Derivatives of Quotients

Example:
P(z) = e3*z / (cos(z) - 2)

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

P(z) = e3*z
----------
cos(z) - 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,

P(z) = e3*z
----------
cos(z) - 2
we can think of as
P(z) = f(z)
----------
g(z)
So the derivative is
P(z)' = ( f(z) )'
----------
g(z)
  = f '(z) g(z) - f(z) g '(z)
----------
( g(z) )2
  = ( e3*z )' ( cos(z) - 2 ) - ( e3*z ) ( cos(z) - 2 )'
----------
( cos(z) - 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
( e3*z )' = 3*e3*z (by the chain rule)
( cos(z) - 2 )' = (-1)*sin(z) - 0 (by the derivative rule for sums, derivative rules for basic functions, and the derivative rule for constants)
so the finished derivative is
P(z)' = ( 3*e3*z ) ( cos(z) - 2 ) - ( e3*z ) ( (-1)*sin(z) - 0 )
----------
( cos(z) - 2 )2
  = 3*e3*z (cos(z) - 2) + e3*z sin(z)
----------
(cos(z) - 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.