Transcript C1: Differentiation from First Principles

C1: Differentiation from First
Principles
Learning Objective: to understand that
differentiation is the process for
calculating the gradient of a curve. The
rate of change can be calculated from
first principles by considering the limit
of the function at any one point.
Rates of change
distance (m)
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
40
with respect to time.
0
time (s)
5
In other words, the gradient tells us
the speed of the car.
change in distance
40
gradient =
=
= 8 m/s
change in time
5
The car in this example is travelling at a constant speed since
the gradient is the same at every point on the graph.
Rates of change
distance (m)
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
0
time (s)
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.
In most situations this involves finding the gradient of a curve.
The gradient of a curve
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:
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.
δx represents a small
change in x and δy
represents a small
change in y.
As point B moves
closer to point A, the
gradient of the chord
AB gets closer to the
gradient of the
tangent at A.
Differentiation from first principles
We can write the gradient of the chord AB as:
y
change in y
=
x
change in x
y
As B gets closer to A, δx gets closer to 0 and
gets closer
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.
Differentiation from first principles
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:
 y (3   x )2  9
=
2
B(3 + δx, (3 + δx) )
x
(3   x )  3
9  6 x  ( x)2  9
=
x
δy
6 x  ( x)2
=
x
A(3, 9)
 x(6   x )
=
δx
x
= 6 + x
Differentiation from first principles
At the limit where δx → 0, 6 + δx → 6.
We write this as:
y
lim
= lim 6 +  x = 6
 x 0  x
 x0
So the gradient of the tangent to the curve y = x2 at the point
(3, 9) is 6.
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 by x2.
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).
Differentiation from first principles
The gradient of chord AB is:
 y ( x   x )2  x 2
=
x
( x   x)  x
x2  2 x x  ( x )2  x 2
=
x
2 x x  ( x )2
=
x
 x(2 x   x )
=
x
= 2x +  x
B(x + δx, (x + δx)2)
δy
A(x, x2)
δx
2
So for y = x ,
y
lim
= 2x
 x 0  x
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
Then:
f ′(x) = 2x
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
Now we shall differentiate y = x3
from first principles:
Using the notation
dy
dx
We have shown that for y = x3
y
lim
= 3 x2
 x 0  x
y
dy
lim
is usually written as
.
 x 0  x
dx
So if y =
x3
then:
dy
= 3 x2
dx
dy
represents the derivative of y with respect to x.
dx
Remember,
y
dy
is the gradient of a chord, while is the gradient
x
dx
of the tangent.
Using the notation
dy
dx
This notation can be adapted for other variables so, for
example:
ds
represents the derivative of s with respect to t.
dt
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.
Also, if we want to differentiate 2x4 with respect to x, for
example, we can write:
d
(2 x 4 )
dx
We could work this out by differentiating from first principles,
but in practice this is unusual.
Task 1
Differentiate from first principles
1. y = x4
2. y = 1/x