How do you find the 1st and 2nd derivative of $g (x) = \frac{1}{2 e^{x} + e^{- x}}$ $g(x) = 1 / (2e^x + e^-x)$ ?

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Answer 1 · 2018-06-01T08:43:42

$color(indigo)(g’(x) = -(2(e^x)^2 - 1) / (2(e^x)^2 + 1)$

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

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

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

$g’(x) = -(2e^x - e^-x) / (2e^x + e^-x)$

$color(indigo)(g’(x) = -(2(e^x)^2 - 1) / (2(e^x)^2 + 1)$

Similarly use division rule to derive the $g”(x)$

Division rule $g”(x) = (v du - u dv) / v^2$

$u = -(2(e^x)^2 - 1) , v = (2(e^x)^2 + 1)$

$du = -2 * (e^x)^2 * 2x = -4x * (e^x)^2$

$dv = 2 * (e^x)^2* 2x = 4x * (e^x)^2$

$g”(x) = ( (2(e^x)^2 + 1) * ( -4x * (e^x)^2)+ ((2(e^x)^2 - 1) * 4x * (e^x)^2)) / ( (2(e^x)^2 + 1))^2$

Further simplification is your homework

How do you find the 1st and 2nd derivative of g(x) = 1 / (2e^x + e^-x)g(x)=12ex+e−xg(x) = 1 / (2e^x + e^-x)?