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 coordinates [(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 +hx
=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^{n1},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^{51}= 5x^{4} when we differentiate it
Delta Notation
 A small increase in x is usually denoted by âx similarly a small increase in y is denoted by ây. Let us consider the points P (x ,y) and Q [(x + âx),(y + ây) on the curve y = x^{2}

Note;
X is a single quantity and not a product of â and x .similarly ây is a single quantity.
The gradient of PQ, ây = (x+ âx)^{2 } x^{2}
âx (x+ âx)  x
= x^{2}+2xâx+(âx)^{2 } x^{2}
x+âxx
= 2x +âx 
As âx tends to zero;
 âx can be ignored
 ^{ây}/_{âx} 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^{n1 }e.g. y = 5x^{2 }= 10x , y = 6x^{3 }= (6 x 3)x^{31 }= 18x^{2}
Derivative of a polynomial.
 A polynomial in x is an expression of the form a_{0}x^{n} +a_{1}x^{n1 }+……………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
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
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^{11 }= 7
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 y4 = 5
x1
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:
y4= ^{1}/_{5}
x1
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
x1
y+2 = 1
x1
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)(x1) = 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.
Application of Differentiation in calculation of velocity and acceleration.
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+4t4) = 0
2(3t2)(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
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 ms1 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 1002x m = (50x)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}.