How do you find the derivative of #(3x)/(2e^x+e^-x)#?

1 Answer
Mia
Nov 28, 2016

#=(6e^x(1-x) + 3e^(-x)(1+x) )/(2e^x + e^(-x))^2#

Explanation:

Differentiating this function is determined by applying the
#" "#
quotient rule.
#" "#
#d/dx(u/v)=(u'v-v'u)/v^2#
#" "#
#" "#
#d/dx((3x)/(2e^x + e^(-x)))#
#" "#
#" "#
#=((3x)'xx(2e^x + e^(-x))-(2e^x + e^(-x))'xx(3x))/(2e^x + e^(-x))^2#
#" "#
#=(3xx(2e^x + e^(-x))-(2e^x - e^(-x))xx(3x))/(2e^x + e^(-x))^2#
#" "#
#=((6e^x + 3e^(-x))-(6xe^x - 3xe^(-x)))/(2e^x + e^(-x))^2#
#" "#
#=(6e^x + 3e^(-x)-6xe^x + 3xe^(-x))/(2e^x + e^(-x))^2#
#" "#
#=(6e^x(1-x) + 3e^(-x)(1+x) )/(2e^x + e^(-x))^2#

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