Derivatives of Quotients

Example:
y = (ex - (e)3) / (x - 4)5

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

y = ex - (e)3
----------
(x - 4)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,

y = ex - (e)3
----------
(x - 4)5
we can think of as
y = f(x)
----------
g(x)
So the derivative is
y' = ( f(x) )'
----------
g(x)
  = f '(x) g(x) - f(x) g '(x)
----------
( g(x) )2
  = ( ex - (e)3 )' ( (x - 4)5 ) - ( ex - (e)3 ) ( (x - 4)5 )'
----------
( (x - 4)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
( ex - (e)3 )' = ex - 0 (by the derivative rule for sums, derivative rules for basic functions, and the derivative rule for constants)
( (x - 4)5 )' = 5*(x - 4)4 (1 - 0) (by the chain rule)
so the finished derivative is
y' = ( ex - 0 ) ( (x - 4)5 ) - ( ex - (e)3 ) ( 5*(x - 4)4 (1 - 0) )
----------
( (x - 4)5 )2
  = ex (x - 4)5 - 5*(ex - (e)3) (x - 4)4
----------
((x - 4)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.