What is the derivative of $y = \frac{e^{2 x}}{e^{2 x} + 1}$ $y=(e^(2x))/(e^(2x)+1)$ ?

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Answer 1 · 2018-04-09T10:50:07

Please look below.

$y = \frac{e^{2 x}}{e^{2 x} + 1}$ $y = e^(2x)/(e^(2x)+1)$

$= \frac{e^{2 x} + 1 - 1}{e^{2 x} + 1}$ $= (e^(2x)+1-1)/(e^(2x)+1)$

$= \frac{e^{2 x} + 1}{e^{2 x} + 1} - \frac{1}{e^{2 x} + 1}$ $= (e^(2x)+1)/(e^(2x)+1)- 1/(e^(2x)+1)$

$= 1 - \frac{1}{e^{2 x} + 1}$ $= 1 - 1/(e^(2x)+1)$

$= - {(e^{2 x} + 1)}^{- 1}$ $= -(e^(2x) + 1)^-1$

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

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

Answer 2 · 2018-04-09T11:04:35

$(dy)/(dx)=(2e^(2x))/((e^(2x)+1)^2)$

We have,

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

$"Using "color(green)"Quotient Rule"$ ,

$(dy)/(dx)=((e^(2x)+1)d/(dx)(e^(2x))-e^(2x)d/(dx) (e^(2x)+1))/((e^(2x)+1)^2)$

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

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

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

What is the derivative of y=(e^(2x))/(e^(2x)+1)y=e2xe2x+1y=(e^(2x))/(e^(2x)+1)?