Home ADVANCED LEVEL TOPIC 9: DIFFERENTIATION ~ ADV MATHEMATICS FORM 5

TOPIC 9: DIFFERENTIATION ~ ADV MATHEMATICS FORM 5

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August 25, 2021

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DIFFERENTIATION

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

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

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
If y= uv

It is the product rule

Examples

Differentiate the following with respect to x

i) y = [ x2+3x] [4x+3]

ii) y = [

+2] [x2+2]

Solution

Y = [x2+3x] [4x+3]

Let u = x2+3x

= 2x+3

V = 4x+3

Therefore

=4x2+12x+8x2+12x+6x+9

=12x2+30x+9

ii)Let u =

→

v = x2+2

=2x

Therefore

DIFFERENTIATION OF A QUOTIENT [QUOTIENT RULE]

Let y =

where u and v are functions of x

Exercise

Differentiate the following with respect to x

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

if y = at2 and x = 2at

Solution

2at

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

∴

2. x2 +y2 + 2xy = 5

Exercise

Find

when x3 + y3 – 3xy2 = 8

Differentiation of trigonometric functions

1) Let y = sin x….. i

…(ii

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

The side of a cube is increasing at the rate of 6cm/s. find the rate of increase of the volume when the length of a side is 9cm

Solution

A hollow right circular cone is held vertex down wards beneath a tap leaking at the rate of 2cm3/s. find the rate of rise of the water level when the depth is 6cm given that the height of the cone is 18cm and its radius is 12cm.

Solution

Volume of the cone

The ratio of corresponding sides

Given

= 20cm3/s

So ,

Then,

A horse trough has triangular cross section of height 25cm and base 30cm and is 2m long. A horse is drinking steadily and when the water level is 5cm below the top is being lowered at the rate of 1cm/min find the rate of consumption in litres per minute

Solution

Volume of horse trough

From the ratio of the corresponding sides

A rectangle is twice as long as it is broad find the rate change of the perimeter when the width of the rectangle is 1m and its area is changing at the rate of 18cm2/s assuming the expansion is uniform

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

= 0 it involves:

∙ 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

changes from being negative to positive i.e.

Increasing as x- increases

Is positive at the minimum point

Is positive for minimum value of the function of (y)

Maximum points

At maximum period the gradient changes from positive to negative

i.e.

Decreases as x- increases

Is negative at the maximum value of the function (y)

Point of inflection

This is the changes of the gradient from positive to positive

Is positive just to the left and just to the left

This is changes of the gradient from negative to negative.

Is negative just to the left and just to the right

Is zero for a point of inflection i.e

Is zero for point of inflection

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

by taking values of x just to the right and left of the turning point

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

= 0,

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