Outcome 3. Higher. Using differentiation (Application). Higher Unit 1. Finding the gradient for a polynomial. Increasing / Decreasing functions. Max / Min and inflexion Points. Differentiating Brackets ( Type 1 ) . Curve Sketching. Differentiating Harder Terms (Type 2).
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Higher
Using differentiation (Application)
Finding the gradient for a polynomial
Increasing / Decreasing functions
Max / Min and inflexion Points
Differentiating Brackets ( Type 1 )
Curve Sketching
Differentiating Harder Terms (Type 2)
Max & Min Values on closed Intervals
Differentiating with Leibniz Notation
Optimization
Equation of a Tangent Line ( Type 3 )
Mind Map of Chapter
www.mathsrevision.com
Outcome 3
Higher
On a straight line the gradient remains constant, however with curves the gradient changes continually, and the gradient at any point is in fact the same as the gradient of the tangent at that point.
The sides of the halfpipe are very steep(S) but it is not very steep near the base(B).
S
Demo
B
Outcome 3
Higher
Gradient of tangent = gradient of curve at A
A
Demo
B
Gradient of tangent = gradient of curve at B
Gradients & Curves
Outcome 3
Higher
For the function y = f(x) we do this by taking the point (x, f(x))
and another “very close point” ((x+h), f(x+h)).
Then we find the gradient between the two.
((x+h), f(x+h))
Approx gradient
(x, f(x))
True gradient
Outcome 3
Higher
The gradient is not exactly the same but is
quite close to the actual value
We can improve the approximation by making the value of h smaller
This means the two points are closer together.
((x+h), f(x+h))
Approx gradient
(x, f(x))
True gradient
Outcome 3
Higher
We can improve upon this approximation by making the value of h even smaller.
So the points are even closer together.
((x+h), f(x+h))
Approx gradient
True gradient
(x, f(x))
Higher
We have seen that on curves the gradient changes continually and is dependant on the position on the curve. ie the xvalue of the given point.
Finding the GRADIENT
Differentiating
The process of finding the gradient is called
Finding the rate of change
DIFFERENTIATING
or
FINDING THE DERIVATIVE (Gradient)
Outcome 3
Higher
If the formula/equation of the curve is given by f(x)
Then the derivative is called f '(x)  “f dash x”
There is a simple way
of finding f '(x) from f(x).
f(x) f '(x)
2x2 4x
4x2 8x
Have guessed the rule yet !
5x10 50x9
6x7 42x6
x3 3x2
x5 5x4
x99 99x98
Rule for Differentiating
Outcome 3
Higher
It can be given by this simple flow diagram ...
multiply by the power
reduce the power by 1
If f(x) = axn
n
n
1
ax
then f '(x) =
NB: the following terms & expressions mean the same
GRADIENT,
DERIVATIVE,
RATE OF CHANGE,
f '(x)
Rule for Differentiating
Outcome 3
Higher
To be able to differentiate
it is VERY IMPORTANT that you are
comfortable using indices rules
Higher
(I) f(x) = ax (Straight line function)
Index Laws
x0 = 1
If f(x) = ax
= ax1
then f '(x) = 1 X ax0
= a X 1 = a
So if g(x) = 12x then g '(x) = 12
Also using y = mx + c
The line y = 12x has gradient 12,
and derivative = gradient !!
Higher
(II) f(x) = a, (Horizontal Line)
Index Laws
x0 = 1
If f(x) = a
= a X 1 = ax0
then f '(x) = 0 X ax1
= 0
So if g(x) = 2 then g '(x) = 0
Also using formula y = c , (see outcome 1 !)
The line y = 2 is horizontal so has gradient 0 !
Differentiate
Differentiate
Differentiate
Outcome 3
Higher
Example 1
A curve has equation f(x) = 3x4
