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

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Answer 1 · 2017-10-17T10:51:50

$(dy)/(dx)=-(2(e^x-e^(-x)))/((e^x+e^(-x))^2$

we will use teh quotient rule

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

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

$u=2=>u'=0$

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

$;.y'=(0-2(e^x-e^(-x)))/(e^x+e^(-x))^2$

$(dy)/(dx)=-(2(e^x-e^(-x)))/((e^x+e^(-x))^2$

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