# Find the derivative of (x^2 + 1) / (x^2 -1) using the quotient rule.

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Here it is specified that the quotient rule has to be used.

Now for finding the derivative of f(x)/g(x) the quotient rule is [f’(x) g(x) – f(x) g’(x)]/ [g(x)] ^2

Here f(x) = x^2+1. So we have f’(x) 2x.

g(x) = x^2 -1, so g’(x) = 2x

Therefore the derivative of (x^2 + 1) / (x^2 -1) is

[(x^2 - 1) * 2x – 2x * (x^2 +1)] / (x^2 -1) ^2

=> 2x [x^2 -1 – x^2 - 1] / (x^2 -1) ^2

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

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

**Therefore the required result is - 4x/(x^2 -1) ^2**

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

To find derivative by quotient rule.

y = u(x)/v(x) form.

So use y' = {u(x)v(x)-u(x)v'(x)]/(v(x))^2 which thee rule to find the derivative of a quotient form .

u(x) = x^2+1.

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

v(x) = x^2-1 .

v'(x) = (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)^2

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

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