© boardworks ltd 2005 1 of 45 differentiation first principles

© Boardworks Ltd 20051 of 45

Differentiation

First Principles


Co

nte

nts


Rates of change

The gradient of the tangent at a point

The gradient of the tangent as a limit

Rates of change


Rates of change

This graph shows the distance that a car travels over a period of 5 seconds.

The gradient of the graph tells us the rate at which the distance changes with respect to time.

In other words, the gradient tells us the speed of the car.

change in distancegradient = =

change in time40

=5

8 m/s

The car in this example is travelling at a constant speed since the gradient is the same at every point on the graph.

time (s)

dist

ance

(m

)

0 5

40


Rates of change

In most situations, however, the speed will not be constant and the distance–time graph will be curved.

For example, this graph shows the distance–time graph as the car moves off from rest.

The speed of the car, and therefore the gradient, changes as you move along the curve.

To find the rate of change in speed we need to find the gradient of the curve.

The process of finding the rate at which one variable changes with respect to another is called differentiation.

time (s)

dist

ance

(m

)

0

In most situations this involves finding the gradient of a curve.


Co

nte

nts


Rates of change



Differentiation of polynomials

Second order derivatives

Tangents and normals

Examination-style questions



The gradient of a curve

The gradient of a curve at a point is given by the gradient of the tangent at that point.

The gradient of a curve at a point is given by the gradient of the tangent at that point.

Look at how the gradient changes as we move along a curve:


Finding the gradient of a curve at a point

Suppose we want to find the gradient of the curve y = x2 at the point (3, 9).

Use graph paper to carefully plot the curve y = x2 for –5 ≤ x ≤ 5. Use a ruler to draw the tangent at the point (3, 9) and find the gradient of the tangent.Find the gradients of the tangents at four other points and write down what you notice.

As you can see, it is quite difficult to do this accurately because we only know a single point that the line passes through.

To find the gradient of a line we need to know two points on the line (x1, y1) and (x2, y2). We can then find its gradient:

y yy

x x x

2 1

2 1

change in =

change in


Co

nte

nts


Rates of change



Differentiation of polynomials

Second order derivatives

Tangents and normals

Examination-style questions



Differentiation from first principles

Suppose we want to find the gradient of a curve at a point A.

We can add another point B on the line close to point A.

As point B moves closer to point A, the gradient of the chord AB gets closer to the gradient of the tangent at A.

δx represents a small change in x and δy represents a small change in y.



We can write the gradient of the chord AB as:

change in =

change in

y

x

y

x

As B gets closer to A, δx gets closer to 0 and gets closery

x

to the value of the gradient of the tangent at A.

δx can’t actually be equal to 0 because we would then have division by 0 and the gradient would then be undefined.

Instead we must consider the limit as δx tends to 0.

This means that δx becomes infinitesimal without actually becoming 0.



Let’s see how this works for the gradient of the tangent to the curve y = x2 at the point A(3, 9).



If A is the point (3, 9) on the curve y = x2 and B is another point close to (3, 9) on the curve, we can write the coordinates of B as (3 + δx, (3 + δx)2).

The gradient of chord AB is:2(3 ) 9

(3 ) 3

x

x

δx

δy

29 6 ( ) 9=

x x

x

26 ( )=

x x

x

(6 )=

x x

x

=y

x

= 6 + x

A(3, 9)

B(3 + δx, (3 + δx)2)



We write this as:

Let’s apply this method to a general point on the curve y = x2.

If we let the x-coordinate of a general point A on the curve y = x2 be x, then the y-coordinate will be x2.

At the limit where δx → 0, 6 + δx → 6.

0lim =x

y

x

0

lim 6 + =x

x

6

So the gradient of the tangent to the curve y = x2 at the point (3, 9) is 6.

So, A is the point (x, x2).

If B is another point close to A(x, x2) on the curve, we can write the coordinates of B as (x + δx, (x + δx)2).



The gradient of chord AB is:2 2( )

=( )

y x x x

x x x x

δx

δy

2 2 22 ( )=

x x x x x

x

22 ( )=

x x x

x

(2 )=

x x x

x

2

0So for = , lim = 2

x

yy x x

x

= 2 +x x

A(x, x2)

B(x + δx, (x + δx)2)


The gradient of y = x2


The gradient function

So the gradient of the tangent to the curve y = x2 at the general point (x, y) is 2x.

2x is often called the gradient function or the derived function of y = x2.

If the curve is written using function notation as y = f(x), then the derived function can be written as f ′(x).

So, if: f(x) = x2

This notation is useful if we want to find the gradient of f(x) at a particular point.

For example, the gradient of f(x) = x2 at the point (5, 25) is:

f ′(5) = 2 × 5 = 10

f ′(x) = 2xThen:


The gradient function of y = x2


Differentiating y = x3 from first principles

For the curve y = x3, we can consider the general point A (x, x3) and the point B (x + δx, (x + δx)3) a small distance away from A.

The gradient of chord AB is then:3 3( )

=( )

y x x x

x x x x

3 2 2 3 33 3 ( ) ( )=

x x x x x x x

x

2 2 33 3 ( ) ( )=

x x x x x

x

2 2(3 3 ( ) )=

x x x x x

x

2 2= 3 3 ( )x x x x

3 2

0So for = , lim = 3

x

yy x x

x



So, the gradient of the tangent to the curve y = x3 at the general point (x, y) is 3x2.

So if, f(x) = x3

What is the gradient of the tangent to the curve f(x) = x3 at the point (–2, –8)?

f ′(x) = 3x2

f ′(–2) = 3(–2)2 = 12

Explain why the gradient of f(x) = x3 will always be positive.

f ′(x) = 3x2Then:


The gradient function of y = x3


Using the notation

We have shown that for y = x3

dydx

represents the derivative of y with respect to x.dy

dx

2

0lim = 3x

yx

x

0limx

y

x

is usually written as .dy

dx

So if y = x3

of the tangent.

Remember, is the gradient of a chord, while is the gradienty

x

dy

dx

2= 3dy

xdx

then:


Using the notation dydx

This notation can be adapted for other variables so, for example:

Also, if we want to differentiate 2x4 with respect to x, for example, we can write:

represents the derivative of s with respect to t.ds

dt

We could work this out by differentiating from first principles, but in practice this is unusual.

4(2 )d

xdx

If s is distance and t is time then we can interpret this as the rate of change in distance with respect to time. In other words, the speed.

© boardworks ltd 2005 1 of 45 differentiation first principles

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