Derivatives of Quotients

Example:
f(q) = sin(2*q - 2) / (4 + ln(3*q))

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

f(q) = sin(2*q - 2)
----------
4 + ln(3*q)

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,

f(q) = sin(2*q - 2)
----------
4 + ln(3*q)
we can think of as
f(q) = g(q)
----------
h(q)
So the derivative is
f(q)' = ( g(q) )'
----------
h(q)
  = g '(q) h(q) - g(q) h '(q)
----------
( h(q) )2
  = ( sin(2*q - 2) )' ( 4 + ln(3*q) ) - ( sin(2*q - 2) ) ( 4 + ln(3*q) )'
----------
( 4 + ln(3*q) )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 - 2) )' = cos(2*q - 2) (2 - 0) (by the chain rule)
( 4 + ln(3*q) )' = 0 + 3*(3*q)-1 (by the derivative rule for sums, derivative rule for constants, and the chain rule)
so the finished derivative is
f(q)' = ( cos(2*q - 2) (2 - 0) ) ( 4 + ln(3*q) ) - ( sin(2*q - 2) ) ( 0 + 3*(3*q)-1 )
----------
( 4 + ln(3*q) )2
  = 2*cos(2*q - 2) (4 + ln(3*q)) - 3*sin(2*q - 2) (3*q)-1
----------
(4 + ln(3*q))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.