Derivatives of Quotients

Example:
Z = (y + 9)9 / sqrt(6*y - 1/4)

The first thing to notice when finding the derivative of this function is that it is a quotient, as shown below:

Z = (y + 9)9
----------
sqrt(6*y - 1/4)

The Derivative Rule for Quotients:

The derivative of a quotient is the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator, all divided by the denominator squared.
If
  z = ( f(x) )
----
g(x)
then the derivative of z is
  z ' = ( f(x) )'
----
g(x)
    = f '(x) g(x) - f(x) g '(x)  
----
( g(x) )2

So our example,

Z = (y + 9)9
----------
sqrt(6*y - 1/4)
we can think of as
Z = f(y)
----------
g(y)
So the derivative is
Z' = ( f(y) )'
----------
g(y)
  = f '(y) g(y) - f(y) g '(y)
----------
( g(y) )2
  = ( (y + 9)9 )' ( sqrt(6*y - 1/4) ) - ( (y + 9)9 ) ( sqrt(6*y - 1/4) )'
----------
( sqrt(6*y - 1/4) )2
and we just need to know each of the derivatives on the right-hand side of the equation. In this case these are
( (y + 9)9 )' = 9*(y + 9)8 (1 + 0) (by the chain rule)
( sqrt(6*y - 1/4) )' = (1/2)*(6*y - 1/4)-1/2 (6 - 0) (by the chain rule)
so the finished derivative is
Z' = ( 9*(y + 9)8 (1 + 0) ) ( sqrt(6*y - 1/4) ) - ( (y + 9)9 ) ( (1/2)*(6*y - 1/4)-1/2 (6 - 0) )
----------
( sqrt(6*y - 1/4) )2
  = 9*(y + 9)8 sqrt(6*y - 1/4) - 3*(y + 9)9 (6*y - 1/4)-1/2
----------
(sqrt(6*y - 1/4))2
[]


additional explanation for the quotient rule
see another quotient rule example
practice gateway test
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glarose@umich.edu
©2001 Gavin LaRose, University of Michigan Math Dept.