- Introduction
- Delta Notation
- Derivative of a Polynomial.
- Equations of Tangents and Normal to a Curve.
- Stationary Points
- Turning Points
- Applications of Differentiation

## Introduction

- Differentiation is generally about rate of change

**Example**

- If we want to get the gradient of the curve y = x
^{2 }at a general point (x ,y ).We note that a general point on the curve y = x^{2 }will have coordinates of the form ( x ,x^{2 }). - The gradient of the curve y= x
^{2 }at a general point ( x, y ) can be established as below. - If we take a small change in x , say h. This gives us a new point on the curve with co-ordinates [(x +h), (x + h)
^{2}]. So point Q is [(x +h), (x + h)^{2}] while point P is ( x,x^{2}). - To find the gradient of PQ = change in Y

change in X

Change in y = (x + h)^{2 }- x^{2}Change in x = (x + h) - x

Gradient = (x + h)^{2}- x^{2}(x + h ) - x

= x^{2}+2xh+ h^{2}-x^{2}x +h-x

=2xh + h^{2}h

= 2x + h

- By moving Q as close to p as possible, h becomes sufficiently small to be ignored. Thus, 2x +h becomes 2x. Therefore, at general point (x,y)on the curve y = x
^{2},the gradient is 2x. - 2x is called the gradient function of the curve y = x
^{2}.We can use the gradient function to determine the gradient of the curve at any point on the curve. - In general, the gradient function of y = x
^{n }is given by nx^{n-1},where n is a positive integer. The gradient function is called the derivative or derived function and the process of obtaining it is called**differentiation**. - The function y=x
^{5 }becomes 5x^{5-1}= 5x^{4}when we differentiate it

## Delta Notation

- A small increase in x is usually denoted by ∆x similarly a small increase in y is denoted by ∆y. Let us consider the points P (x ,y) and Q [(x + ∆x),(y + ∆y) on the curve y = x
^{2} **Note;**

X is a single quantity and not a product of ∆ and x .similarly ∆y is a single quantity.

The gradient of PQ, ∆y = (x+ ∆x)^{2 }- x^{2}

∆x (x+ ∆x) - x

= x^{2}+2x∆x+(∆x)^{2 }- x^{2}

x+∆x-x

= 2x +∆x- As ∆x tends to zero;
- ∆x can be ignored
^{∆y}/_{∆x}gives the derivative which is denoted by^{dy}/_{dx}- thus
^{dy}/_{dx}= 2x

- When we find
^{dy}/_{dx}, we say we are differentiating with respect to x, For example given y = x^{4}; then = 4x^{3} ^{}In general the derivatives of y =ax^{n }is nax^{n-1 }e.g. y = 5x^{2 }= 10x , y = 6x^{3 }= (6 x 3)x^{3-1 }= 18x^{2}

## Derivative of a Polynomial.

- A polynomial in x is an expression of the form a
_{0}x^{n}+a_{1}x^{n-1 }+……………a_{n}; where a_{0},a_{1}………. a_{n}are constants - To differentiate a polynomial function, all you have to do is multiply the coefficients of each variable by their corresponding exponents/powers, subtract each exponent/powers by one , and remove any constants.

### Steps Involved in Solving Polynomial

#### Identify the variable terms and constant terms in the equation.

- A variable term is any term that includes a variable and a constant term is any term that has only a number without a variable.
- Find the variable and constant terms in this polynomial function: y = 5x
^{3 }+ 9x^{2 }+ 7x + 3- The variable terms are 5x
^{3}, 9x^{2}, and 7x - The constant term is 3

- The variable terms are 5x

#### Multiply the coefficients of each variable term by their respective powers.

- Their products will form the new coefficients of the differentiated equation. Once you find their products, place the results in front of their respective variables. For example:
- 5x
^{3}= 5x^{3}= 1 5 - 9x
^{2}= 9x^{2}= 1 8 - 7x = 7x
^{1}= 7

- 5x

#### Lower each exponent by one.

- To do this, simply subtract 1 from each exponent in each variable term. Here's how you do it:
- 5x
^{3 }= 15x2 - 9x
^{2 }= 18x - 7x
^{1-1 }= 7

- 5x

#### Replace the old coefficients and old exponents/powers with their new counterparts.

- To finish differentiating the polynomial equation, simply replace the old coefficients with their new coefficients and replace the old powers with their values lowered by one.
- The derivative of constants is zero so you can omit 3, the constant term, from the final result.
- The derivative of the polynomial y = 15x
^{2 }+ 18x + 7 - In general, the derivative of the sum of a number of terms is obtained by differentiating each term in turn.

