Differentiation
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Transcript of Differentiation
Higher Maths 1 3 Differentiation
1
Higher Maths 1 3 Differentiation
The History of DifferentiationDifferentiation is part of the science of Calculus, and was first developed in the 17th century by two different Mathematicians.
Gottfried Leibniz(1646-1716)
Germany
Sir Isaac Newton(1642-1727)
England
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Differentiation, or finding the instantaneous rate of change, is an essential part of: • Mathematics and Physics
• Chemistry
• Biology
• Computer Science
• Engineering
• Navigation and Astronomy
Higher Maths 1 3 Differentiation
Calculating Speed
2 4 6 8
4
8
2
6
10
00
Time (seconds)
Dis
tan
ce (m
)
DS T×
÷÷
Example
Calculate the speed for each section of the journey opposite.
A
B
C
speed in A =
43
speed in B = 51 5
m/s
=
speed in C =
25 0.4
m/s
=
average speed =
9 1.22 m/s
≈
11≈
1.33 m/s
Notice the following things:• the speed at each instant is not the same as the average• speed is the same as gradientD
TS
=
yx
= m=
3
Instantaneous Speed
Higher Maths 1 3 Differentiation
Time (seconds)
Dis
tan
ce (m
)
Time (seconds)
Dis
tan
ce (m
)
In reality speed does not often change instantly. The graph on the right is more realistic as it shows a gradually changing curve.The journey has the same average speed, but the instantaneous speed is different at each point because the gradient of the curve is constantly changing. How can we find the instantaneous speed?
DT
S
=
yx
= m=
4
Introduction to Differentiation
Higher Maths 1 3 Differentiation
Differentiate means
D
Tspeed
=
‘rate of change ofdistance with respect to time’
S
Tacceleration
=
‘find out how fast something is changing in comparison with something else at any one instant’.
gradient
=
yx
‘rate of change ofspeed with respect to time’‘rate of change of -coordinate with respect to -coordinate’y x
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Estimating the Instantaneous Rate of Change
Higher Maths 1 3 Differentiation
The diagrams below show attempts to estimate the instantaneous gradient (the rate of change of with respect to ) at the point A.
A yAxx
yy
A
x
Notice that the accuracy improves as gets closer to zero.
x
y x
The instantaneous rate of change is written as:
yx
dydx = as approaches 0.
x
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Basic Differentiation
The instant rate of change of y with respect to x is written as .
Higher Maths 1 3 Differentiation
dydx
By long experimentation, it is possible to prove the following:
dydx =
y = x n
nxn–1
If
then • multiply by the power
• reduce the power by
one
How to Differentiate:
Note that describes both the rate of change and the gradient.
dydx
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Differentiation of Expressions with Multiple Terms
Higher Maths 1 3 Differentiation
y =ax m + bx
n + …
• multiply every x-term by the power
How to Differentiate:
amx m + bnx
n + …–1 –1dy
dx =
• reduce the power of every x-term by one
The basic process of differentiation can be applied to
every x-term in an algebraic expession.
ImportantExpressions must be written as the sum of individual terms before differentiating.
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Examples of Basic Differentiation
Higher Maths 1 3 Differentiation
Find for
dydx
y = 3 x 4 – 5 x
3 + + 9
7x
2
y = 3 x 4 – 5 x
3 + 7 x -2 +
9dydx = 12 x
3 – 15 x 2 – 14
x -3
9 =9 x 0
this disappears because
= 12 x 3 – 15 x
2 –14x
3
9
(multiply by zero)
Example 1
Examples of Basic Differentiation
Higher Maths 1 3 Differentiation
Find the gradient of the curve
Example 2
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y =( x + 3 ) ( x – 5 )
y =x ² – 2 x – 15
x 2
= 1 2x – 15
x 2
x 2
–
= 1 – 2 x -1 – 15 x
- 2
dydx = 2 x
- 2 + 30 x
-
3
disappears (multiply by zero)
=2 + 30
x 3x
2
At x =
5,
at the point
(5,0).
dydx =
2 + 3025 125
=8
25
The Derived Function
The word ‘derived’ means ‘produced from’, for example orange juice is derived from oranges.
Higher Maths 1 3 Differentiation
It is also possible to express differentiation using function notation.
=
f (x)
nxn–1
If
then f ′(x)
= x n f ′(x) dy
dxand
mean exactly the same thing written in different ways.
