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 graph
The gradient of the line between two points (x1, y1) and (x2, y2) is
change in y  y2  y1  m
change in x x2  x1
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 curve
A curve does not have a constant gradient. Its direction is continuously changing, so its
gradient will continuously change too.
y = f(x)
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.
A tangent is a straight line, which touches, but does not cut, the curve.
y
A
O
Tangent to
the curve
at A.
x
We cannot calculate the gradient of a tangent directly, as we know only one point
on the tangent and we require two points to calculate the gradient of a line.
Using geometry to approximate to a gradient
Look at this curve.
y
B1
B2
B3
Tangent to
the curve
at A.
A
O
x
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 change
Here 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).
x
changes
from
y
changes
from
gradient
AB1
2 to 3
4 to 9
94
=5
3 2
AB2
2 to 2.5
4 to 6.25
6.25  4
= 4.5
2.5  2
AB3
2 to 2.1
4 to 4.41
4.41  4
= 4.1
2.1  2
AB4
2 to 2.001
4 to 4.004001
AB5
2 to
2.00001
4 to
4.0000400001
B1 (3, 9)
Chord
B2 (2.5, 6.25)
B3 (2.1, 4.41)
B4 (2.001, 4.004001)
Complete
the table
A (2, 4)
The gradient of the chord AB1 is
94  5
3 2
4.004001  4
= 4.001
2.001  2
4.00001
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.
y = x2
y
It looks right to me.
4
2
x
Example (1)
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.
The gradient of the chord is
1.0012 1  1.0020011
1.0011
1.0011
 0.002001
0.001
(1.001, 1.0012)
= 2.001
I’d guess 2.
(1, 1))
Example (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.
The gradient of the chord is
8.00012  64  64.00160001 64
8.0001 8
8.00018
 0.00160001
0.0001
(8.0001, 8.00012 )
= 16.0001
I’d guess 16.
(8, 64)
Let’s make a table of the results so far:
I think I can sees a pattern
but can I prove it?
x-coordinate
gradient
1
2
2
4
3
4
5
6
7
8
16
I need to consider what happens
when I increase x by a general
increment. I will call it h.
I will call it ∆x.
(2 + h, (2 + h)2)
(2, 4)
h
Let y = x2 and let A be the point (2, 4)
Let B be the point (2 + h, (2 + h)2)
y
Here we have increased x by a very small
amount h. In the early days of calculus h
was referred to as an infinitesimal.
y = x2
B(2 + h, (2 + h)2)
Draw the chord AB.
A(2, 4)
2
2
(
2

h
)

4
4

4
h

h
4

Gradient of AB 
2 h2
2 h2
 4h  h
h
 h( 4  h)
h
=4+h
2
O
x
If h ≠ 0 we can
cancel the h’s.
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)
We can now add to our table:
It looks like the gradient is
simply 2x.
x-coordinate
gradient
1
2
2
4
3
6
4
8
5
6
7
8
16
Let’s check this result.
y
y = x2
15
10
5
x
−4
−3
−2
−1
1
2
3
4
Let’s check this result.
y
y = x2
15
10
Gradient at
(3, 9) = 6
5
x
−4
−3
−2
−1
1
2
3
4
Let’s check this result.
y
y = x2
15
10
Gradient at
(2, 4) = 4
5
x
−4
−3
−2
−1
1
2
3
4
Let’s check this result.
y
y = x2
15
10
Gradient at
(1, 1) = 2
5
x
−4
−3
−2
−1
1
2
3
4
Let’s check this result.
y
y = x2
15
10
Gradient at
(0, 0) = 0
5
x
−4
−3
−2
−1
1
2
3
4
Let’s check this result.
y
y = x2
15
10
Gradient at
(–1, 1) = –2
5
x
−4
−3
−2
−1
1
2
3
4
Let’s check this result.
y
y = x2
15
10
Gradient at
(–2, 4) = –4
5
x
−4
−3
−2
−1
1
2
3
4
Let’s check this result.
y
y = x2
15
10
Gradient at
(–3, 9) = –6
5
x
−4
−3
−2
−1
1
2
3
4
Another way of seeing what the gradient is at the point (2, 4) is to plot an accurate graph
and ‘zoom in’.
9
y
y = x2
8
7
6
ZOOM IN
5
4
3
2
1
x
−4
−2
2
4
5
y
4.5
When we zoom in the curve
starts to look like a straight
line which makes it easy to
estimate the gradient.
0.8
4
Gradient  0.8  4
0.2
0.2
3.5
x
3
1
1.5
2
2.5
3
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”.
Leibniz
Thus the “increase in x” is written as ∆x, and the “increase in y” is written as ∆y.Gottfried
1646 - 1716
So when considering the gradient of a straight line, ∆x is the same as x2 – x1 and
∆y is the same as y2 – y1.
y
Note
∆x is the same as h
used on the previous
slide show.
Q (x2, y2)
∆y y2  y1
P (x1, y1)
O
gradient =
Dy
Dx
∆x
x2  x1
x
Gradient of the curve y = x2 at the point P(x, y)
Suppose that the point Q(x + ∆x, y + ∆y) y
is very close to the point P on the curve.
Q(x + ∆x, y + ∆y)
The small change from P in the value
of x is ∆x and the corresponding small
change in the value of y is ∆y.
∆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:
(y + Dy)- y = Dy
(x + Dx)- x Dx
P(x, y)
y
y = x2
∆x
y
x
O
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:
(x + Dx)2 - x 2= x 2 + 2xDx +(Dx)2 - x 2= 2xDx +(Dx)2 = 2x + ∆x
Dx
Dx
(x + Dx)- x
x
So Dy = 2x + ∆x
Dx
As ∆x gets smaller Dy approaches a limit and we start to refer to it in theoretical
Dx
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.
This is denoted by
dy
or d y .
dx
dx
dy  lim  δy 
dx δx  0 δx 
For the curve y = x2,
dy  lim 2x  δx
dx δx  0
= 2x
This is the result we obtained previously.
Gradient of the curve y = f(x) at the point P(x, y)
For any function y = f(x) the gradient
of the chord PQ is:
y
Q(x + δx, y + δy)
( y  δy)  y  δy
( x  δx)  x δx
δy
The coordinates of P can also be written
as (x, f(x)) and the coordinates of Q as
[(x + δx), f(x + δx)].
P(x, y)
So the gradient of the chord PQ can be
written as:
f ( x  δx)  f ( x)
( x  δx)  x
dy  lim  δy 
dx δx  0 δx  
O


