Derivatives of Quotients

Example:
z = 2y / (sin(y) + 1)

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

z = 2y
----------
sin(y) + 1

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 = 2y
----------
sin(y) + 1
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
  = ( 2y )' ( sin(y) + 1 ) - ( 2y ) ( sin(y) + 1 )'
----------
( sin(y) + 1 )2
and we just need to know each of the derivatives on the right-hand side of the equation. In this case these are
( 2y )' = (ln(2))*2y (by the derivative rules for basic functions)
( sin(y) + 1 )' = cos(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' = ( (ln(2))*2y ) ( sin(y) + 1 ) - ( 2y ) ( cos(y) + 0 )
----------
( sin(y) + 1 )2
  = (ln(2))*2y (sin(y) + 1) - 2y cos(y)
----------
(sin(y) + 1)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.