Derivatives of Compositions

Example:
W = cos(sqrt(y) - y3/2)

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

W = cos([])
  where  [] = sqrt(y) - y3/2

The Chain Rule (Derivative Rule for Compositions):

The derivative of a composition is the product of the derivative of the outer function (with the inner function plugged in) and the derivative of the inner function.
If
  z = f( g(x) )
then the derivative of z is
  z' = ( f(g(x)) )'
    = f '(g(x)) g '(x)
Or, if
  z = f( [] ), where [] = g(x)
then the derivative of z is
  z' = ( f( [] ) )'
    = f '( [] ) ( [] )'
    = f '( [] ) g '(x)

So our example,

W = cos([])
  where  [] = sqrt(y) - y3/2
we can think of as
W = f( [] ) , where  [] = g(y) = sqrt(y) - y3/2
So the derivative is
W ' = ( f( [] ) )'
  = f '( [] ) ( [] )'  
  = ( cos([]) )' ( sqrt(y) - y3/2 )'
and we just need to know each of the derivatives on the right-hand side of the equation. In this case these are
( cos([]) )' = (-1)*sin([]) (by the derivative rules for basic functions)
( sqrt(y) - y3/2 )' = ( (1/2)*y-1/2 - (3/2)*y1/2 ) (by the derivative rule for sums, power rule, with exponent 1/2, and the power rule (again))
so the finished derivative is
W ' = (-1)*sin([]) ( (1/2)*y-1/2 - (3/2)*y1/2 )
  = (-1)*sin(sqrt(y) - y3/2) ( (1/2)*y-1/2 - (3/2)*y1/2 )
  = (-1)*sin(sqrt(y) - y3/2) ((1/2)*y-1/2 - (3/2)*y1/2)
[]


additional explanation for the chain rule
see another chain rule example
practice gateway test
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Comments to Gavin LaRose
glarose@umich.edu
©2001 Gavin LaRose, University of Michigan Math Dept.