How do you differentiate #y=((x^2+1)/(x^2-1))^3#?

1 Answer
Tamir E.
Nov 14, 2017

# y'=-12x(((x^2+1)^2)/(x^2-1)^4)#

Explanation:

#y=((x^2+1)/(x^2-1))^3#
#y'=3((x^2+1)/(x^2-1))^2*((x^2+1)/(x^2-1))'#

# ((x^2+1)/(x^2-1))'=((2x(x^2-1)-(x^2+1)2x)/(x^2-1)^2)#
#=((2x((x^2-1)-(x^2+1)))/(x^2-1)^2)#
#=((2x((x^2-1-x^2-1)))/(x^2-1)^2)#
#=((2x((-2)))/(x^2-1)^2)#
#=((-4x)/(x^2-1)^2)#

#y'=3((x^2+1)/(x^2-1))^2*((x^2+1)/(x^2-1))'#
#=> y'=3((x^2+1)/(x^2-1))^2*((-4x)/(x^2-1)^2)#
#=> y'=3((x^2+1)^2/(x^2-1)^2)*((-4x)/(x^2-1)^2)#
#=> y'=-12x(((x^2+1)^2)/(x^2-1)^4)#

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