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Basic mathematics differentiation application | PPTX
Worked examples and exercises are in the text
DIFFERENTIATION
APPLICATIONS
Worked examples and exercises are in the text
Tangents and normals to a curve at a given point
Tangent
The gradient of a curve, y = f
(x), at a point P with
coordinates (x1, y1) is given by
the derivative of y (the gradient
of the tangent) at the point:
The equation of the tangent can
then be found from the
equation:
1 1at ( , )
dy
x y
dx
1 1( ) where
dy
y y m x x m
dx
   
Worked examples and exercises are in the text
Tangents and normals to a curve at a given point
Example
Worked examples and exercises are in the text
Tangents and normals to a curve at a given point
Normal
The gradient of a curve, y = f (x), at a point P with
coordinates (x1, y1) is given by the derivative of y (the
gradient of the tangent) at the point:
The equation of the normal (perpendicular to the
tangent) can then be found from the equation:
1 1at ( , )
dy
x y
dx
1 1
1
( ) where
/
y y m x x m
dy dx
    
Worked examples and exercises are in the text
Tangents and normals to a curve at a given point
Example
Found the normal of the last exercise!
Worked examples and exercises are in the text
Tangents and normals to a curve at a given point
Exercise
Worked examples and exercises are in the text
Try!
Worked examples and exercises are in the text
Maximum and Minimum Value
See the figure below
Points A, B and C are called stationary points on the graph.
From the first derivative curve, we see that for stationary
points
Worked examples and exercises are in the text
The second derivative
Example
Worked examples and exercises are in the text
L’H 𝒐pital’s Rule for Forms of Type
𝟎
𝟎
and
∞
∞
Example
Both numerator and denominator have limit 0. Hence
Try! , ,
Worked examples and exercises are in the text
Total Revenue of an economic function
Exercise
*Diketahui fungsi total revenue TR = - Q2 + 15Q, pada
jumlah produk Q berapa akan mencapai revenue
maksimum ? Dan berapakah revenue maksimum ?
TRmax  MR = 0 , MR =
dTR
dQ
= -2Q + 15
-2Q + 15 = 0  Q = 7.5
Q = 7.5  TR = -(7.5)2 + 15.(7.5) = 56,25
Worked examples and exercises are in the text
Profit of an economic function
Exercise
*Diketahui fungsi total revenue TR = -0.75Q2 + 13.4Q dan
TC = 2/15Q3 – 2Q2 + 14Q +8. Tentukan besarnya produksi
agar diperoleh laba !
MR =
dTR
dQ
= -1.5Q + 13.4
MC =
dTC
dQ
= 0.4Q2 + 4Q + 14
Profit  MR = MC
Worked examples and exercises are in the text
MR = MC
-1.5Q + 13.4 = 0.4Q2 – 4Q + 14
0.4Q2 – 2.5Q + 0.6 = 0
Q =
−b± b2−4a.c
2a
=
−(−2.5)± (−2.5)2−4(0.4).(0.6)
2.(0.4)
=
2.5±2.3
0.8
 Q1 = 6 , Q2 = 0.25
Worked examples and exercises are in the text
Inflection Points
As you might guess, points where 𝑓′′
𝑥 = 0 𝑜𝑟 𝑤ℎ𝑒𝑟𝑒 𝑓′′ 𝑥 doesn’t exist are
candidates for points of inflection. We use the word candidate deliberately. Just as
a candidate for political office, may fail to be elected, so, for example, may a point
where 𝑓′′
𝑥 = 0 fail to be a point of inflection. Consider 𝑓 𝑥 = 𝑥4
, which has
the graph shown in figure below. It is true that 𝑓′′ 0 = 0 ; yet the origin isn’t a
point of inflection. Therefore, in searching for inflection points, we begin by
identifying those point where 𝑓′′ 𝑥 = 0 (and where 𝑓′′ 𝑥 does not exist). Then
we check to see if they really are inflection points.
Example
Find all the points of the inflection of 𝐹 𝑥 = 𝑥
1
3 + 2
Worked examples and exercises are in the text
Try!
Find the stationary points and points of inflexion on
the following curve

