Question

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

Answer 1

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

Explanation:

We are dealing with the quotient rule inside the chain rule

Chain rule for cosine

`cos(s) rArr s'*-sin(s)`

Now we have to do the quotient rule

`s=( 1-e^(2x))/(1+e^(2x))`

`dy/dxu/v=(u'v-v'u)/v^2`

Rule for deriving e

Rule: `e^u rArr u'e^u`

Derive both the top and bottom functions

`1-e^(2x) rArr 0-2e^(2x)`

`1+e^(2x) rArr 0+2e^(2x)`

Put it into the quotient rule

`s'=(u'v-v'u)/v^2=(-2e^(2x)(1+e^(2x))-2e^(2x)( 1-e^(2x)))/(1+e^(2x))^2`

Simply
`s'=(-2e^(2x)((1+e^(2x))+( 1-e^(2x))))/(1+e^(2x))^2`

`s'=(-2e^(2x)(2))/(1+e^(2x))^2=(-4e^(2x))/(1+e^(2x))^2`

Now put it back into the derivative equation for `cos(s)`

`cos(s) rArr s'*-sin(s)`

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