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)))#