Derivatives of Sums

Example:
f(t) = (6*t)3 + (sqrt(3))*t3 - 3*t - 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:

f(t) = ( (6*t)3 ) + ( (sqrt(3))*t3 ) - ( 3*t ) - ( 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,

f(t) = ( (6*t)3 ) + ( (sqrt(3))*t3 ) - ( 3*t ) - ( 4 )
we can think of as
f(t) = g(t) + h(t) - p(t) - q(t)
So the derivative is
f '(t) = ( g(t) + h(t) - p(t) - q(t) )'
  = g '(t) + h '(t) - p '(t) - q '(t)  
  = ( (6*t)3) ' + ( (sqrt(3))*t3) ' - ( 3*t) ' - ( 4) '  
and we just need to know each of the derivatives on the right-hand side of the equation. In this case these are
( (6*t)3 )' = 6*3*(6*t)2 (by the chain rule)
( (sqrt(3))*t3 )' = (sqrt(3))*3*t2 (by the rule for constant multiples, and the power rule)
( 3*t )' = 3 (by the rule for constant multiples, and the derivative rule for variables)
( 4 )' = 0 (by the derivative rule for constants)
so the finished derivative is
f '(t) = 6*3*(6*t)2 + (sqrt(3))*3*t2 - 3 - 0  
  = 18*(6*t)2 + (3(sqrt(3)))*t2 - 3
[]

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