Question

How do you find the derivative of the function `y=cos((1-e^(2x))/(1+e^(2x)))`?

Answer 1

`y'=4*e^(2*x)/(1+e^(2+x))^2*sin((1-e^(2x))/(1+e^(2*x)))`

Explanation:

Using the chain rule, then we get
`y'=-sin((1-e^(2x))/(1+e^(2x)))*(-2e^(2x)(1+e^(2x))-(1-e^(2x))*e^(2x))/(1+e^(2x))^2`

`y=-sin((1-e^(2x))/(1+e^(2x)))*(2e^(2x)(-1-e^(2x)-1+e^(2x))/(1+e^(2x))^2)`
`y'=4e^(2x)/(1+e^(2x))^2*sin((1-e^(2x))/(1+e^(2x)))`