Derivatives of Sums

Example:
g(y) = ((-2)*y)5 + y - ln(2)

The first thing to notice when finding the derivative of this function is that it is the sum of several terms, as shown in color below:

g(y) = ( ((-2)*y)5 ) + ( y ) - ( ln(2) )

The Derivative Rule for Sums:

The derivative of a sum is the sum of the derivatives.
If
  z = ( f(x) + g(x) )
then the derivative of z is
  z' = ( f(x) + g(x) )'
    = f '(x) + g'(x)

So our example,

g(y) = ( ((-2)*y)5 ) + ( y ) - ( ln(2) )
we can think of as
g(y) = f(y) + h(y) - p(y)
So the derivative is
g '(y) = ( f(y) + h(y) - p(y) )'
  = f '(y) + h '(y) - p '(y)  
  = ( ((-2)*y)5) ' + ( y) ' - ( ln(2)) '  
and we just need to know each of the derivatives on the right-hand side of the equation. In this case these are
( ((-2)*y)5 )' = (-2)*5*((-2)*y)4 (by the chain rule)
( y )' = 1 (by the derivative rule for variables)
( ln(2) )' = 0 (by the derivative rule for constants)
so the finished derivative is
g '(y) = (-2)*5*((-2)*y)4 + 1 - 0  
  = (-10)*((-2)*y)4 + 1
[]

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