Derivatives of Quotients

Example:
G(x) = 2x / sin(2*x + 1)

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

G(x) = 2x
----------
sin(2*x + 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,

G(x) = 2x
----------
sin(2*x + 1)
we can think of as
G(x) = f(x)
----------
g(x)
So the derivative is
G(x)' = ( f(x) )'
----------
g(x)
  = f '(x) g(x) - f(x) g '(x)
----------
( g(x) )2
  = ( 2x )' ( sin(2*x + 1) ) - ( 2x ) ( sin(2*x + 1) )'
----------
( sin(2*x + 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
( 2x )' = (ln(2))*2x (by the derivative rules for basic functions)
( sin(2*x + 1) )' = cos(2*x + 1) (2 + 0) (by the chain rule)
so the finished derivative is
G(x)' = ( (ln(2))*2x ) ( sin(2*x + 1) ) - ( 2x ) ( cos(2*x + 1) (2 + 0) )
----------
( sin(2*x + 1) )2
  = (ln(2))*2x sin(2*x + 1) - 2*2x cos(2*x + 1)
----------
(sin(2*x + 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.