Graphically, the derivative is the slope of a tangent line at a particular point. We can represent the derivatives of graphs as the slope of any function.
To find the derivative, let’s first use the original definition of the slope using the slope formula.
x1 and y1 are the x and y coordinates for P1 and x2 and y2 are the x and y coordinates for P2.
To find the derivative of a function, we are going to 1) take the slope formula and change the variables of x1, y1, x2, and y2 and 2) compute the limit of the modified slope formula.
Consider the figure above. Look at the x axis and focus on x1 and x2. The distance between x1 and x2 is called the change in x, or ∆x. Therefore, to go from x1 to x2, we need to move ∆x units along the x axis to the right. Therefore, x2 can also be written as x + ∆x. Also, we can rewrite x1 as simply x.
Look at the y axis and focus on y1 and y2. If our original function is F(x), and we want to find y1, we can plug in x1 (now called x) into the function to obtain y1. Therefore, F(x1) = F(x) = y1. So, we can rewrite y1 as F(x).
To find y2, plug in x2 (which is now x + ∆x) into F(x). Therefore, y2 = F(x2) = F(x + ∆x). Therefore, we can rewrite y2 as F(x + ∆x).
We have just discovered the following:
x1 = x
x2 = x + ∆x
y1 = F(x)
y2 = F( x + ∆x)
We can rewrite the slope equation as the following:
The line that connects the two points, P and Q, is called the secant line. The slope of this line is called the average rate of change. To find the derivative of the function, which is the instantaneous rate of change, we need the slope of the tangent line at a single point. This requires that ∆x, or the distance between x1 and x2, must become as close to zero as possible. When ∆x is zero, we will have the tangent line.
Therefore, the formal definition of the derivative is as follows:
Limit Definition of The Derivative:
The derivative of a function F(x) is the limit of the slope of the secant line as x approaches zero: The steps to find the derivative using the limit definition are as follows:

Example 1:
Find the derivative of F(x) = x^{2}
Solution:
Find the derivative of the function by using the limit definition:
First, find F( x + ∆x) and combine like terms if necessary.
F( x + ∆x) = ( x + ∆x)^{2} = ( x + ∆x)( x + ∆x) = x^{2} + x * ∆x + x * ∆x + ∆x^{2}
Then combine like terms:
x^{2}+ x * ∆x + x * ∆x + ∆x^{2}= x^{2}+ 2(x * ∆x) + ∆x^{2}
Second, find F(x). Fortunately, the problem already states that F(x)= x^{2}
Third, find F( x + ∆x) – F(x) and combine like terms if necessary.
F( x + ∆x) – F(x) = x^{2} + 2(x * ∆x) + ∆x^{2} – (x^{2 }) = 2(x * ∆x) + ∆x^{2} (the x^{2} and x^{2} cancelled out)
Fourth, factor out the ∆x from both terms.
2(x * ∆x) + ∆x^{2} = ∆x (2x + ∆x)
Plug in the results into the formula:
The ∆x term on the numerator and denominator can be cancelled out. When cancelled out, the result is:
Since we are finding limit of ∆x as it approaches 0, substitute 0 for ∆x:
The derivative of F(x) = x^{2} is F’(x) = 2x
Example 2:
Find the derivative of F(x) = 3x + 2
Solution:
Find the derivative of the function by using the limit definition:
First, find F( x + ∆x) and combine like terms if necessary.
F( x + ∆x) = 3(x + ∆x) +2 = 3x + 3∆x + 2
Second, find F(x). Fortunately, the problem already states that F(x)= 3x + 2
Third, find F( x + ∆x) – F(x) and combine like terms if necessary.
F( x + ∆x) – F(x) = 3x + 3∆x + 2 – (3x + 2) = 3x + 3∆x + 2 – 3x – 2
3x + 3∆x + 2 – 3x – 2 = 3∆x
Fourth, factor out the ∆x.
In this case, factoring the ∆x is not necessary because 3∆x is the only term left.
Plug in the results into the formula:
The ∆x term on the numerator and denominator can be cancelled out. When cancelled out, the result is:
In this case, the ∆x term has been eliminated entirely.
The derivative of F(x) = 3x + 2 is F’(x) = 3
Example 3:
Find the derivative of F(x) =
Solution:
Find the derivative of the function by using the limit definition:
First, find F( x + ∆x) and combine like terms if necessary.
F( x + ∆x) =
Second, find F(x).
Fortunately, the problem already states that F(x)=
Third, find F( x + ∆x) – F(x) and combine like terms if necessary.
F( x + ∆x) – F(x) = –
Fourth, factor out the ∆x from both terms.
In this case, we cannot factor the ∆x from the expression because it is embedded in the square root.
Fifth, plug in the results into the formula:
Because we have a square root expression in the numerator, , we need to multiply both the numerator and denominator by the conjugate of the same expression.
To find the conjugate, simply change the subtraction sign to an addition sign:
You do not have to FOIL the denominator. Leave it in this form because it will be easier to cancel the ∆x term on the top and bottom.
Summary of Section
The average rate of change, or the slope of the secant line between two points is The instantaneous rate of change, also known as the slope of the tangent line, occurs when x becomes increasingly smaller until it reaches zero. In other words, the tinier x becomes, the closer this is to the true instantaneous rate of change: 
References:
https://www.mathsisfun.com/calculus/derivativesintroduction.html
http://scidiv.bellevuecollege.edu/dh/Calculus_all/CC_2_1_IntroDerivative
http://www.opentextbookstore.com/appcalc/Chapter22.pdf
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