Derivatives of Quotients

Example:
z = (tan(x) - 1) / (x + 1)2

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

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