Transcript C1.7 Differentiation - Heathcote School & Science

AS-Level Maths:
Core 1
for Edexcel
C1.7 Differentiation
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Rates of change
Rates of change
Contents
The gradient of the tangent at a point
The gradient of the tangent as a limit
Differentiation of polynomials
Second order derivatives
Tangents and normals
Examination-style questions
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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.
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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.
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The gradient of the tangent at a point
Rates of change
Contents
The gradient of the tangent at a point
The gradient of the tangent as a limit
Differentiation of polynomials
Second order derivatives
Tangents and normals
Examination-style questions
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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:
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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:
change in y y2  y1
=
change in x x2  x1
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The gradient of the tangent as a limit
Rates of change
Contents
The gradient of the tangent at a point
The gradient of the tangent as a limit
Differentiation of polynomials
Second order derivatives
Tangents and normals
Examination-style questions
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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.
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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.
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Differentiation from first principles
Let’s see how this works for the gradient of the tangent to the
curve y = x2 at the point A(3, 9).
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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
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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 be 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).
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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 ,
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y
lim
= 2x
 x 0  x
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The gradient of y = x2
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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
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The gradient function of y = x2
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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:
 y ( x   x )3  x 3
=
x
( x   x)  x
x3  3 x 2 x  3 x( x )2  ( x )3  x3
=
x
3 x 2 x  3 x( x )2  ( x )3
=
x
 x(3 x2  3 x x  ( x)2 )
=
x
2
= 3 x  3 x x  ( x )2
y
3
So for y = x ,
lim
= 3 x2
 x 0  x
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The gradient of y = x3
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The gradient of y = x3
So, the gradient of the tangent to the curve y = x3 at the
general point (x, y) is 3x2.
So if,
f(x) = x3
Then:
f ′(x) = 3x2
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.
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The gradient function of y = x3
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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.
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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.
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Differentiation of polynomials
Contents
Rates of change
The gradient of the tangent at a point
The gradient of the tangent as a limit
Differentiation of polynomials
Second order derivatives
Tangents and normals
Examination-style questions
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Differentiation
If we continued the process of differentiating from first
principles we would obtain the following results:
y
x
x2
x3
x4
x5
x6
dy
dx
1
2x
3x2
4x3
5x4
6x5
What pattern do you notice?
In general:
dy
If y  x then
 nx n1
dx
n
and when xn is preceded by a constant multiplier a we have:
dy
If y = ax then
= anx n1
dx
n
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Differentiation
y = 2x4
For example, if
dy
= 2× 4 x 41 = 8x3
dx
Suppose we have a function of the form y = c, where c is a
constant number.
This corresponds to a horizontal line through the point (0, c).
y
The gradient of this line is always
equal to 0.
c
 If
0
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x
y=c
dy
=0
dx
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Differentiation
Find the gradient of the curve
y = 3x4 at the point (–2, 48).
Differentiating:
dy
= 12 x3
dx
At the point (–2, 48) x = –2 so:
dy
= 12( 2)3
dx
= 12 × 8
= 96
The gradient of the curve y = 3x4 at the point (–2, 48) is –96.
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Differentiating polynomials
Suppose we want to differentiate a polynomial function.
For example:
Differentiate y = x4 + 3x2 – 5x + 2 with respect to x.
dy
We can differentiate each term in the function to give
.
dx
dy
= 4x3 + 6x – 5
So:
dx
In general, if y is made up of the sum or difference of any given
number of functions, its derivative will be made up of the
derivative of each of the functions.
dy
If y = f ( x ) ± g ( x ) then
= f '( x ) ± g '( x )
dx
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Differentiating polynomials
Find the point on y = 4x2 – x – 5 where the gradient is 15.
dy
= 8x  1
dx
dy
= 15 when 8x – 1 = 15
dx
8x = 16
x=2
We now substitute this value into the equation y = 4x2 – x – 5
to find the value of y when x = 2
y = 4(2)2 – 2 – 5
=9
The gradient of y = 4x2 – x – 5 is 15 at the point (2, 9).
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Differentiating polynomials
In some cases, a function will have to be written as separate
terms containing powers of x before differentiating.
For example:
Given that f(x) = (2x – 3)(x2 – 5) find f ′(x).
Expanding:
f(x) = 2x3 – 3x2 – 10x + 15
f ′(x) = 6x2 – 6x –10
Find the gradient of f(x) at the point (–3, –36).
When x = –3 we have f ′(–3) = 6(–3)2 – 6(–3) –10
= 54 + 18 – 10
= 62
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Differentiating polynomials
3 x5 + 4 x 2  8 x
Differentiate y =
with respect to x.