 f ( x  δx)  f ( x) 


x  0 ( x  δx)  x 

lim


lim  f ( x  δx)  f ( x) 
x  0
y = f(x)
x
δx

y
δx
y
x
DEFINITION OF THE DERIVATIVE OF A FUNCTION
dy
The symbol dx is called the derivative or the differential coefficient of y with respect to x.
It is defined by
dy  lim  f ( x  δx)  f ( x) 

dx δx  0
δx

In words we say:
δy
“dee y by dee x is the limit of δx as δx tends to zero”
“tends to” is another way of saying “approaches”
DEFINITION OF THE DERIVATIVE OF A FUNCTION
Sometimes we write h instead of ∆x and so the derivative of f(x) can be written as
dy  lim  f ( x  h)  f ( x) 

dx h  0
h

If y = f(x) we can also use the notation:
dy
= f’ (x)
dx
In this case f’ is often called the derived function of f. This is also called f-prime.
dy
The procedure used to find dx from y is called differentiating y with respect to x.
Example (1)
Find dy for the function y = x3.
dx
Results so far
f(x)
f‘(x)
x2
2x
x3
3x2
In this case, f(x) = x3.
dy
dx
 f ( x  h)  f ( x) 





h

h  0
lim

3
3
 ( x  h)  x 



h

h  0
lim
 3
2
2
3
3
 x  3x h  3xh  h  x 



h

h  0
lim

2
2
3
 3x h  3xh  h 



h

h  0
lim
 lim 3x 2  3xh h 2 
h  0
= 3x2

y = x3
y
15
10
5
x
−4
−3
−2
−1
1
2
3
4
y = x3
y
15
10
Gradient at
(2, 8) = 12
5
x
−4
−3
−2
−1
1
2
3
4
Example (2)
Find dy for the function y = x4.
dx
Results so far
f(x)
f‘(x)
x2
2x
x3
3x2
x4
4x3
In this case, f(x) = x4.
dy
dx
 f ( x  h)  f ( x) 





h

h  0
lim

4
4
 ( x  h)  x 



h

h  0
lim
 4
3
2 2
3
4
4
 x  4 x h  6 x h  4 xh  h  x 



h

h  0
lim

3
2 2
3
4
 4 x h  6 x h  4 xh  h 



h

h  0
lim
 lim 4x3  6x 2h  4xh2  h3 
h  0
= 4x3

Example (3)
Find dy for the function y = 1x .
dx
In this case, f(x) = 1x .
dy  lim  f ( x  h)  f ( x) 

dx h  0
h


 h 




h  0 xh( x  h) 
lim

1 




h  0 x( x  h) 
lim
  12
x





h  0

lim

 x  ( x  h) 
 h 
1
1 








x  h x   lim  x( x  h)   lim  x( x  h) 




