Question

pls check out if my solution is correct.

Answer 1

?

Answer 2

derivatives
f(x)=x^2+1/x^2-1
=x^2-1(2x)-x^2+1(2x)/(x^2-1)^2
=-4x/(x^2-1)2

Answer 3

i cancelled x^2-1 and x^2+1 is it correct?

Answer 4

was this the original equation?
\[f(x) = \frac{x^2+1}{x^2-1}\]

Answer 5

yup

Answer 6

So you want to use the quotient rule:
\[\bigg(\frac{f(x)}{g(x)}\bigg)' = \frac{f'(x)g(x)-f(x)g(x)'}{g(x)^2}\]
we have that:
\[f(x) = x^2+1 \Longrightarrow f'(x) = 2x\]\[g(x) = x^2-1 \Longrightarrow g'(x) = 2x\]
So following the formula we obtain:
\[\frac{(2x)(x^2-1)-(x^2+1)(2x)}{(x^2-1)^2} = 2x*\frac{x^2-1-x^2-1}{(x^2-1)^2} = \frac{-4x}{(x^2-1)^2}\]

Answer 7

am all negative?

Answer 8

yeah the result is going to be negative.