Derivatives of Quotients

Example:
p = sin(2*q + 3) / (q4 - 3*q - 5)

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

p = sin(2*q + 3)
----------
q4 - 3*q - 5

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 = sin(2*q + 3)
----------
q4 - 3*q - 5
we can think of as
p = f(q)
----------
g(q)
So the derivative is
p' = ( f(q) )'
----------
g(q)
  = f '(q) g(q) - f(q) g '(q)
----------
( g(q) )2
  = ( sin(2*q + 3) )' ( q4 - 3*q - 5 ) - ( sin(2*q + 3) ) ( q4 - 3*q - 5 )'
----------
( q4 - 3*q - 5 )2
and we just need to know each of the derivatives on the right-hand side of the equation. In this case these are
( sin(2*q + 3) )' = cos(2*q + 3) (2 + 0) (by the chain rule)
( q4 - 3*q - 5 )' = 4*q3 - 3 - 0 (by the derivative rule for sums, power rule, rule for constant multiples, and the derivative rule for constants)
so the finished derivative is
p' = ( cos(2*q + 3) (2 + 0) ) ( q4 - 3*q - 5 ) - ( sin(2*q + 3) ) ( 4*q3 - 3 - 0 )
----------
( q4 - 3*q - 5 )2
  = 2*cos(2*q + 3) (q4 - 3*q - 5) - sin(2*q + 3) (4*q3 - 3)
----------
(q4 - 3*q - 5)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.