differential calculus is concerned with the rate at which things change. for example, the speed of a...

Differential calculus is concerned with the rate at which things change.

For example, the speed of a car is the rate at which the distance it travels changes with time.

First we shall review the gradient of a straight line graph, which represents a rate of change.

Gradient of a straight line graphThe gradient of the line between two points (x1, y1) and (x2, y2) is

mxxyy

xy

12

12in changein change

where m is a fixed number called a constant.

A gradient can be thought of as the rate of change of y with respect to x.

Gradient of a curveA curve does not have a constant gradient. Its direction is continuously changing, so its gradient will continuously change too.

The gradient of a curve at any point on the curve is defined as being the gradient of the tangent to the curve at this point.

y = f(x)

, as we know only one point on the tangent and we require two points to calculate the gradient of a line.

A tangent is a straight line, which touches, but does not cut, the curve.

x

y

O

A

Tangent to the curve at A.

We cannot calculate the gradient of a tangent directly

Using geometry to approximate to a gradient

xO

y

A

Tangent to the curve at A.

B1

B2

B3

Look at this curve.

Look at the chords AB1, AB2, AB3, . . .

For points B1, B2, B3, . . . that are closer and closer to A the sequence of chords AB1, AB2, AB3, . . . move closer to becoming the tangent at A.

The gradients of the chords AB1, AB2, AB3, . . . move closer to becoming the gradient of the tangent at A.

A numerical approach to rates of changeHere is how the idea can be applied to a real example. Look at the section of the graph of y = x2 for 2 > x > 3. We want to find the gradient of the curve at A(2, 4).

A (2, 4)

B1 (3, 9)

B2 (2.5, 6.25)

B3 (2.1, 4.41)

B4 (2.001, 4.004001)

The gradient of the chord AB1 is

52349

Chord

xchanges

from

ychanges

from

gradient

AB1

AB2

AB3

AB4

AB5

2349

25.2425.6

21.2441.4

2001.24004001.4

2 to 3 4 to 9 = 5

2 to 2.5 4 to 6.25 = 4.5

2 to 2.1 4 to 4.41 = 4.1

2 to 2.001 4 to 4.004001 = 4.001

2 to 2.00001

4 to 4.0000400001

4.00001

Complete the table

As the points B1, B2, B3, . . . get closer and closer to A the gradient is getting closer to 4.

This suggests that the gradient of the curve y = x2 at the point (2, 4) is 4.

x

y

2

4

y = x2

It looks right to me.

Find the gradient of the chord joining the two points with x-coordinates 1 and 1.001 on the graph of y = x2. Make a guess about the gradient of the tangent at the point x = 1.

Example (1)

(1, )

(1.001, )

1001.11001.1 2

1.0012

1)

= 2.001

1001.11002001.1

001.0002001.0

The gradient of the chord is

I’d guess 2.

Find the gradient of the chord joining the two points with x-coordinates 8 and 8.0001 on the graph of y = x2. Make a guess about the gradient of the tangent at the point x = 8.

Example (2)

(8, )

(8.0001, )

The gradient of the chord is

80001.8640001.8 2

8.00012

64

= 16.0001

80001.86400160001.64

0001.000160001.0

I’d guess 16.

Let’s make a table of the results so far: x-coordinate gradient

1 2

2 4

3

4

5

6

7

8 16

I think I can sees a pattern but can I prove it?

(2, 4)

(2 + h, (2 + h)2)

I need to consider what happens when I increase x by a general increment. I will call it h.

h

I will call it ∆x.

Let y = x2 and let A be the point (2, 4)

x

y

A(2, 4)

y = x2

O

B(2 + h, (2 + h)2)

Let B be the point (2 + h, (2 + h)2)

Here we have increased x by a very small amount h. In the early days of calculus h was referred to as an infinitesimal.

Draw the chord AB.

Gradient of AB 224)2( 2

hh

22444 2

hhh

hhh 24

hhh )4(

= 4 + h

As h approaches zero, 4 + h approaches 4.

So the gradient of the curve at the point (2, 4) is 4.

Use a similar method to find the gradient of y = x2 at the points (i) (3, 9)(ii) (4, 16)

If h ≠ 0 we can cancel the h’s.

We can now add to our table: x-coordinate gradient

1 2

2 4

3

4

5

6

7

8 16

It looks like the gradient is simply 2x.

