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Quotient Rule (PowerPoint)

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									Quotient Rule

   Robert Veasey & Joe Kelly
When to use the Quotient Rule

  When you need to find the derivative of an
   equation that has a complex numerator and
   denominator
  An example equation that would use the
   quotient rule is:

                        ( x  2)  5
                              2
               h( x ) 
                            3x  2
Generalized Quotient Rule

  The original function is f / g
  g is the denominator of the original function
  f is the numerator of the original function

  In English:
  It’s the bottom times the derivative of the top, minus the top times
  the derivative of the bottom, all over the bottom squared



             g  ( x) 
               f               g ( x) f ( x)  f ( x) g ( x)
                                            g ( x)
                                                     2
Sample Problem

                  ( x  2)  5   2           f is the numerator
         h( x ) 
                      3x  2                 g is the denominator



 Step 1: Bottom times the derivative of the top    g ( x) f ( x)


                        (3x  2)(2( x  2))

                      g ( x)
                                                      f ( x)
Sample Problem Cont.

  Step 2: Top times the derivative of the bottom     f ( x) g ( x)


                    (( x  2)  5)(3)
                                   2



                f ( x)                               g ( x)
Sample Problem Cont.

Step 3: Subtract the second step from the first step    g ( x) f ( x)  f ( x) g ( x)


       Step 1 was:     g ( x) f ( x)     (3x  2)(2( x  2))


       Step 2 was:      f ( x) g ( x)    (( x  2) 2  5)(3)

                                            Therefore Step 3 is


 g ( x) f ( x)  f ( x) g ( x)  (3x  2)(2( x  2))  (( x  2) 2  5)(3)
Sample Problem Cont.

                                g ( x) 
                                        2
 Step 4: Square the bottom 




                     g ( x)        (3x  2)
                                2                2
Finally the answer
                                              g ( x) f ( x)  f ( x) g ( x)
 Step 5: Divide step 3 by step 4 
                                                         g ( x) 
                                                                 2




   Step 3 was:  g ( x) f ( x)  f ( x) g ( x)           (3x  2)(2( x  2))  (( x  2) 2  5)(3)


                            g ( x)      (3 x  2) 2
                                    2
   Step 4 was: 



                                                        Therefore step 5 is


    g ( x) f ( x)  f ( x) g ( x)       (3x  2)(2( x  2))  (( x  2)2  5)(3)
                                        
               g ( x)                                 (3x  2)2
                       2
Answer




   
  f
    g
      
      ( x) 
             (3x  2)(2( x  2))  (( x  2) 2  5)(3)
                           (3x  2)  2
Summary

  It’s the bottom times the derivative of the top, minus the top times
  the derivative of the bottom, all over the bottom squared




     g  ( x) 
       f                g ( x) f ( x)  f ( x) g ( x)
                                      g ( x)
                                               2
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