Derivatives of Sums

Example:
y = e*p-1 + 2*((1)/(2))p

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:

y = ( e*p-1 ) + ( 2*((1)/(2))p )

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,

y = ( e*p-1 ) + ( 2*((1)/(2))p )
we can think of as
y = f(p) + g(p)
So the derivative is
y ' = ( f(p) + g(p) )'
  = f '(p) + g '(p)  
  = ( e*p-1) ' + ( 2*((1)/(2))p) '  
and we just need to know each of the derivatives on the right-hand side of the equation. In this case these are
( e*p-1 )' = e*(-1)*p-2 (by the rule for constant multiples, and the power rule)
( 2*((1)/(2))p )' = 2*(ln(((1)/(2))))*((1)/(2))p (by the rule for constant multiples, and the derivative rules for basic functions)
so the finished derivative is
y ' = e*(-1)*p-2 + 2*(ln(((1)/(2))))*((1)/(2))p  
  = (-e)*p-2 + (2((ln(((1)/(2))))))*((1)/(2))p
[]

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.