6

8

−4 −3 −2 −1 1 2 3 4

5

10

15

x

yLet’s check this result.

y = x2

−4 −3 −2 −1 1 2 3 4

5

10

15

x


y = x2

Gradient at (3, 9) = 6

−4 −3 −2 −1 1 2 3 4

5

10

15

x


y = x2


−4 −3 −2 −1 1 2 3 4

5

10

15

x


y = x2

Gradient at (–1, 1) = –2

−4 −3 −2 −1 1 2 3 4

5

10

15

x


y = x2


−4 −2 2 4

1

2

3

4

5

6

7

8

9

x

y

−4 −2 2 4

1

2

3

4

5

6

7

8

9

x

y

Another way of seeing what the gradient is at the point (2, 4) is to plot an accurate graph and ‘zoom in’.

ZOOM IN

y = x2

1 1.5 2 2.5 33

3.5

4

4.5

5

x

y

When we zoom in the curve starts to look like a straight line which makes it easy to estimate the gradient.

0.2

0.8

42.08.0Gradient

Using a similar approach to that in the previous slide show it is possible to find the gradient at any point (x, y) on the curve y = f(x).

At this point it is useful to introduce some “new” notation due to Leibniz.

The Greek letter ∆(delta) is used as an abbreviation for “the increase in”.

Thus the “increase in x” is written as ∆x, and the “increase in y” is written as ∆y.

P (x1, y1)

Q (x2, y2)

x

y

O

12 yy

12 xx

So when considering the gradient of a straight line, ∆x is the same as x2 – x1 and ∆y is the same as y2 – y1.

∆x

∆y

Gottfried Leibniz1646 - 1716

gradient

Note∆x is the same as h used on the previous slide show.

Gradient of the curve y = x2 at the point P(x, y)Suppose that the point Q(x + ∆x, y + ∆y) is very close to the point P on the curve.

The small change from P in the value of x is ∆x and the corresponding small change in the value of y is ∆y.

It is important to understand that ∆x is read as “delta x” and is a single symbol.

The gradient of the chord PQ is:

xO

y

P(x, y)

Q(x + ∆x, y + ∆y)

y = x2

x

y y∆x

∆y

The coordinates of P can also be written as (x, x2) and the coordinates of Q as [(x + ∆x), (x + ∆x)2].

So the gradient of the chord PQ can be written as:

= 2x + ∆x

So = 2x + ∆x

As ∆x gets smaller approaches a limit and we start to refer to it in theoretical terms.

This limit is the gradient of the tangent at P which is the gradient of the curve at P.

It is called the rate of change of y with respect to x at the point P.

x

yxy

x δ

δdd

0δlim

For the curve y = x2,

xxxy

xδ2

dd

0δlim

= 2x

This is the result we obtained previously.

This is denoted by or .xy

dd

Gradient of the curve y = f(x) at the point P(x, y)

xO

y

P(x, y)

Q(x + δx, y + δy)

y = f(x)

x

y yδx

δy

For any function y = f(x) the gradient of the chord PQ is:

xy

xxxyyy

δδ

)δ()δ(

The coordinates of P can also be written as (x, f(x)) and the coordinates of Q as [(x + δx), f(x + δx)].

So the gradient of the chord PQ can be written as:

xxxxxx

)δ(

)(f)δ(f

x

yxy

x δ

δdd

0δlim

xxx

xxx

x )δ(

)(f)δ(f

0lim

x

xxx

x δ

)(f)δ(f

0lim

It is defined by

In words we say:

The symbol is called the derivative or the differential coefficient of y with respect to x.xy

dd

x

xxxxy

x δ

)(f)δ(fdd

0δlim

“dee y by dee x is the limit of as δx tends to zero”xy

δδ

“tends to” is another way of saying “approaches”

DEFINITION OF THE DERIVATIVE OF A FUNCTIONDEFINITION OF THE DERIVATIVE OF A FUNCTION

Sometimes we write h instead of ∆x and so the derivative of f(x) can be written as

h

xhxxy

h

)(f)(fdd

0lim

DEFINITION OF THE DERIVATIVE OF A FUNCTIONDEFINITION OF THE DERIVATIVE OF A FUNCTION

If y = f(x) we can also use the notation:

xy

dd = f’ (x)

In this case f’ is often called the derived function of f. This is also called f-prime.

The procedure used to find from y is called differentiating y with respect to x.xy

dd

Find for the function y = x3.

