Question #0e708

1 Answer
sente
Nov 26, 2016

#d/dx(e^x-e^-x)/(e^x+e^-x)=4/(e^x+e^-x)^2#

Explanation:

Using the quotient rule and the sum rule along with the following:

  • #d/dxe^x = e^x#
  • #d/dxe^-x = -e^-x#
  • #a^2-b^2=(a+b)(a-b)#

we have

#d/dx(e^x-e^-x)/(e^x+e^-x)#

#=[(e^x+e^-x)(d/dx(e^x-e^-x))-(e^x-e^-x)(d/dx(e^x+e^-x))]/(e^x+e^-x)^2#

#=((e^x+e^-x)(e^x+e^-x)-(e^x-e^-x)(e^x-e^-x))/(e^x+e^-x)^2#

#=((e^x+e^-x)^2-(e^x-e^-x)^2)/(e^x+e^-x)^2#

#=([(e^x+e^-x)+(e^x-e^-x)][(e^x+e^-x)-(e^x-e^-x)])/(e^x+e^-x)^2#

#=((2e^x)(2e^-x))/(e^x+e^-x)^2#

#=4/(e^x+e^-x)^2#

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