Derivatives of Products

Example:
g(t) = (6*t + cos(2)) (10*t + 3)4

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

g(t) = ( 6*t + cos(2)) ( (10*t + 3)4)

The Derivative Rule for Products:

The derivative of a product is the derivative of the first term times the second (and third, etc.) term(s), plus the first (and third, etc.) term(s) times the derivative of the second, etc.
If
  z = (f(x) g(x))
then the derivative of z is
  z ' = (f(x) g(x))'
    = f '(x) g(x) + f(x) g '(x)

So our example,

g(t) = ( 6*t + cos(2)) ( (10*t + 3)4)
we can think of as
g(t) = f(t) h(t)    
So the derivative is
g '(t) = ( f(t) h(t) )'    
  = f '(t) h(t) + f(t) h '(t)
  = ( 6*t + cos(2) )' ( (10*t + 3)4 ) + ( 6*t + cos(2) ) ( (10*t + 3)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 + cos(2) )' = 6 + 0 (by the derivative rule for sums, rule for constant multiples, and the derivative rule for constants)
( (10*t + 3)4 )' = 4*(10*t + 3)3 (10 + 0) (by the chain rule)
so the finished derivative is
g '(t) = ( 6 + 0 ) ( (10*t + 3)4 ) + ( 6*t + cos(2) ) ( 4*(10*t + 3)3 (10 + 0) )
  = 6*(10*t + 3)4 + 40*(6*t + cos(2)) (10*t + 3)3
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additional explanation for the product rule
see another product rule example
practice gateway test
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glarose@umich.edu
©2001 Gavin LaRose, University of Michigan Math Dept.