Example (1)

In this case, f(x) = x3.

xy

dd

h

xhxxy

h

)(f)(fdd

0lim

h

xhx

h

33

0

)(lim

h

xhxhhxx

h

33223

0

33lim

h

hxhhx

h

322

0

33lim

22

033lim hxhx

h

= 3x2

f(x) f ‘ (x)

x2 2x

x3 3x2

Results so far

−4 −3 −2 −1 1 2 3 4

5

10

15

x

yy = x3


Example (2)

In this case, f(x) = x4.

xy

dd

h

xhxxy

h

)(f)(fdd

0lim

h

xhx

h

44

0

)(lim

h

xhxhhxhxx

h

4432234

0

464lim

3223

0464lim hxhhxx

h

= 4x3

Results so far

h

hxhhxhx

h

43223

0

464lim

f(x) f ‘(x)

x2 2x

x3 3x2

x4 4x3

Find for the function y = .

Example (3)

xy

dd

x1

In this case, f(x) = .x1

h

xhxxy

h

)(f)(fdd

0lim

h

xhxh

11

0lim

h

hxxhxx

h

)()(

0lim

h

hxxh

h

)(

0lim

)(0lim

hxxh

h

h

)(

1

0lim

hxxh

21x

So we now have the following results:

We also know that if y = x = x1 then )1or ( 1dd 0x

xy

and if y = 1 = x0 then )0or ( 0dd 1 x

xy

These results suggest that if y = xn then 1dd nnx

xy

It can be proven that this statement is true for all values of n.

n xn

–1

2 x2 2x

3 x3 3x2

4 x4 4x3

xy

dd

)or ( 1 1xx)or ( 1 2

2 x

x

Find for the function y = 1.

Example (1)

xy

dd

y = 1 can be thought of as y = x0.

1dd nnx

xyUsing

100dd x

xy

= 0

But of course we knew this already.

y = 11

y

xO

Find for the function y = x.

Example (2)

xy

dd

y = x can be thought of as y = x1.

1dd nnx

xyUsing

111dd x

xy

= 1

= x0

But of course we knew this already.

y = xy

xO


Example (3)

xy

dd

1dd nnx

xyUsing

122dd x

xy

= 2x


Example (4)

xy

dd

1dd nnx

xyUsing

133dd x

xy

= 3x2


Example (5)

xy

dd

1dd nnx

xyUsing

144dd x

xy

= 4x3


Example (6)

xy

dd

x1

1dd nnx

xyUsing

111dd x

xy

= –x–2

y = can be written as y = x–1.x1

21x


Example (7)

xy

dd

21x

1dd nnx

xyUsing

122dd x

xy

= –2x–3

y = can be written as y = x–2.21x

32x


Example (8)

xy

dd x

1dd nnx

xyUsing

x21

y = can be written as .x 21

xy

121

21

dd x

xy

21

21 x


Example (9)

xy

dd

x1

1dd nnx

xyUsing

y = can be written as .x1 2

1xy

121

21

dd x

xy

23

21 x

32

1

x

Most of the functions we meet are not powers of the single variable x but consist of a number of terms such as y = 2x2 + 3x + 5.

The following rules can be proven:

If you multiply a function by a constant, you multiply its derivative by the same constant.

If f(x) = ag(x), then f’ (x) = ag’ (x).

This is because multiplying a function by a constant a has the effect of stretching the function in the y direction by a scale factor a which increases the gradient by the factor a.

If you add two functions, then the derivative of the sum is the sum of the derivatives.

If f(x) = g(x) + h(x), then f’ (x) = g’ (x) + h’ (x).

However if you multiply two functions, then the derivative of the product is NOT the product of the derivatives.

Find for the function y = 3x2.

Example (1)

xy

dd

xxy 23

dd

= 6x

Find for the function y = 8x3.

Example (2)

xy

dd

238dd x

xy

= 24x2

Find for the function y = x4 + x3.

Example (3)

xy

dd

23 34dd xx

xy

Find for the function y = x2 + x + 1

Example (4)

xy

dd

012dd x

xy

= 2x + 1

Find the gradient of the curve y = 2x2 + 3x – 7 at the point (2, 7).