Find the formula for its gradient and find the gradient when x = 2
Its gradient is f '(x) = 12x3
f '(2) = 12 X 23 =
12 X 8 =
96
Example 2
A curve has equation f(x) = 3x2
Find the formula for its gradient and find the gradient when x = 4
Its gradient is f '(x) = 6x
At the point where x = 4 the gradient is
f '(4) = 6 X 4 =
24
Outcome 3
Higher
Example 3
If g(x) = 5x4  4x5 then find g '(2) .
g '(x) = 20x3  20x4
g '(2) = 20 X 23  20 X 24
= 160  320
= 160
Outcome 3
Higher
Example 4
h(x) = 5x2  3x + 19
so h '(x) = 10x  3
and h '(4) = 10 X (4)  3
= 40  3 = 43
Example 5
k(x) = 5x4  2x3 + 19x  8, find k '(10) .
k '(x) = 20x3  6x2 + 19
So k '(10) = 20 X 1000  6 X 100 + 19
= 19419
Outcome 3
Higher
Example 6 : Find the points on the curve
f(x) = x3  3x2 + 2x + 7 where the gradient is 2.
NB: gradient = derivative = f '(x)
Now using original formula
We need f '(x) = 2
ie 3x2  6x + 2 = 2
f(0) = 7
or 3x2  6x = 0
ie 3x(x  2) = 0
f(2) = 8 12 + 4 + 7
ie 3x = 0 or x  2 = 0
= 7
Points are (0,7) & (2,7)
so x = 0 or x = 2
Differentiate
Outcome 3
Higher
Example
f(x) = 3x3  x + 2 x2
= 3x3  x + 2 x2 x2x2
= 3x  x1 + 2x2
f '(x) = 3 + x2  4x3
= 3 + 1  4 x2 x3
Higher
Leibniz Notation is an alternative way of expressing derivatives to f'(x) , g'(x) , etc.
If y is expressed in terms of x then the derivative is written as dy/dx .
eg y = 3x2  7x
so dy/dx = 6x  7 .
Example 19
Q = 9R2  15 R3
Find dQ/dR
= 18R + 45 R4
NB: Q = 9R2  15R3
So dQ/dR = 18R + 45R4
Outcome 3
Higher
Example 20
A curve has equation y = 5x3  4x2 + 7 .
Find the gradient where x = 2 ( differentiate ! )
gradient = dy/dx = 15x2  8x
if x = 2 then
gradient = 15 X (2)2  8 X (2)
= 60  (16) = 76
Physics
Outcome 3
Higher
Newton’s 2ndLaw of Motion
s = ut + 1/2at2 where s = distance & t = time.
Finding ds/dt means “diff in dist” “diff in time”
ie speed or velocity
so ds/dt = u + at
but ds/dt = v so we get
v = u + at
and this is Newton’s 1st Law of Motion
y = mx +c
Outcome 3
Higher
y = f(x)
A(a,b)
tangent
NB: at A(a, b) gradient of line = gradient of curve
gradient of line = m (from y = mx + c )
gradient of curve at (a, b) = f (a)
it follows that m = f (a)
Equation of Tangents
Outcome 3
Higher
Example 21
Find the equation of the tangent line to the curve
y = x3  2x + 1 at the point where x = 1.
Point: if x = 1 then y = (1)3  (2 X 1) + 1
= 1  (2) + 1
= 2 point is (1,2)
Gradient:dy/dx = 3x2  2
when x = 1 dy/dx = 3 X (1)2  2
m = 1
= 3  2 = 1
Outcome 3
Higher
Now using y  b = m(x  a)
point is (1,2)
m = 1
we get y  2 = 1( x + 1)
or y  2 = x + 1
or y = x + 3
Outcome 3
Higher
Example 22
Find the equation of the tangent to the curve y = 4 x2 at the point where x = 2. (x 0)
Also find where the tangent cuts the Xaxis and Yaxis.
Point: when x = 2 then y = 4 (2)2
= 4/4 = 1
point is (2, 1)
Gradient: y = 4x2 so dy/dx = 8x3
= 8 x3
when x = 2 then dy/dx = 8 (2)3
= 8/8 = 1
m = 1
Outcome 3
Higher
Now using y  b = m(x  a)
we get y  1 = 1( x + 2)
or y  1 = x + 2
or y = x + 3
Axes
Tangent cuts Yaxis when x = 0
so y = 0 + 3 = 3
at point (0, 3)
Tangent cuts Xaxis when y = 0
so 0 = x + 3 or x = 3
at point (3, 0)
Outcome 3
Higher
Example 23  (other way round)
Find the point on the curve y = x2  6x + 5 where the gradient of the tangent is 14.
gradient of tangent = gradient of curve
dy/dx =
2x  6
so 2x  6 = 14
2x = 20
x = 10
Put x = 10 into y = x2  6x + 5
Point is (10,45)
Giving y = 100  60 + 5
= 45
Higher
Consider the following graph of y = f(x) …..
y = f(x)
+
0
0
+

