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

1 Answer
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)