{d}{dx}\big(kx^n\big)=knx^{n-1}$

Differentiate one term at a time, then add the results.
We write it two ways. $\tfrac{dy}{dx}$ (Leibniz) and $f'(x)$ (Lagrange) mean the same thing: the gradient function.
$f'(x)$ , said “f-prime”, just means “how steep is the curve here”.

Worked example

Warm-up: one term at a time

A single clean power. Multiply by the power, then subtract one from the power: $\dfrac{d}{dx}\big(x^5\big)=5x^{5-1}=5x^4$ .
A coefficient just rides along. Multiply the power out front: $\dfrac{d}{dx}\big(3x^2\big)=3\cdot 2x^{2-1}=6x$ .

Answer

\dfrac{d}{dx}(x^5)=5x^4,\qquad \dfrac{d}{dx}(3x^2)=6x

Worked example

Differentiate $y = 3x^4 - \dfrac{1}{2x^2} + \dfrac{4}{\sqrt{x}} + 5$

Rewrite every term as a power of $x$ before you touch it: $y=3x^4-\tfrac12 x^{-2}+4x^{-1/2}+5$ .
Apply $nx^{n-1}$ term by term; the $+5$ goes to $0$ : $\tfrac{dy}{dx}=12x^3-\tfrac12(-2)x^{-3}+4(-\tfrac12)x^{-3/2}$ .
Tidy the numbers and translate back to roots and fractions: $12x^3+x^{-3}-2x^{-3/2}$ .

Answer

\dfrac{dy}{dx}=12x^3+\dfrac{1}{x^3}-\dfrac{2}{\sqrt{x^3}}

Using it to find a gradient

The derivative is a function. To get the gradient at one point, put that point's $x$ -value in.
If the curve is brackets multiplied together, expand first. You can only differentiate powers, not brackets times brackets.

Worked example

Find the gradient of $y = x(2x-1)(x+3)$ at $(1, 4)$

Expand to a sum of powers: $y=x(2x^2+5x-3)=2x^3+5x^2-3x$ .
Differentiate term by term: $\tfrac{dy}{dx}=6x^2+10x-3$ .
Put $x=1$ in: $6+10-3=13$ .

Answer

\text{gradient}=13

Where the marks go

Common mistake

You cannot differentiate

\sqrt{x}

\tfrac1x

or brackets times brackets as they are. Turn them into powers first (or expand first). This is the most common slip in this topic.

Common mistake

Negative powers also go down by one:

\tfrac{d}{dx}(x^{-2})=-2x^{-3}

, not

-2x^{-1}

, because

-2-1=-3

. Fractions cause the same mistake:

\tfrac{1}{2}-1=-\tfrac12

Now you try

The curve $y=2x^2-3$ has points $A$ and $B$ . Their $x$ -coordinates are $2$ and $2+h$ . Find a simple expression for the gradient of chord $AB$ . Then work out the gradient of the curve at $A$ . (9709/12 Oct/Nov 2024 Q3)

Your turn— tap to reveal the worked answer (9709/12 Oct/Nov 2024 Q3)

Chord gradient $=\dfrac{2(2+h)^2-3-(2(2)^2-3)}{h}=\dfrac{8h+2h^2}{h}=8+2h$ .

As $B\to A$ , $h\to 0$ . So the chord gradient gets closer to the gradient of the curve:

Answer

\text{gradient at }A=8

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Derivatives and the gradient function

Differentiation gives you a formula called the derivative. It tells you the gradient of a curve at any point.
Use it to measure how steep a curve is. A curve bends, so its steepness keeps changing.
Almost everything in this chapter is the same two steps: differentiate, then put a number in.

The power rule: $\dfrac{d}{dx}\big(x^n\big)=nx^{n-1}$ . Multiply by the power, then take one off it. A number in front stays in front. A plain number on its own differentiates to $0$ .

Where the rule comes from

Pick a point on the curve. Pick a second point a tiny step $h$ away. The line joining them is a chord. Its gradient is rise over run.
Now move the second point toward the first. The chord turns. Its gradient gets closer and closer to one value.
That value is the limit. It is the gradient of the curve at the point.

