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

Mar 9, 2016

$- \frac{4 \pi}{5} \sin \left(\left(\frac{2 \pi}{5}\right) \left(t - 3\right)\right)$

#### Explanation:

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

Then the derivative of $g \left(t\right)$ 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 $- \frac{4 \pi}{5} \sin \left(\left(\frac{2 \pi}{5}\right) \left(t - 3\right)\right)$