DIFFERENTIATION
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DERIVATIVES
Slope of a curve
A curve has different slopes at each point. Let A, B, and C be different points of a curve f (x)
Where ðx is the small increase in x
ðy is the small increase in y
The slope of chord AC =
If C moves right up to A the chord AC becomes the tangent to the curve at A and the slope at A is the limiting value of
Therefore
The gradient at A is
or
This is known as differentiating by first principle
From the first principle
i) f(x)= x
ii) f(x)= x2
=
∴
iii) f(x) =x3
iv) f(x)= xn
By binomial series
In general
If
Example
Differentiate the following with respect to x
i)y = x2+3x
Solution
y =x2+3x
ii) 2x4+5
iii)
Differentiation of products functions [ product rule]
Let y =uv
Where u and v are functions of x
If x → x+ðx
u → u +ðu
v → v+ðv
y → ðy +y
y= uv ……i)
Therefore
y+ðy = [ u+ðu][v+ðv]
y+ðy = uv +uðv+vðu +ðuðv….ii
Subtract (i) from (ii)
δy =uðv +vðu +ðuðv
Therefore
Therefore
Therefore
If y= uv
Examples
Differentiate the following with respect to x
i) y = [ x2+3x] [4x+3]
ii) y = [
Solution
Y = [x2+3x] [4x+3]
Let u = x2+3x
V = 4x+3
Therefore
=4x2+12x+8x2+12x+6x+9
=12x2+30x+9
ii)Let u =
v = x2+2
Therefore
DIFFERENTIATION OF A QUOTIENT [QUOTIENT RULE]
Let y =
As
Exercise
Differentiate the following with respect to x
I.
II.
DIFFERENTIATION OF A FUNCTION [CHAIN RULE]
If y = f(u), where u = f(x)
Then
Therefore
PARAMETRIC EQUATIONS
Let y = f(t) , and x = g (t)
Example
I. Find
Solution
IMPLICIT FUNCTION
Implicit function is the one which is neither x nor y a subject e.g.
1) x2+y2 = 25
2) x2+y2+2xy=5
One thing to remember is that y is the function of x
Then
1.
∴
2. x2 +y2 + 2xy = 5
Exercise
Find
Differentiation of trigonometric functions
1) Let y = sin x….. i
Provided that x is measured in radian [small angle]
2. Let y = cos x …… (i)
3. Let
From the quotient rule
4. Let y = cot x
∴
5. y =
∴
Let y =cosec x
Therefore
Differentiation of inverses
1) Let y = sin-1x
x = sin y
2) Let y = cos-1 x
x= cos y
3) Let y = tan -1 x
x = tan y
Let y =
X =
∴
Exercises
Differentiate the following with respect to x.
i) Sin 6x
ii) Cos (4x2+5)
iii) Sec x tan 2 x
Differentiate sin2 (2x+4) with respect to x
Differentiate the following from first principle
i) Tan x
ii)
Differentiation of logarithmic and exponential functions
1- Let y = ln x
Example
Find the derivative of
Solution
By quotient rule
2- Let y =
Differentiation of Exponents
1) Let y = ax
If a function is in exponential form apply natural logarithms on both sides
i.e. ln y = ln ax
ln y =x ln a
2) Let
Since â„®x does not depend on h,then
Therefore
Example
Find the derivative of y = 105x
Solution
Y = 105x
Iny = In105x
Therefore
Exercise
Find the derivatives of the following functions
a) a) Y =
b) b) Y =
c) c)Y=
APPLICATION OF DIFFERENTIATION
Differentiation is applied when finding the rates of change, tangent of a curve, maximum and minimum etc
i) The rate of change
Example
Solution
Solution
Volume of the cone
V
The ratio of corresponding sides
Given
V=
So ,
Then,
Solution
Volume of horse trough
From the ratio of the corresponding sides
Solution
TANGENTS AND NORMALS
From a curve we can find the equations of the tangent and the normal
Example
i. Find the equations of the tangents to the curve y =2x2 +x-6 when x=3
Solution
(x. y)= (3, 5) is the point of contact of the curve with the tangent
But
Gradient of the tangent at the curve is
Example
ii. Find the equation of the tangent and normal to the curve y = x2 – 3x + 2 at the point where it cuts y axis
Solution
The curve cuts y – axis when x = 0
Slope of the tangent [m] = -3
Equation of the tangent at (0, 2) is
Slope of the normal
From; m1m2 = -1,Given m1=-3
Equation of the normal is
Exercise
Find the equation of the tangent to 2x2 – 3x which has a gradient of 1
Find the equations of the normal to the curve y = x2-5x +6 at the points where the curve cuts the x axis
Stationary points [turning points]
A stationary point is the one where by
Minimum turning point
Maximum turning point
Point of inflection
Nature of the curve of the function
At point A, a maximum value of a function occurs
At point B, a minimum value of a function occurs
At point C, a point of inflection occurs
At the point of inflection is a form of S bend
Note that
Points A, B and C are called turning points on the graph or stationary values of the function
Investigating the nature of the turning point
Minimum points
At turning points the gradient
Maximum points
At maximum period the gradient changes from positive to negative
i.e.
Point of inflection
This is the changes of the gradient from positive to positive
This is changes of the gradient from negative to negative.
Examples
Find the stationary points of the and state the nature of these points of the following functions
Y = x4 +4x3-6
Solution
At stationary points
Therefore,
Then the value of a function
At x = 0, y = -6
X= -3, y =-33
Stationary point at (0,-6) and (-3,-33)
At (0,-6)
Point (0,-6) is a point of inflection
At (-3,-33)
At (-3, -33) is a minimum point
Alternatively,
You test
Exercise
1) 1. Find and classify the stationary points of the following curves
a) (i) y = 2x-x2
b) (ii) y =
c) (iii) y= x2(x2– 8x)
2) 2. Determine the smallest positive value of x at which a point of inflection occurs on the graph of y = 3â„®2x cos (2x-3)
3) 3. If 4x2 + 8xy +9y2 8x – 24y +4 =0 show that when
x + y = 1. Hence find the maximum and minimum values of y
Example
1. A farmer has 100m of metal railing with which to form two adjacent sides of a rectangular enclosure, the other two sides being two existing walls of the yard meeting at right angles, what dimensions will give the maximum possible area?
Solution
Where, W is the width of the new wall
L is the length of the new wall
The length of the metal railing is 100m
2. An open card board box width a square base is required to hold 108cm3 what should be the dimensions if the area of cardboard used is as small as possible
Solution
Exercise
The gradient function of y = ax2 +bx +c is 4x+2. The function has a maximum value of 1, find the values of a, b, and c
MACLAURIN’S SERIES [from power series ]
Let f(x) = a1 +a2x+a3x2 +a4x3 +a5x4+ a6x5…….i
In order to establish the series we have to find the values of the constant co efficient a1, a2, a3, a4, a5, a6 etc
Put x = 0 in …i
Putting the expressions a1,a2,a3,a4,a5,………back to the original series and get
Examples
Expand the following
i) â„®x
ii) f(x) = cos x
Solution
i.
ii.
Exercise
Write down the expansion of
If x is so small that x3 and higher powers of x may be neglected, show that
TAYLOR’S SERIES
Taylor’s series is an expansion useful for finding an approximation for f(x) when x is close to a
By expanding f(x) as a series of ascending powers of (x-a)
f(x) = a0 +a1(x-a) +a2(x-a)2+a3(x-a)3 +……..
This becomes
Example
Expand
1.7321 And 5.50 as 0.09599c find the value of cos 54.5 to three decimal places
Solution
Obtain the expansion of
Introduction to partial derivative
Let f (x, y) be a differentiable function of two variables. If y kept constant and differentiates f (assuming f is differentiable with respect to x)
Keeping x constant and differentiate f with respect to y
Example
find the partial derivatives of fx and fy
If f(x, y) = x2y +2x+y
Solution
Find fx and fy if f (x,y) is given by
f(x, y) = sin(xy) +cos x
Solution
Exercise
1. find fx and fy if f(x,y) is given by
a)
b)
c)
d)
Suppose
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