LeibnizNewton
The derived function is the rate of change of the function with respect to .
f ′(x) f (x)x
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A tangent to a function is a straight line which intersects
the function in only one place, with the same gradient as
the function.
Tangents to Functions
Higher Maths 1 3 Differentiation
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The gradient of any tangent to the function can be
found by substituting the x-coordinate of intersection into .
f (x)
A
B
f ′(x)
f (x)
mAB = f ′(x)
f ′(x)
Equations of Tangents
Higher Maths 1 3 Differentiation
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y – b = m ( x – a ) To find the equation of a
tangent:
• substitute gradient and point of intersection intoy – b = m ( x – a )
• substitute -coordinate to find gradient at point of intersection
x
• differentiate Straight Line Equation
Example
Find the equation of the tangent to the function
at the point (2,4).
f (x)= x 31
2
= x 23
2
f ′(2) = ×
(2)
2
32
= 6
m =
y – 4 = 6 ( x – 2 )
6 x – y – 8 = 0
substitute:
Increasing and Decreasing Curves
Higher Maths 1 3 Differentiation
The gradient at any point on a curve can be found by differentiating.
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uphill slopePositive
downhill slopeNegative
Gradient
Ifdydx
> 0 then y is
increasing.
Ifdydx
< 0 then y is
decreasing. Alternatively,If > 0f ′(x)is increasing.
f (x)then
If < 0f ′(x)is decreasing.
f (x)then
dydx < 0
dydx > 0
dydx < 0
If a function is neither increasing or decreasing, the gradient is zero and the function can be described as stationary.
Stationary Points
Higher Maths 1 3 Differentiation
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There are two main types of stationary point.
Minimum
Maximum
Turning Points
Points of Inflection
At any stationary point, or alternatively
dydx
= 0 f ′(x)= 0
Rising
Falling
Investigating Stationary Points
Higher Maths 1 3 Differentiation
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ExampleFind the stationary
point off (x) = x 2 – 8 x + 3
Stationary point
given by
f ′(x) = 2 x – 8
2 x – 8= 0
and determine its
nature.
x= 4
f ′(x) = 0
x 4 – 4+4
slope
The stationary point atis a minimum turning point.
– 0 +f ′(x)
x= 4
Use a nature table to reduce the amount of working.
‘slightly less than
four’
‘slightly more’
gradientis positive
Investigating Stationary Points
Higher Maths 1 3 Differentiation
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Example 2
Investigate the stationary
points ofy =4 x 3 – x
4
dydx
=12 x 2 – 4 x
3 = 0
4 x 2 (3 – x ) = 0
4 x 2= 0
x = 0
3 – x= 0
x = 3
or
y = 0 y = 27
x 0 – 0+0
slope
dydx
rising point of inflection
+ 0 +
stationary point at (0,0):
x 3 – 3+3
slope
dydx
maximum turning point
+ 0 –
stationary point at (3, 27):
Positive and Negative Infinity
Higher Maths 1 3 Differentiation
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Example
∞The symbol is used for infinity.+∞–∞
‘positive infinity’
‘negative infinity’
The symbol means approaches’.
y =5 x 3 + 7 x
2
For very large , the value
of
becomes insignificant
compared with the value of
.
5 x 3
± x
as x ,∞ y 5 x 3
5×( )3+∞ = +∞5×( )3–∞ = –∞
as x +∞
∞+1 = ∞
y +∞and as x –∞
y –∞
7 x 2
Higher Maths 1 3 Differentiation
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x 7 – 7+7
slope
– 0 +dydx
Curve SketchingTo sketch the graph of any function,
the following basic information is
required:• the stationary points and their nature
• the x-intercept(s) and y-intercept
• the value of y as x
approachespositive and negative
infinity
solve for y = 0 and
x = 0
solve for = 0 and use nature table
dydx
y = x 3 – 2 x
4
–∞y +∞
as x ∞y -2 x
4
+∞y –∞
as
x
and as x
Example
Graph of the Derived Function
Higher Maths 1 3 Differentiation
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f ′(x)
f (x)The graph of can be thought of
as the graph of the gradient of .
Example
y = f (x)
y = f ′(x)f ′(x) = 0
f ′(x) = 0
Positive
Negative
Gradient
f ′(x) > 0
f ′(x) < 0
The roots of
are
f (x)
f ′(x)
given by the
stationarypoints of
.