Difference (Newton) Quotient Made Simple

To set up a difference quotients for a given function requires an understanding of a function notation. Given the function f(x)=4x^2-3x-7. This notation is read as “f of x equals..” This implies that the value of the function, that is the y-value depends upon the replacement for x. We get the numerical value for the function by substituting a number for ‘x’. If a non-numerical quantity is substituted for ‘x’, we get an expression rather than a numerical value. One important point to be remembered is careful use of parenthesis which is essential. For instance, f(x)=4x^2-3x-7; f(3)=4(3)^2 – 3(3)-7=36-9-7=20

Derivative Quotients  at x for a function f is given by, [f(x+h)-f(x)]/h. Sometimes it is written using delta(x) for the change in x and delta(y) for change in y; delta(x)=h and delta(y)=f(x+delta(x))- f(x). The Difference Quotient is so called as it involves the operations subtraction and division. The common forms of Difference Quotient are as follows:
1. [f(x+h) – f(x)]/h
2. [f(a+h) – f(a)]/h
3. [f(x+delta(x)) – f(x)]/delta(x)

Simplifying Difference Quotient
The difference quotient is simplified to get a h or delta(x) in the denominator which can be canceled to get the final value. Let us consider a difference quotient example to understand the step involved in simplifying difference quotient ; f(x) = 4x^2-3x-7. First we need to find the function f(x+h),which we can get by substituting (x+h) in all x in the given function.  f(x+h)= 4(x+h)^2 – 3(x+h) – 7 = 4(x^2+2xh+h^2) – 3x – 3h – 7= 4x^2+8xh + 4h^2-3x – 3h – 7. Subsituting f(x+h) in the difference quotient, we get,
Difference Quotient = [f(x+h) – f(x)]/h
       = {[4x^2+8xh+4h^2-3x-3h-7] – [4x^2-3x-7]}/h
        = 4x^2+8xh+4h^2-3x-3h-7 -4x^2+3x+7]/h  (opening the parenthesis)
On simplification, we get
        = [8xh+4h^2-3h]/h  
        = h[8x +4h-3]/h            (taking h common)
        = (8x +4h-3)     (canceling h)

Difference Quotient Example
Given function, f(x) = 2x^2-1
First we need to calculate f(x+h) which is got by substituting (x+h) in all x of the function
f(x+h) = 2(x+h)^2 -1 = 2(x^2+2xh+h^2) -1 = 2x^2+ 4xh+2h^2 -1
Next we substitute f(x+h) and f(x) in difference quotient
 [f(x+h) – f(x)]/h ={[2x^2 +4xh +2h^2-1]- [2x^2 -1]}/h
= [2x^2+4xh +2h^2-2x^2+1]/h
= [4xh+2h^2]/h  (combining like terms)
= h[4x +2h]/h      (taking h common)
= [4x +2h] (canceling h)

Newton Quotient
The difference quotient is attributed to Sir Isaac Newton and hence given the name Newton Quotient. The slope of a line through the points [(x+h),f(x+h)] and [x, f(x)] is given by [f(x+h) – f(x)]/h. This expression is the Newton Quotient or Newton’s difference quotient.
Newton Quotient  =[f(x+h) – f(x)]/h