Derivatives of Quotients

Example:
y = sqrt(x) / (x3 - x - 2)

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

y = sqrt(x)
----------
x3 - x - 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,

y = sqrt(x)
----------
x3 - x - 2
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
  = ( sqrt(x) )' ( x3 - x - 2 ) - ( sqrt(x) ) ( x3 - x - 2 )'
----------
( x3 - x - 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
( sqrt(x) )' = (1/2)*x-1/2 (by the power rule, with exponent 1/2)
( x3 - x - 2 )' = 3*x2 - 1 - 0 (by the derivative rule for sums, power rule, derivative rule for variables, and the derivative rule for constants)
so the finished derivative is
y' = ( (1/2)*x-1/2 ) ( x3 - x - 2 ) - ( sqrt(x) ) ( 3*x2 - 1 - 0 )
----------
( x3 - x - 2 )2
  = (1/2)*x-1/2 (x3 - x - 2) - sqrt(x) (3*x2 - 1)
----------
(x3 - x - 2)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.