**Examples**

Find the derived function of each of the following

- S=2t
^{3 }- 3t^{2 }+ 4t - A = V
^{2 }- 2V + 10

**Solution**

- dS = 6t
^{2 }- 6t + 4

dt - dA = 2v - 2

dv

## Equations of Tangents and Normal to a Curve.

- The gradient of a curve is the same as the gradient of the tangent to the curve at that point. We use this principle to find the equation of the tangent to a curve at a given point.

**Example**

Find the equation of the tangent to the curve;

y = x^{3 }+ 2x + 1 at (1,4)

**Solution**

dy = 3x^{2}+ 2

dx

At the point (1,4), the gradient is 3 x 1 2+ 2 = 5(we have substituted the value of x with 1)

We want the equation of straight line through (1 , 4) whose gradient is 5.

Thus y-4 = 5

x-1

y - 4 = 5x - 5

y = 5x - 1 (this is the equation of the tangent)

- A normal to a curve at appoint is the line perpendicular to the tangent to the curve at the given point.
- In the example above the gradient of the tangent of the tangent to the curve at (1 , 4) is 5. Thus the gradient of the normal to the curve at this point is -
^{1}/_{5}. - Therefore, equation of the normal is:

y-4= -^{1}/_{5}

x-1

5(y – 4) = - 1(x – 1)

y =-x+21

5

**Example**

Find the equation of the normal to the curve y =x^{3 }- 2x - 1 at (1, - 2)

**Solution**

y = x^{3 }- 2x - 1

dy = 3x^{2} - 2

dx

At the point ( 1 ,-2) gradient of the tangent line is 1 .Therefore the gradient of the normal is -1 .the required equation is

y-(-2) = - 1

x-1

y+2 = -1

x-1

y + 2 = - x + 1

The equation of the normal is y = -x -1

## Stationary Points

**Note;**

- In each of the points A ,B and C the tangent is horizontal meaning at these points the gradient is zero. So
^{dy}/_{dx}= 0 at points A,B, C. - Any point at which the tangent to the graph is horizontal is called a stationary point. We can locate stationary points by looking for points at which
^{dy}/_{dx}= 0.

## Turning Points

- The point at which the gradient changes from negative through zero to positive is called
**minimum point**while the point which the gradient changes from positive through zero to negative is called**maximum point** - In the figure above A is the maximum while B is the minimum.

**Minimum point**

- Gradient moves from negative through zero to positive.

**Maximum point**

- Gradient moves from positive through zero to negative.

The maximum and minimum points are called **turning points**.

A point at which the gradient changes from positive through zero to positive or from negative zero to negative is called **point of inflection**.

**Example**

Identify the stationary points on the curve y =x^{2 }- 3x + 2.for each point, determine whether it is a maximum, minimum or a point of inflection.

**Solution**

y = x^{3 }- 3x + 2

dy =3x^{2 }- 3

dx

At stationary point,^{dy}/_{dx} = 0

Thus 3x^{2 }- 3 = 0

3(x^{2}-1) = 0

3(x+1)(x-1) = 0

x = - 1 or x = 1

when = -1 ,y = 4

when x = 1 , y = 0

Therefore, stationary points are (-1, 4) and (1 ,0).

Consider the sign of the gradient to the left and right of x = 1

x | 0 | 1 | 2 |

dy dx |
-3 | 0 | 9 |

Diagrammaticrepresentation | \ | ___ | / |

Therefore (1, 0) is a minimum point.

Similarly, sign of gradient to the left and right of x = -1 gives

x | -2 | -1 | 0 |

dy dx |
9 | 0 | -3 |

Diagrammatic representation | / | ___ | \ |

Therefore (-1 , 4) is a maximum point.

