To solve this equation first we need to obtain f'(x)

while .......... f(x) = 3x^2 / x^2 - 1

lets assume that f(x) = g(x)/r(x)

then f'(x) = [g'(x) r(x) - g(x) r'(x)] / r(x)^2

==> f'(x) = [(6x) (x^2-1) - (3x^2)( 2x)]/ (x^2-1)^2

==> f'(x) = (6x^3 - 6X - 6x^3)/ (x^2-1)^2

==> f'(x) = -6x / (x^2-1)^2

Now we need to prove that :

f'(x) + 6x/x^2-1 = 0

==> -6x/(x^2-1)^2 + 6x/(x^2 -1)^2

then f'(x) + 6x/(x^2-1)^2 = 0

If we re-write what we have to verify, we'll come to the conclusion that we have to verify if

f'(x)=-6x/(x^2 - 1)

To find out the first derivative, we'll use the rule of the ratio:

(u/v)' = (u'*v-u*v')/v^2

In our case, u=3x^2 and v=(x^2 - 1)

u'=6x

v'=2x

We'll substitute u,v,u' and v' in the rule of derivative above:

(u/v)' = [6x(x^2 - 1)-3x^2*2x]/(x^2 - 1)^2

(u/v)' = (6x^3 - 6x - 6x^3)/(x^2 - 1)^2

After reducing similar terms, we'll get:

f'(x) = (u/v)' = - 6x/(x^2 - 1)^2, exactly what we've searched for.

- 6x/(x^2 - 1)^2 + 6x/(x^2 - 1)^2 = 0 q.e.d.

f(x) =3x^2/(x^2-1). To verify f'(x0 +6x/(x^2-1) = 0.

Solution:

f(x) = 3x^2/(x^2-1). Differentiating, we get:

f'(x) = -[3x^2/(x^2-1)^2](x^2-1)'+ 3*2x/(x^2-1)

= -3x^2/(x^2-1)^2](2x) +6x(x^2-1)/(x^2-1)^2

= (-6x^3+6x)/(x^2-1)^2

=-6x(x^2-1)/(x^2-1)^2

f'(x) = -6x/(x^2-1). