We'll differentiate applying quotient rule.

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

y = u(x)/v(x)

We'll differentiate:

y' = [u'(x)*v(x)-u(x)*v'(x)]/v(x)^2

v(x) = (x^2+1)

v'(x) = 2x

u(x) = x^2-1 .

u'(x) = (x^2-1)'

u'(x) = 2x

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

We'll remove the brackets:

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

We'll eliminate like terms:

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

**The result of differentiation is: y' = (4x)/(x^2+1)^2**

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

We use {u(x)/v(x)}' = {u'(x)v(x)-u(x)v(x)}/(v(x))^2 to fifferentiate the given expression.

u(x) = (x^2-1). u'(x) = (x^2-1)' = 2x

v(x) = (x^2+1). So (x^2+1' = 2x.

Therefore {x^2-1)/(x^2+1)}' = {(x^2-1)(x^2+1) -(x^2-1)(x^2+1)'}/(x^2+1)^2

{(x^2-1)/(x^2+1) }'= {2x(x^2+1)-(x^2-1)*2x}/(x^2+1)

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