Example (5)

34dd x

xy

At the point (2, 7), 324dd

xy

So the gradient = 11

−4 −3 −2 −1 1 2 3 4

−5

5

10

15

x

y

2

23 4462

xxxxxy Find for the function

Example (6)

xy

dd

2

23 4462

xxxxxy

22

23 4462

xx

xxxx

xxx 4462 3

13 4462 21

xxx

22 436dd 2

1

xxx

xy

22 436

xxx

Find the coordinates of the points on the graph of y = 2x3 – 3x2 – 36x + 10 at which the gradient is zero.

Example (7)

Let f(x) = 2x3 – 3x2 – 36x + 10

−6 −4 −2 2 4 6

−50

50

x

y

Then f’ (x) = 6x2 – 6x – 36

The gradient is zero when f’ (x) = 0

That is when 6x2 – 6x – 36 = 0

This simplifies to x2 – x – 6 = 0

In factor form this is (x – 3)(x + 2) = 0

So x = –2 or x = 3Substituting these values into y = 2x3 – 3x2 – 36x + 10 to find the y-coordinates gives y = 54 and y = –71

The coordinates of the required points are therefore (–2, 54) and (3, –71)

x

y

O

A

TangentNormal

The tangent at the point A(a, f(a)) has gradient f’ (a). We can use the formula for the equation of a straight line, y – y1 = m(x – x1) to obtain the equation of the tangent at (a, f(a)).

The equation of the tangent to a curve at a point (a, f(a)) is

y – f(a) = f’(a)(x – a)

The normal to the curve at the point A is defined as being the straight line through A which is perpendicular to the tangent at A.

The gradient of the normal is as the product of the gradients of perpendicular lines is –1. )(f

1a

The equation of the normal to a curve at a point (a, f(a)) is

)()(f

1)(f axa

ay

12

Example (1)The curve C has equation y = 2x3 + 3x2 + 2. The point A with coordinates (1, 7) lies on C. Find the equation of the tangent to C at A, giving your answer in the form y = mx + c, where m and c are constants.

xy

dd

= 6x2 + 6x

At x = 1, the gradient of C =

Equation of the tangent at A is

y = 12x – 5

which simplifies to

y – 7 = 12(x – 1)

−3 −2 −1 1 2 3

−5

5

10

15

x

y

Example (2)

A B x

y

O

Cl

The diagram shows the curve C with the equation y = x3 + 3x2 – 4x and the straight line l.The curve C crosses the x-axis at the origin, O, and at the points A and B.

(a) Find the coordinates of A and B.

The line l is the tangent to C at O.

(b) Find an equation for l.

(c) Find the coordinates of the point where l intersects C again.

(a) y = x3 + 3x2 – 4x

So A is (–4, 0) and B is (1, 0)

(b) 463dd 2 xx

xy

4dd

xy

At O,

So an equation for l is y = –4x

(c) x3 + 3x2 – 4x = –4x

= x(x + 4)(x – 1) x3 + 3x2 = 0

x2(x + 3) = 0

So l intersects C again when x = –3 and y = 12

So the coordinates of the point of intersection are (–3, 12)

= x(x2 + 3x – 4)

Example (3)

For the curve C with equation y = x4 – 9x2 + 2,

(a) find .dd

xy

The point A, on the curve C, has x-coordinate 2. (b) Find an equation for the normal to C at A, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(a) y = x4 – 9x2 + 2

xy

dd = 4x3 – 18x

(b) When x = 2, y = –18.

So the coordinates of A are (2, –18).

When x = 2, gradient of C = 4 • 8 – 18 • 2 = –4

So the gradient of the normal = 41

Equation of the normal to C at A is

which simplifies to x – 4y – 74 = 0

)2(4118 xy

A curve C has equation y = 2x2 – 6x + 5. Example (4)

The point A, on the curve C, has x-coordinate 1.

(a) Find an equation for the normal to C at A, giving your answer in the form ax + by + c = 0, where a, b and c are integers. The normal at A cuts C again at the point B.(b) Find the coordinates of the point B.

When x = 1, y = 2 – 6 + 5 = 1

Gradient of the tangent to the curve at A = 4 – 6 = –2

which simplifies to x – 2y + 1 = 0

(a)xy

dd = 4x – 6

4x2 – 12x + 10 = x + 1

4x2 – 13x + 9 = 0

(x – 1)(4x – 9) = 0

Gradient of the normal to the curve at A = 21

Equation of the normal to the curve at A is )1(121 xy

(b) 2x2 – 6x + 5 = 21x

So x-coordinate of B = 49

y-coordinate of B = 813

differential calculus is concerned with the rate at which things change. for example, the speed of a...

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