+
+
a
b
c
d
e
f

X
+
0
Outcome 3
Higher
In the graph of y = f(x)
The function is increasing if the gradient is positive
i.e. f (x) > 0 when x < b or d < x < f or x > f .
The function is decreasing if the gradient is negative
and f (x) < 0 when b < x < d .
The function is stationary if the gradient is zero
and f (x) = 0 when x = b or x = d or x = f .
These are called STATIONARY POINTS.
At x = a, x = c and x = e
the curve is simply crossing the Xaxis.
Outcome 3
Higher
Example 24
For the function f(x) = 4x2  24x + 19 determine the intervals when the function is decreasing and increasing.
f (x) = 8x  24
so 8x  24 < 0
f(x) decreasing when f (x) < 0
8x < 24
Check: f (2) = 8 X 2 – 24 = 8
x < 3
f(x) increasing when f (x) > 0
so 8x  24 > 0
8x > 24
Check: f (4) = 8 X 4 – 24 = 8
x > 3
Outcome 3
Higher
Example 25
For the curve y = 6x – 5/x2
Determine if it is increasing or decreasing when x = 10.
y = 6x  5 x2
= 6x  5x2
so dy/dx = 6 + 10x3
= 6 + 10 x3
when x = 10 dy/dx = 6 + 10/1000
= 6.01
Since dy/dx > 0 then the function is increasing.
Outcome 3
Higher
Example 26
Show that the function g(x) = 1/3x3 3x2 + 9x 10
is never decreasing.
g (x) = x2  6x + 9
= (x  3)(x  3)
= (x  3)2
Squaring a negative or a positive value produces a positive value, while 02 = 0. So you will never obtain a negative by squaring any real number.
Since (x  3)2 0 for all values of x
then g (x) can never be negative
so the function is never decreasing.
Outcome 3
Higher
Example 27
Determine the intervals when the function
f(x) = 2x3 + 3x2  36x + 41
is (a) Stationary (b) Increasing (c) Decreasing.
f (x) = 6x2 + 6x  36
Function is stationary when f (x) = 0
= 6(x2 + x  6)
ie 6(x + 3)(x  2) = 0
= 6(x + 3)(x  2)
ie x = 3 or x = 2
Outcome 3
Higher
We now use a special table of factors to determine when f (x) is positive & negative.
x
3
2

+
+
0
0
f’(x)
Function increasing when f (x) > 0
ie x < 3 or x > 2
Function decreasing when f (x) < 0
ie 3 < x < 2
Higher
y = f(x)
Consider this graph of y = f(x) again
0
+
0

+
+

c
+
a
b
X
+
0
Outcome 3
Higher
This curve y = f(x) has three types of stationary point.
When x = a we have a maximum turning point (max TP)
When x = b we have a minimum turning point (min TP)
When x = c we have a point of inflexion (PI)
Each type of stationary point is determined by the gradient ( f(x) ) at either side of the stationary value.
Outcome 3
Higher
Maximum Turning point
Minimum Turning Point
x
a
x
b
 0 +
f(x)
f(x)
+ 0 
Outcome 3
Higher
Rising Point of inflexion
Other possible type of inflexion
x
c
x
d
f(x)
+ 0 +
f(x)
 0 
Outcome 3
Higher
Example 28
Find the coordinates of the stationary point on the curve y = 4x3 + 1 and determine its nature.
SP occurs when dy/dx = 0
Using y = 4x3 + 1
so 12x2 = 0
if x = 0 then y = 1
x2 = 0
SP is at (0,1)
x = 0
Outcome 3
Higher
Nature Table
x
0
+
+
dy/dx
0
dy/dx = 12x2
So (0,1) is a rising point of inflexion.
Outcome 3
Higher
Example 29
Find the coordinates of the stationary points on the curve y = 3x4  16x3 + 24 and determine their nature.
Using y = 3x4  16x3 + 24
SP occurs when dy/dx = 0
So 12x3  48x2 = 0
if x = 0 then y = 24
12x2(x  4) = 0
if x = 4 then y = 232
12x2 = 0 or (x  4) = 0
x = 0 or x = 4
SPs at (0,24) & (4,232)
Outcome 3
Higher
Nature Table
4
x
0
dy/dx
 0  0 +
dy/dx=12x3  48x2
So (0,24) is a Point of inflexion
and (4,232) is a minimum Turning Point
Outcome 3
Higher
Example 30
Find the coordinates of the stationary points on the curve y = 1/2x4  4x2 + 2 and determine their nature.
Using y = 1/2x4  4x2 + 2
SP occurs when dy/dx = 0
if x = 0 then y = 2
So 2x3  8x= 0
if x = 2 then y = 6
2x(x2  4) = 0
if x = 2 then y = 6
2x(x + 2)(x  2) = 0
x = 0 or x = 2 or x = 2
SP’s at(2,6), (0,2) & (2,6)
Outcome 3
Higher
Nature Table
x
2
0
2
dy/dx
 0 + 0  0 +
So (2,6) and (2,6) are Minimum Turning Points
and (0,2) is a Maximum Turning Points
Higher
Note: A sketch is a rough drawing which includes important details. It is not an accurate scale drawing.
Process
(a) Find where the curve cuts the coordinate axes.
for Yaxis put x = 0
for Xaxis put y = 0 then solve.
(b) Find the stationary points & determine their nature as done in previous section.
(c) Check what happens as x +/ .
This comes automatically if (a) & (b) are correct.
Outcome 3
Higher
Dominant Terms
Suppose that f(x) = 2x3 + 6x2 + 56x  99
As x +/ (ie for large positive/negative values)
The formula is approximately the same as f(x) = 2x3
Graph roughly
As x + then y 
As x  then y +
Outcome 3
Higher
Example 31
Sketch the graph of y = 3x2 + 12x + 15
(a) Axes
If x = 0 then y = 15
If y = 0 then 3x2 + 12x + 15 = 0
( 3)
x2  4x  5 = 0
(x + 1)(x  5) = 0
x = 1 or x = 5
Graph cuts axes at (0,15) , (1,0) and (5,0)
Outcome 3
Higher
(b) Stationary Points
occur where dy/dx = 0
so 6x + 12 = 0
If x = 2
then y = 12 + 24 + 15 = 27
6x = 12
x = 2
Stationary Point is (2,27)
Nature Table
x
2
dy/dx
+ 0 
So (2,27)
is a Maximum Turning Point
Outcome 3
Higher
Summarising
as x + then y 
(c) Large values
as x  then y 
using y = 3x2
Y
Sketching
5
Cuts xaxis at 1 and 5
1
15
Cuts yaxis at 15
Max TP (2,27)
(2,27)
X
y = 3x2 + 12x + 15
Outcome 3
Higher
Example 32
Sketch the graph of y = 2x2 (x  4)
(a) Axes
If x = 0 then y = 0 X (4) = 0
If y = 0 then 2x2 (x  4) = 0
2x2 = 0 or (x  4) = 0
x = 0 or x = 4
(b) SPs
Graph cuts axes at (0,0) and (4,0) .
y = 2x2 (x  4)
= 2x3 + 8x2
SPs occur where dy/dx = 0
so 6x2 + 16x = 0
Outcome 3
Higher
2x(3x  8) = 0
2x = 0 or (3x  8) = 0
x = 0 or x = 8/3
If x = 0 then y = 0 (see part (a) )
If x = 8/3 then y = 2 X (8/3)2X (8/3 4) =512/27
nature
x
0
8/3

+

0
0
dy/dx
Outcome 3
Higher
Summarising
(c) Large values
as x + then y 
using y = 2x3
as x  then y +
Y
Sketch
Cuts x – axis at 0 and 4
4
0
Max TP’s at (8/3, 512/27)
(8/3, 512/27)
X
y = 2x2 (x – 4)
Outcome 3
Higher
Example 33
Sketch the graph of y = 8 + 2x2  x4
(a) Axes
If x = 0 then y = 8 (0,8)
If y = 0 then 8 + 2x2  x4 = 0
Let u = x2 so u2 = x4
Equation is now 8 + 2u  u2 = 0
(4  u)(2 + u) = 0
(4  x2)(2 + x2) = 0
or (2 + x) (2  x)(2 + x2) = 0
So x = 2 or x = 2 but x2 2
Graph cuts axes at (0,8) , (2,0) and (2,0)
Outcome 3
Higher
SPs occur where dy/dx = 0
(b) SPs
So 4x  4x3 = 0
4x(1  x2) = 0
4x(1  x)(1 + x) = 0
x = 0 or x =1 or x = 1
Using y = 8 + 2x2  x4
when x = 0 then y = 8
when x = 1 then y = 8 + 2  1 = 9 (1,9)
when x = 1 then y = 8 + 2  1 = 9 (1,9)
Outcome 3
Higher
nature
x
1
0
1
+

