Question 943836
<pre>
When doing the difference quotient, the best way to
begin is by finding
<b> 
           1
h(x+&#916;x) = ————
          x+&#916;x
</b>
Then find 
<b>
                  1      1
h(x+&#916;x) - h(x) = ———— - ———
                 x+&#916;x    x
</b>
Simplify that by getting an LCD:
<b>                 
                 x-(x+&#916;x)      
h(x+&#916;x) - h(x) = ——————————
                  x(x+&#916;x)    
</b>
Simplify further:
<b>
                  x-x-&#916;x      
h(x+&#916;x) - h(x) = ————————
                  x(x+&#916;x)
</b>
Simplify further:
<b<
                    -&#916;x      
h(x+&#916;x) - h(x) = ————————
                  x(x+&#916;x)
</b>
Next divide by &#916;x on the left by putting &#916;x 
under the left side.  Divide by &#916;x on the
right by multiplying by the reciprocal 1/(&#916;x): 
 
<b>
h(x+&#916;x) - h(x)        -&#916;x       1
——————————————  =  ———————— • —————— 
      &#916;x            x(x+&#916;x)     &#916;x
</b>
Cancel the &#916;x's on the right:
<b>
h(x+&#916;x) - h(x)        -<s>&#916;x</s>       1
——————————————  =  ———————— • —————— 
      &#916;x            x(x+&#916;x)     <s>&#916;x</s>
</b>
and you have:

<b>
h(x+&#916;x) - h(x)        -1       
——————————————  =  ————————  
      &#916;x            x(x+&#916;x)
</b>
And you can move the negative sign out in front:
<b>
h(x+&#916;x) - h(x)           1       
——————————————  =  - ————————  
      &#916;x              x(x+&#916;x)    
</b>
Edwin</pre>