Differentiation
Differential
Calculus concerns the rate of change of a function
with respect to the
change in the variable on which it depends.
For
example,
Speed is
a measure of distance travelled
by an object in a
unit period of time.
The distance travelled
totally depends on the time taken.
Instantaneous speed is the
speed of an object at a particular time.
Average speed = total
distance divided by total time taken.
Velocity
is a measure of
the distance travelled
by an object in a
particular direction over a unit period
of time.
Acceleration
is a measure of
how the velocity of an object
changes with time.
The derivative of a
function for some particular value
is a measure of
the rate at which the function is changing
at that
particular value.
The
derivative of a function for some particular value
is also the gradient of the graph of the
function at that point.
· Finding
Gradients of Curves
The
gradient of a straight line is found by dividing the change
in the y-axis ( the y-step) by the change in
the x-axis ( the x-step).
But
what happens if we need to find the gradient of a curve ?
The
points A(2,4) and B(3,9) lie on the curve y = x2 .
The average gradient of the curve from A to B
is the gradient of
the chord AB.
As B moves towards A, the gradient of the curve changes.
The closer B is to A,the
nearer it reaches a limit.
|
x |
f (x) |
Δy |
Δx |
m =
Δy/Δx |
|
2.001 |
4.004001 |
0.004001 |
0.001 |
4.001 |
|
2.0009 |
4.003601 |
0.003601 |
0.0009 |
4.0009 |
|
2.0008 |
4.003201 |
0.003201 |
0.0008 |
4.0008 |
|
2.0007 |
4.0028 |
0.0028 |
0.0007 |
4.0007 |
|
2.0006 |
4.0024 |
0.0024 |
0.0006 |
4.0006 |
|
2.0005 |
4.002 |
0.002 |
0.0005 |
4.0005 |
|
2.0004 |
4.0016 |
0.0016 |
0.0004 |
4.0004 |
|
2.0003 |
4.0012 |
0.0012 |
0.0003 |
4.0003 |
|
2.0002 |
4.0008 |
0.0008 |
0.0002 |
4.0002 |
|
2.0001 |
4.0004 |
0.0004 |
1E-04 |
4.0001 |
This limit is
the gradient of the curve
and occurs at the
tangent to the curve at the point A.
When
A is the point ( 2, 4), the gradient of the curve is
4.
Other
points have different gradients.
Is there an easier way to find the gradient
of the curve at other points ?
Let A(x, f(x)) and B(x+h,(f(x+h)) be points on the graph y = f(x)
·
Differentiation by First Principles – deriving f’(x) from f(x)
using 
If f(x) = x2 ,
find f’ (x)
The
derived function of x is f’(x) = 2x
The
derivative when
x = 3 is 6
This means that the rate of
change of the function is 2x,
and that the
gradient of the tangent to the graph of the
curve y = x2
at any point is found by doubling the x co-ordinate.
f(-1) = 1 f’(-1) =-2 has tangent
y-1 =-2(x-(-1)) i.e. y = -2x-1
f(0) = 0 f’(0) =0 has tangent
y-0 =0(x-0) i.e. y =0
f(1) = 1 f’(1) =2 has tangent
y-1 =2(x-1) i.e. y = 2x-1
f(2) = 4 f’(2) =4 has tangent
y-4 =4(x-2) i.e. y = 4x-4
f(3) = 9 f’(3) =6 has tangent
y-9 =6(x-3) i.e. y = 6x-9
f(4)
= 16 f’(4) =8 has tangent
y-16 =8(x-4) i.e. y = 8x-16
The
gradient of the tangent to the graph at the point (3, 9) is 6.
The
gradient of the tangent to the graph at the point (4, 16) is 8.
In General
If f(x) = xn then f’(x)
= nxn-1
(n is a rational number)
Example
If f(x) = x3 ,
find f’ (x)
