Difference quotients were introduced in an earlier lesson:
Prerequisites: function review, difference quotients.
Evaluating difference quotients requires a high degree of comfort working with
functions and function notation,
and this earlier lesson offers a thorough review.
Then, this current lesson gives practice
with difference quotient problems involving reciprocal functions and
square roots.
DIFFERENCE QUOTIENTS
Let
[beautiful math coming... please be patient]$\,f\,$ be a
function and let
[beautiful math coming... please be patient]$\,h\,$ be a nonzero number.
A difference quotient is an expression of the form [beautiful math coming... please be patient] $$\frac{f(x+h)f(x)}h\ ,$$ which is a simplified version of: $$\frac{f(x+h)f(x)}{(x+h)x}$$ This expression gives the slope of the line through the points: Since the expression is a quotient (division) of differences (subtractions), the name difference quotient is appropriate. 

$\displaystyle\frac{f(x+h)f(x)}h$  the desired difference quotient 
$\displaystyle = \frac{\frac 1{x+h}  \frac 1x}h$  function evaluation; use definition of $\,f\,$ 
$\displaystyle = \frac{\frac 1{x+h}\cdot\frac xx  \frac 1x\cdot\frac{x+h}{x+h}}h$  get a common denominator 
$\displaystyle = \frac{\frac x{x(x+h)}  \frac {x+h}{x(x+h)}}h$  simplify 
$\displaystyle = \frac{\frac {x  (x+h)}{x(x+h)}}h$  write the numerator as a single fraction 
$\displaystyle = {\frac {xxh}{x(x+h)}}\cdot\frac 1h$ 
distributive law; and, dividing by $\,h\,$ is the same as multiplying by $\,\frac 1h\,$ 
$\displaystyle = {\frac {h}{x(x+h)h}}$  $x  x = 0\,$; and multiply across 
$\displaystyle = {\frac {1}{x(x+h)}}$ 
cancel: since $\,h\ne 0\,$, $\,\frac hh = 1\,$ Now, the factor of $\,h\,$ is gone in the denominator! 
As $\,h\,$ approaches zero, $\,\displaystyle {\frac {1}{x(x+h)}}\,$ approaches
$\,\displaystyle\frac{1}{x(x+0)}\,$, which equals
$\,\displaystyle \frac 1{x^2}\,$. The formula $\displaystyle\,\frac{1}{x^2}\,$ gives the slopes of the tangent lines to the graph of $\displaystyle\,f(x) = \frac 1x\,$! For example, the slope of the tangent line to $\displaystyle\,f(x) = \frac 1x\,$ at the point $\displaystyle\,(3,\frac 13)\,$ is $\displaystyle\,\frac{1}{3^2} = \frac 19\,$. Click ‘Submit’ on the WolframAlpha widget below to explore the power of Wolfram Alpha! You can create your own widgets by clicking here! 
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On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
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