How do you find the 1st and 2nd derivative of #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