How do you find the derivative of #g(t)=1+2cos(((2pi)/5)(t-3))# using the chain rule?

1 Answer
Zanzen Z.
Mar 9, 2016

#-(4pi)/5sin(((2pi)/5)(t-3))#

Explanation:

First to simplify this we're going to eliminate the constant #1# because it's derivative is zero, and then expand inner function #(((2pi)/5)(t-3))# function into #("(2pit)/5-6pi/5)#

Then the derivative of #g(t)# is the derivative of the outer function #2cos("(2pit)/5-6pi/5)prime# #times# the derivative of the inner function #("(2pit)/5-6pi/5)prime#

Which is equal to #-2sin("(2pit)/5-6pi/5)times(2pi/5-0)#

Simplifying this gives us #-(4pi)/5sin(((2pi)/5)(t-3))#

Related questions
See all questions in Chain Rule
Impact of this question
1625 views around the world
You can reuse this answer
Creative Commons License