+

0
0
0
dy/dx
So (0,8) is a min TP while (1,9) & (1,9) are max TPs .
Outcome 3
Higher
Summarising
(c) Large values
Using y =  x4
Sketch is
as x + then y 
Y
as x  then y 
Cuts x – axis at 2 and 2
2
2
Cuts y – axis at 8
8
(1,9)
Max TP’s at
(1,9)
(1,9)
(1,9)
X
y = 8 + 2x2  x4
Higher
In the previous section on curve sketching we dealt with the entire graph.
In this section we shall concentrate on the important details to be found in a small section of graph.
Suppose we consider any graph between the points where x = a and x = b (i.e. a x b)
then the following graphs illustrate where we would expect to find the maximum & minimum values.
Outcome 3
Higher
y =f(x)
(b, f(b))
max = f(b) end point
(a, f(a))
min = f(a) end point
X
a b
Outcome 3
Higher
(c, f(c))
max = f(c ) max TP
y =f(x)
(b, f(b))
min = f(a) end point
(a, f(a))
x
a b
c
NB: a < c < b
Outcome 3
Higher
y =f(x)
max = f(b) end point
(b, f(b))
(a, f(a))
(c, f(c))
min = f(c) min TP
x
NB: a < c < b
c
a b
Outcome 3
Higher
From the previous three diagrams we should be able to see that the maximum and minimum values of f(x) on the closed interval a x b can be found either at the end points or at a stationary point between the two end points
Example 34
Find the max & min values of y = 2x3  9x2 in the interval where 1 x 2.
End points
If x = 1 then y = 2  9 = 11
If x = 2 then y = 16  36 = 20
Outcome 3
Higher
Stationary points
dy/dx = 6x2  18x
= 6x(x  3)
SPs occur where dy/dx = 0
6x(x  3) = 0
6x = 0 or x  3 = 0
x = 0 or x = 3
not in interval
in interval
If x = 0 then y = 0  0 = 0
Hence for 1 x 2 , max = 0 & min = 20
Outcome 3
Higher
Extra bit
Using function notation we can say that
Domain = {xR: 1 x 2 }
Range = {yR: 20 y 0 }
Higher
Note: Optimum basically means the best possible.
In commerce or industry production costs and profits can often be given by a mathematical formula.
Optimum profit is as high as possible so we would look for a max value or max TP.
Optimum production cost is as low as possible so we would look for a min value or min TP.
We can have for the given dimensions
Optimization
Outcome 3
Higher
Example 35
A rectangular sheet of foil measuring 16cm X 10 cm has four small squares each x cm cut from each corner.
16cm
x cm
10cm
x cm
NB: x > 0 but 2x < 10 or x < 5
ie 0 < x < 5
This gives us a particular interval to consider !
Outcome 3
Higher
By folding up the four flaps we get a small cuboid
x cm
(10  2x) cm
(16  2x) cm
The volume is now determined by the value of x so we can write
V(x) = x(16  2x)(10  2x)
= x(160  52x + 4x2)
= 4x3  52x2 +160x
We now try to maximize V(x) between 0 and 5
Outcome 3
Higher
End Points
Considering the interval 0 < x < 5
V(0) = 0 X 16 X 10 = 0
V(5) = 5 X 6 X 0 = 0
SPs
V '(x) = 12x2  104x + 160
= 4(3x2  26x + 40)
= 4(3x  20)(x  2)
Outcome 3
Higher
SPs occur when V '(x) = 0
ie 4(3x  20)(x  2) = 0
3x  20 = 0 or x  2 = 0
ie x = 20/3or x = 2
not in interval
in interval
When x = 2 then
V(2) = 2 X 12 X 6 = 144
We now check gradient near x = 2
Outcome 3
Higher
Example 36
When a company launches a new product its share of the market after x months is calculated by the formula
(x 2)
So after 5 months the share is
S(5) = 2/5 – 4/25
= 6/25
Find the maximum share of the market
that the company can achieve.
Outcome 3
Higher
End points
S(2) = 1 – 1 = 0
There is no upper limit but as x S(x) 0.
SPs occur where S (x) = 0
Outcome 3
Higher
rearrange
8x2 = 2x3
8x2  2x3 = 0
2x2(4 – x) = 0
x = 0 or x = 4
In interval
Out with interval
We now check the gradients either side of 4
Outcome 3
Higher
Nature
S (3.9 ) = 0.00337…
x 4
S (4.1) = 0.0029…

+
0
S (x)
Hence max TP at x = 4
And max share of market = S(4)
= 2/4 – 4/16
= 1/2 – 1/4
= 1/4
Equation of tangent line
Leibniz Notation
x
1
2
5

+
f’(x)
0
Straight Line
Theory
Max
Gradient at a point
f’(x)=0
Stationary Pts
Max. / Mini Pts
Inflection Pt
Graphs
f’(x)=0
Derivative
= gradient
= rate of change
Differentiation
of Polynomials
f(x) = axn
then f’x) = anxn1