Question

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

Answer 1

`-(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))`