Derivative help

Question

y= (x²+1/x²-1)²

 

What is the simplest way of doing it. 

Answer 1

Apply the chain rule to the whole thing, letting u = (x²+1)/(x²-1):

 

     dy/dx = du²/du * du/dx = 2u * du/dx = 2(x²+1)/(x²-1) * du/dx

 

Then apply the quotient rule to differentiate du/dx

 

 

 

 

Answer 2

You have to use the chain rule and the quotient rule. 

 

y= ((x^2+1)/(x^2-1))^2

 

To make it a little easier let u = (x^2+1)/(x^2-1).

y = u^2
y' = 2u * u'

 

f = x^2 +1

f' = 2x

g = x^2-1

g' = 2x

 

u' = (f'*g - g'* f )/ g^2

= [2x(x^2-1) - (2x)(x^2+1)]/(x^2-1)^2

= (2x^3 - 2x - 2x^3 - 2x)/(x^2-1)^2

= -4x/(x^2-1)^2

 

y' = 2u * u'

y' = 2[(x^2+1)/(x^2-1)] * [-4x/(x^2-1)^2]

y' = 2[(-4x)(x^2+1)/(x^2-1)^3]

y' = (-8x)(x^2+1)/(x^2-1)^3