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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
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

 
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
powerpoint.close();

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