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

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Answer 1 · 2018-04-01T09:28:32

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

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 $cos(s)$

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

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

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