2x
3 x5 4 x 2 8 x
y=
+

2x
2x 2x
y = 32 x 4 + 2 x  4
We can now differentiate:
dy 12 3
= 2 x +2
dx
= 6x3 + 2
= 2(3x3 + 1)
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Differentiating axn for all rational n
If y = ax n then
dy
= anx n1
dx
This is true for all negative or fractional values of n.
For example:
2
Differentiate y = with respect to x.
x
Start by writing this as
y = 2x–1
dy
= 1 × 2 x 11
dx
= 2x 2
So:
=
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2
x2
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Differentiating axn for all rational n
Differentiate y = 4 x with respect to x.
Start by writing this as
y = 4x
1
2
1 1
dy 1
2
= 2 × 4x
dx
 21
= 2x
So:
=
2
x
Remember, in some cases, a function will have to be written as
separate terms containing powers of x before differentiating.
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Differentiating axn for all rational n
Given that f (x) = (1+ x )2 find f '( x).
f ( x) = (1+ x )(1+ x )
= 1+ 2 x + x
1
2
= 1+ 2 x + x
 21
f '( x) = x +1
=
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1
x
+1
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Second order derivatives
Contents
Rates of change
The gradient of the tangent at a point
The gradient of the tangent as a limit
Differentiation of polynomials
Second order derivatives
Tangents and normals
Examination-style questions
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Second order derivatives
Differentiating a function y = f(x) gives us the derivative
dy
dx
or
f ′(x)
Differentiating the function a second time gives us the second
order derivative. This can be written as
d2y
dx 2
or
f ′′(x)
The second order derivative gives us the rate of change of the
gradient of a function.
We can think of it as the gradient of the gradient.
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Using second order derivatives
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Tangents and normals
Contents
Rates of change
The gradient of the tangent at a point
The gradient of the tangent as a limit
Differentiation of polynomials
Second order derivatives
Tangents and normals
Examination-style questions
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Tangents and normals
Remember, the tangent to a curve at a point is a straight line
that just touches the curve at that point.
The normal to a curve at a point is a straight line that is
perpendicular to the tangent at that point.
We can use differentiation to find the equation of the tangent
or the normal to a curve at a given point. For example:
Find the equation of the tangent and the normal
to the curve y = x2 – 5x + 8 at the point P(3, 2).
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Tangents and normals
y = x2 – 5x + 8
dy
= 2x  5
dx
At the point P(3, 2) x = 3 so:
dy
= 2(3)  5 = 1
dx
The gradient of the tangent at P is therefore 1.
Using y – y1 = m(x – x1), give the equation of the tangent at the
point P(3, 2):
y–2=x–3
y=x–1
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Tangents and normals
The normal to the curve at the point P(3, 2) is perpendicular to
the tangent at that point.
The gradient of the tangent at P is 1 and so the gradient of the
normal is –1.
Using y – y1 = m(x – x1) give the equation of the tangent at the
point P(3, 2):
y  2 = ( x  3)
y+ x 5 =0
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Examination-style questions
Contents
Rates of change
The gradient of the tangent at a point
The gradient of the tangent as a limit
Differentiation of polynomials
Second order derivatives
Tangents and normals
Examination-style questions
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Examination-style question
The curve f(x) = x2 + 2x – 8 cuts the x-axes at the points A(–a,
0) and B(b, 0) where a and b are positive integers.
a) Find the coordinates of points A and B.
b) Find the gradient function of the curve.
c) Find the equations of the normals to the curve at points A
and B.
d) Given that these normals intersect at the point C, find the
coordinates of C.
a) The equation of the curve can be written in factorized form
as y = (x + 4)(x – 2) so the coordinates of A are (–4, 0) and
the coordinates of B are (2, 0).
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Examination-style question
f(x) = x2 + 2x – 8  f’(x) = 2x + 2
b)
c) At (–4, 0) the gradient of the curve is –8 + 2 = –6
 The gradient of the normal at (–4, 0) is
1
6
.
Using y – y1 = m(x – x1), the equation of the normal at (–4, 0) is:
y–0=
1
6 (x
+ 4)
6y = x + 4
1
At (2, 0) the gradient of the curve is 4 + 2 = 6
 The gradient of the normal at (2, 0) is –
1
6
.
Using y – y1 = m(x – x1), the equation of the normal at (2, 0) is:
y–0=–
1
6
(x – 2)
6y = 2 – x
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Examination-style question
d) Equating
1
and
When x = –1, y =
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1
2
2
:
x+4=2–x
2x = –2
x=1
. So the coordinates of C are (–1, 21 ).
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