Derivatives of Sums

Example:
y = sqrt((sqrt(3))*x2 + 2) - x2 + 4

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 = ( sqrt((sqrt(3))*x2 + 2) ) - ( x2 ) + ( 4 )

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 = ( sqrt((sqrt(3))*x2 + 2) ) - ( x2 ) + ( 4 )
we can think of as
y = f(x) - g(x) + h(x)
So the derivative is
y ' = ( f(x) - g(x) + h(x) )'
  = f '(x) - g '(x) + h '(x)  
  = ( sqrt((sqrt(3))*x2 + 2)) ' - ( x2) ' + ( 4) '  
and we just need to know each of the derivatives on the right-hand side of the equation. In this case these are
( sqrt((sqrt(3))*x2 + 2) )' = (1/2)*((sqrt(3))*x2 + 2)-1/2 ((sqrt(3))*2*x + 0) (by the chain rule)
( x2 )' = 2*x (by the power rule)
( 4 )' = 0 (by the derivative rule for constants)
so the finished derivative is
y ' = (1/2)*((sqrt(3))*x2 + 2)-1/2 ((sqrt(3))*2*x + 0) - 2*x + 0  
  = (1/2(2(sqrt(3))))*x ((sqrt(3))*x2 + 2)-1/2 - 2*x
[]

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