Basic mathematics differentiation application

  • 1.
    Worked examples andexercises are in the text DIFFERENTIATION APPLICATIONS
  • 2.
    Worked examples andexercises are in the text Tangents and normals to a curve at a given point Tangent The gradient of a curve, y = f (x), at a point P with coordinates (x1, y1) is given by the derivative of y (the gradient of the tangent) at the point: The equation of the tangent can then be found from the equation: 1 1at ( , ) dy x y dx 1 1( ) where dy y y m x x m dx    
  • 3.
    Worked examples andexercises are in the text Tangents and normals to a curve at a given point Example
  • 4.
    Worked examples andexercises are in the text Tangents and normals to a curve at a given point Normal The gradient of a curve, y = f (x), at a point P with coordinates (x1, y1) is given by the derivative of y (the gradient of the tangent) at the point: The equation of the normal (perpendicular to the tangent) can then be found from the equation: 1 1at ( , ) dy x y dx 1 1 1 ( ) where / y y m x x m dy dx     
  • 5.
    Worked examples andexercises are in the text Tangents and normals to a curve at a given point Example Found the normal of the last exercise!
  • 6.
    Worked examples andexercises are in the text Tangents and normals to a curve at a given point Exercise
  • 7.
    Worked examples andexercises are in the text Try!
  • 8.
    Worked examples andexercises are in the text Maximum and Minimum Value See the figure below Points A, B and C are called stationary points on the graph. From the first derivative curve, we see that for stationary points
  • 9.
    Worked examples andexercises are in the text The second derivative Example
  • 10.
    Worked examples andexercises are in the text L’H 𝒐pital’s Rule for Forms of Type 𝟎 𝟎 and ∞ ∞ Example Both numerator and denominator have limit 0. Hence Try! , ,
  • 11.
    Worked examples andexercises are in the text Total Revenue of an economic function Exercise *Diketahui fungsi total revenue TR = - Q2 + 15Q, pada jumlah produk Q berapa akan mencapai revenue maksimum ? Dan berapakah revenue maksimum ? TRmax  MR = 0 , MR = dTR dQ = -2Q + 15 -2Q + 15 = 0  Q = 7.5 Q = 7.5  TR = -(7.5)2 + 15.(7.5) = 56,25
  • 12.
    Worked examples andexercises are in the text Profit of an economic function Exercise *Diketahui fungsi total revenue TR = -0.75Q2 + 13.4Q dan TC = 2/15Q3 – 2Q2 + 14Q +8. Tentukan besarnya produksi agar diperoleh laba ! MR = dTR dQ = -1.5Q + 13.4 MC = dTC dQ = 0.4Q2 + 4Q + 14 Profit  MR = MC
  • 13.
    Worked examples andexercises are in the text MR = MC -1.5Q + 13.4 = 0.4Q2 – 4Q + 14 0.4Q2 – 2.5Q + 0.6 = 0 Q = −b± b2−4a.c 2a = −(−2.5)± (−2.5)2−4(0.4).(0.6) 2.(0.4) = 2.5±2.3 0.8  Q1 = 6 , Q2 = 0.25
  • 14.
    Worked examples andexercises are in the text Inflection Points As you might guess, points where 𝑓′′ 𝑥 = 0 𝑜𝑟 𝑤ℎ𝑒𝑟𝑒 𝑓′′ 𝑥 doesn’t exist are candidates for points of inflection. We use the word candidate deliberately. Just as a candidate for political office, may fail to be elected, so, for example, may a point where 𝑓′′ 𝑥 = 0 fail to be a point of inflection. Consider 𝑓 𝑥 = 𝑥4 , which has the graph shown in figure below. It is true that 𝑓′′ 0 = 0 ; yet the origin isn’t a point of inflection. Therefore, in searching for inflection points, we begin by identifying those point where 𝑓′′ 𝑥 = 0 (and where 𝑓′′ 𝑥 does not exist). Then we check to see if they really are inflection points. Example Find all the points of the inflection of 𝐹 𝑥 = 𝑥 1 3 + 2
  • 15.
    Worked examples andexercises are in the text Try! Find the stationary points and points of inflexion on the following curve