Derivatives of Sums

Example:
C = ((sqrt(3))*z + 5)3 + (sqrt(3))*z2 + 5

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:

C = ( ((sqrt(3))*z + 5)3 ) + ( (sqrt(3))*z2 ) + ( 5 )

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,

C = ( ((sqrt(3))*z + 5)3 ) + ( (sqrt(3))*z2 ) + ( 5 )
we can think of as
C = f(z) + g(z) + h(z)
So the derivative is
C ' = ( f(z) + g(z) + h(z) )'
  = f '(z) + g '(z) + h '(z)  
  = ( ((sqrt(3))*z + 5)3) ' + ( (sqrt(3))*z2) ' + ( 5) '  
and we just need to know each of the derivatives on the right-hand side of the equation. In this case these are
( ((sqrt(3))*z + 5)3 )' = 3*((sqrt(3))*z + 5)2 ((sqrt(3)) + 0) (by the chain rule)
( (sqrt(3))*z2 )' = (sqrt(3))*2*z (by the rule for constant multiples, and the power rule)
( 5 )' = 0 (by the derivative rule for constants)
so the finished derivative is
C ' = 3*((sqrt(3))*z + 5)2 ((sqrt(3)) + 0) + (sqrt(3))*2*z + 0  
  = (3(sqrt(3)))*((sqrt(3))*z + 5)2 + (2(sqrt(3)))*z
[]

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.