h 
h
h
h  0
h  0











So we now have the following results:
n
xn
dy
dx
–1
1 (or x1)
x
 12 (or  x 2 )
x
2
x2
2x
3
x3
3x2
4
x4
4x3
We also know that if y = x = x1 then
dy 1 (or 1x0 )
dx
dy
and if y = 1 = x0 then dx  0 (or 0 x 1)
dy
These results suggest that if y = xn then dx  nx n  1
It can be proven that this statement is true for all values of n.
Example (1)
Find dy for the function y = 1.
dx
But of course we knew this already.
y
y = 1 can be thought of as y = x0.
Using dy  nx n  1
dx
1
y=1
dy  0 x 0  1
dx
=0
O
x
Example (2)
Find dy for the function y = x.
dx
But of course we knew this already.
y
y=x
y = x can be thought of as y = x1.
Using dy  nx n  1
dx
dy 1x11
dx
= x0
=1
O
x
Example (3)
Find dy for the function y = x2.
dx
Using dy  nx n  1
dx
dy  2 x 2  1
dx
= 2x
Example (4)
Find dy for the function y = x3.
dx
Using dy  nx n  1
dx
dy  3 x 3  1
dx
= 3x2
Example (5)
Find dy for the function y = x4.
dx
Using dy  nx n  1
dx
dy  4 x 4  1
dx
= 4x3
Example (6)
Find dy for the function y = 1x .
dx
y = 1x can be written as y = x–1.
Using dy  nx n  1
dx
dy  1x11
dx
= –x–2
  12
x
Example (7)
1
Find dy for the function y = 2 .
dx
x
y=
1
can be written as y = x–2.
2
x
Using dy  nx n  1
dx
dy  2x 2 1
dx
= –2x–3
  23
x
Example (8)
Find dy for the function y = x .
dx
y = x can be written as y 
Using dy  nx n  1
dx
dy  1 x 12 1
dx 2
1
1x 2
2
 1
2 x
1
x2
.
Example (9)
Find dy for the function y = 1 .
x
dx
1
1
y = x can be written as y  x 2 .
Using dy  nx n  1
dx
dy   1 x121
2
dx
3
1 x 2
2
 1
2 x3
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.
Example (1)
Find dy for the function y = 3x2.
dx
dy  3 2 x
dx
= 6x
Example (2)
Find dy for the function y = 8x3.
dx
dy  8  3 x 2
dx
= 24x2
Example (3)
Find dy for the function y = x4 + x3.
dx
dy  4x3  3x2
dx
Example (4)
Find dy for the function y = x2 + x + 1
dx
dy  2 x  1  0
dx
= 2x + 1
Example (5)
Find the gradient of the curve y = 2x2 + 3x – 7 at the point (2, 7).
15 y
dy  4 x  3
dx
At the point (2, 7),
dy  4  2  3
dx
10
So the gradient = 11
5
x
−4
−3
−2
−1
1
−5
2
3
4
Example (6)
2
Find dy for the function y  2x3  6 x  4x 2 4x
dx
x
2
y  2x3  6 x  4x 2 4x
x
2
4
x
 2x  6 x  2  4x2
x
x
3
 2 x3  6 x  4  4
x
 2x
3
1
 6 x 2  4  4 x1
dy  6 x2  3x12  4 x2
dx
 6x2  3  42
x x
Example (7)
Find the coordinates of the points on the graph of y = 2x3 – 3x2 – 36x + 10 at which
the gradient is zero.
Let f(x) = 2x3 – 3x2 – 36x + 10
Then f’ (x) =
6x2
y
50
– 6x – 36
The gradient is zero when f’ (x) = 0
That is when 6x2 – 6x – 36 = 0
This simplifies to
x2
–x–6=0
x
−6
−4
−2
2
In factor form this is (x – 3)(x + 2) = 0
So x = –2 or x = 3
−50
Substituting 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)
4
6
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)).
y
Tangent
Normal
The equation of the tangent to a curve at a point
(a, f(a)) is
A
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.
O
x
1
The gradient of the normal is   as the product of the gradients of perpendicular lines is –1.
f (a )
The equation of the normal to a curve at a point (a, f(a)) is
y  f (a)   1 ( x  a)
f (a)
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.
15
dy
2
dx = 6x + 6x
y
At x = 1, the gradient of C = 12
10
Equation of the tangent at A is
y – 7 = 12(x – 1)
which simplifies to
5
y = 12x – 5
x
−3
−2
−1
1
−5
2
3
Example (2)
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.
y
(a) Find the coordinates of A and B.
l
C
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
= x(x2 + 3x – 4)
= x(x + 4)(x – 1)
So A is (–4, 0) and B is (1, 0)
(b) dy  3x2  6x  4
dx
dy
At O,  4
dx
So an equation for l is y = –4x
A
O
B
x
(c) x3 + 3x2 – 4x = –4x
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)
Example (3)
For the curve C with equation y = x4 – 9x2 + 2,
dy
(a) find dx .
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
dy
= 4x3 – 18x
dx
(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
1
So the gradient of the normal = 4
1
Equation of the normal to C at A is y  18  4 ( x  2)
which simplifies to x – 4y – 74 = 0
Example (4)
A curve C has equation y = 2x2 – 6x + 5.
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.
(a) dy = 4x – 6
dx
When x = 1, y = 2 – 6 + 5 = 1
Gradient of the tangent to the curve at A = 4 – 6 = –2
Gradient of the normal to the curve at A = 12
Equation of the normal to the curve at A is y 1 12 ( x 1)
which simplifies to
x – 2y + 1 = 0
x 1
2
4x2 – 12x + 10 = x + 1
(b) 2x2 – 6x + 5 =
4x2 – 13x + 9 = 0
(x – 1)(4x – 9) = 0
9
So x-coordinate of B = 4
13
y-coordinate of B = 8