See it on $y=2x^2-3$ at $x=2$ . The chord to the point at $x=2+h$ has gradient

$\dfrac{\big[2(2+h)^2-3\big]-\big[2(2)^2-3\big]}{h}=\dfrac{8h+2h^2}{h}=8+2h$

As $h\to 0$ the $2h$ goes to zero. The gradient is exactly 8. This is differentiation from first principles. The power rule below is the short way to get the same answer.
The exam will not ask you to do this. But it does want you to know that the gradient at a point is this limit of chord gradients. (9709/12 Oct/Nov 2024 Q3)

The gradient at P is the limit of chord gradients: as Q moves toward P, the chord lines up with the gold tangent (slope 8).

Why the h goes away

The power rule, term by term

That limit always gives the same rule:

$\dfrac{d}{dx}\big(x^n\big)=nx^{n-1}\qquad \dfrac{d}{dx}\big(kx^n\big)=knx^{n-1}$

Differentiate one term at a time, then add the results.
We write it two ways. $\tfrac{dy}{dx}$ (Leibniz) and $f'(x)$ (Lagrange) mean the same thing: the gradient function.
$f'(x)$ , said “f-prime”, just means “how steep is the curve here”.

Worked example

Warm-up: one term at a time

A single clean power. Multiply by the power, then subtract one from the power: $\dfrac{d}{dx}\big(x^5\big)=5x^{5-1}=5x^4$ .
A coefficient just rides along. Multiply the power out front: $\dfrac{d}{dx}\big(3x^2\big)=3\cdot 2x^{2-1}=6x$ .

Answer

\dfrac{d}{dx}(x^5)=5x^4,\qquad \dfrac{d}{dx}(3x^2)=6x

Worked example

Differentiate $y = 3x^4 - \dfrac{1}{2x^2} + \dfrac{4}{\sqrt{x}} + 5$

Rewrite every term as a power of $x$ before you touch it: $y=3x^4-\tfrac12 x^{-2}+4x^{-1/2}+5$ .
Apply $nx^{n-1}$ term by term; the $+5$ goes to $0$ : $\tfrac{dy}{dx}=12x^3-\tfrac12(-2)x^{-3}+4(-\tfrac12)x^{-3/2}$ .
Tidy the numbers and translate back to roots and fractions: $12x^3+x^{-3}-2x^{-3/2}$ .

Answer

\dfrac{dy}{dx}=12x^3+\dfrac{1}{x^3}-\dfrac{2}{\sqrt{x^3}}

Using it to find a gradient

The derivative is a function. To get the gradient at one point, put that point's $x$ -value in.
If the curve is brackets multiplied together, expand first. You can only differentiate powers, not brackets times brackets.

Worked example

Find the gradient of $y = x(2x-1)(x+3)$ at $(1, 4)$

Expand to a sum of powers: $y=x(2x^2+5x-3)=2x^3+5x^2-3x$ .
Differentiate term by term: $\tfrac{dy}{dx}=6x^2+10x-3$ .
Put $x=1$ in: $6+10-3=13$ .

Answer

\text{gradient}=13

Where the marks go

Common mistake

You cannot differentiate

\sqrt{x}

\tfrac1x

or brackets times brackets as they are. Turn them into powers first (or expand first). This is the most common slip in this topic.

Common mistake

Negative powers also go down by one:

\tfrac{d}{dx}(x^{-2})=-2x^{-3}

, not

-2x^{-1}

, because

-2-1=-3

. Fractions cause the same mistake:

\tfrac{1}{2}-1=-\tfrac12

Now you try

Your turn— tap to reveal the worked answer (9709/12 Oct/Nov 2024 Q3)

Chord gradient $=\dfrac{2(2+h)^2-3-(2(2)^2-3)}{h}=\dfrac{8h+2h^2}{h}=8+2h$ .

As $B\to A$ , $h\to 0$ . So the chord gradient gets closer to the gradient of the curve:

Answer

\text{gradient at }A=8