Derivatives of Quotients

Example:
z = (ln(5*y) - 1/3) / (cos(y) + 1/2)

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

z = ln(5*y) - 1/3
----------
cos(y) + 1/2

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 = ln(5*y) - 1/3
----------
cos(y) + 1/2
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
  = ( ln(5*y) - 1/3 )' ( cos(y) + 1/2 ) - ( ln(5*y) - 1/3 ) ( cos(y) + 1/2 )'
----------
( cos(y) + 1/2 )2
and we just need to know each of the derivatives on the right-hand side of the equation. In this case these are
( ln(5*y) - 1/3 )' = 5*(5*y)-1 - 0 (by the derivative rule for sums, chain rule, and the derivative rule for constants)
( cos(y) + 1/2 )' = (-1)*sin(y) + 0 (by the derivative rule for sums, derivative rules for basic functions, and the derivative rule for constants)
so the finished derivative is
z' = ( 5*(5*y)-1 - 0 ) ( cos(y) + 1/2 ) - ( ln(5*y) - 1/3 ) ( (-1)*sin(y) + 0 )
----------
( cos(y) + 1/2 )2
  = 5*(5*y)-1 (cos(y) + 1/2) + (ln(5*y) - 1/3) sin(y)
----------
(cos(y) + 1/2)2
[]

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