**Example**

Identify the stationary points on the curve y =1 + 4x^{3 }- x^{4}.Determine the nature of each stationary point.

**Solution**

y =1 + 4x^{3 }− x^{4}

dy= 12x^{2 }−4x^{3}

dx

At stationary points,^{dy}/_{dx} = 0

12x^{2 }- 4x^{3 }= 0

4x^{2}(3−x) = 0

x = 0 or x = 3

Stationary points are (0, 1) and (3, 28)

Therefore (0, 1 ) is a point of inflection while (3, 28) is a maximum point.

## Applications of Differentiation

### Calculation of Velocity

If the displacement, S is expressed in terms of time t, then the velocity is v =^{dS}/_{dt}

**Example**

The displacement, S metres, covered by a moving particle after time, t seconds, is given by S =2t^{3 }+ 4t^{2 }- 8t + 3.Find

- Velocity at :
- t = 2
- t= 3

- Instant at which the particle is at rest.

**Solution**

**S =2t**^{3 }+ 4t^{2 }- 8t + 3

The gradient function is given by;

V =dS

dt

= 6t^{2 }+ 8t - 8

- velocity
- at t = 2 is ;

v = 6 x 22 + 8 x 2 - 8

= 24 + 16 – 8

=32m/s - at t = 3 is ;

v = 6 x 3 2 + 8 x 3 - 8

= 54 + 24 – 8

=70m/s

- at t = 2 is ;
- the particle is at rest when v is zero

6t2 + 8t - 8 = 0

2(3t2+4t-4) = 0

2(3t-2)(t+2) = 0

t =^{2}/_{3}or t = - 2

It is not possible to have t = -2

The particle is therefore at rest at t = ^{2}/_{3} seconds

### Calculation of Acceleration

Acceleration is found by differentiating an equation related to velocity. If velocity v , is expressed in terms of time, t , then the acceleration, a, is given by a = ^{dv}/_{d}

**Example**

A particle moves in a straight line such that is its velocity v ms-1 after t seconds is given by v = 3 + 10t − t^{2}.

Find

- the acceleration at :
- t =1 sec
- t =3 sec

- t =1 sec
- the instant at which acceleration is zero

**Solution**

- v = 3 + 10t + t
^{2}

a = dv = 10 - 2t

dt- At t = 1 sec a = 10 – 2 x 1

= 8 ms^{-2} ^{}At t = 3 sec a = 10 – 2 x 3

= 4 ms^{-2}

- At t = 1 sec a = 10 – 2 x 1
^{}Acceleration is zero when^{dy}/_{dt}= 0

Therefore, 10 – 2t = 0 hence t = 5 seconds

**Example**

A closed cylindrical tin is to have a capacity of 250π ml. if the area of the metal used is to be minimum, what should the radius of the tin be?

**Solution**

**Let the total surface area of the cylinder be A cm**^{2},radius r cm and height h cm.

Then, A = 2πr^{2 }+ 2 πrh

Volume = 2πr^{2}h = 250πcm^{2}

πr^{2}h = 250π

Making h the subject,

h = 250π

πr^{2}

= 250

r^{2}

Put h =^{250}/r_{2} in the expression for surface area to get;

A = 2 πr^{2 }+ 2 πr×250r^{2}

=2πr^{2 }+ 500πr^{−}^{1}

dA = 4πr - 500πr^{−2}

dr

For minimum surface area,^{dA}/_{dr} = 0

4πr - ^{500π}/_{r2} = 0

4πr^{3 }- 500π = 0

4r^{3 }= 500

r^{3 }= ^{500}/_{4} = 125

r = ^{3}√125

= 5

Therefore the minimum area when r = 5 cm

**Example**

A farmer has 100 metres of wire mesh to fence a rectangular enclosure. What is the greatest area he can enclose with the wire mesh?

**Solution**

Let the length of the enclosure be x m. Then the width is 100-2x m = (50-x)m

2

Then the area A of the rectangle is given by;

A = x(50 –x)m^{2}

= 50x – x^{2} m^{2}

For maximum or minimum area,

dA = 0

dx

Thus, 50 – 2x = 0

x = 25

The area is maximum when x = 25 m

That is A = 50 X 25 – (25)^{2}

= 625 m^{2}.

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