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

1 Answer
Jun 1, 2018

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

Explanation:

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