Derivatives of Quotients

Example:
h(z) = (tan(z) - 1) / (z3 + 1)

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

h(z) = tan(z) - 1
----------
z3 + 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,

h(z) = tan(z) - 1
----------
z3 + 1
we can think of as
h(z) = f(z)
----------
g(z)
So the derivative is
h(z)' = ( f(z) )'
----------
g(z)
  = f '(z) g(z) - f(z) g '(z)
----------
( g(z) )2
  = ( tan(z) - 1 )' ( z3 + 1 ) - ( tan(z) - 1 ) ( z3 + 1 )'
----------
( z3 + 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
( tan(z) - 1 )' = 1 / (cos(z))2 - 0 (by the derivative rule for sums, derivative rules for basic functions, and the derivative rule for constants)
( z3 + 1 )' = 3*z2 + 0 (by the derivative rule for sums, power rule, and the derivative rule for constants)
so the finished derivative is
h(z)' = ( 1 / (cos(z))2 - 0 ) ( z3 + 1 ) - ( tan(z) - 1 ) ( 3*z2 + 0 )
----------
( z3 + 1 )2
  = (1 / (cos(z))2) (z3 + 1) - 3*z2 (tan(z) - 1)
----------
(z3 + 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.