How do you differentiate $y=(1+e^(2x))/(2-e^(2x))$ ?

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Answer 1 · 2016-11-07T09:57:27

$:.y'=(6e^(2x))/(2-e^(2x))^2$

$y=u/v=>y'=(u'v-uv')/v^2$

$y=(1+e^(2x))/(2-e^(2x))$

$u=1+e^(2x)=>u'=2e^(2x)$

$v=2-e^(2x)=>v'=-2e^(2x)$

$:.y'=(2e^(2x)(2-e^(2x))-(1+e^(2x))(-2e^(2x)))/(2-e^(2x))^2$

$:.y'=[2e^(2x)[(2-e^(2x))+(1+e^(2x))]]/(2-e^(2x))^2$

$:.y'=[2e^(2x)[(2-cancel(e^(2x)))+(1+cancel(e^(2x)))]]/(2-e^(2x))^2$

$:.y'=[2e^(2x)[2+1]]/(2-e^(2x))^2$

$:.y'=(6e^(2x))/(2-e^(2x))^2$

How do you differentiate y=(1+e^(2x))/(2-e^(2x))y=(1+e^(2x))/(2-e^(2x))?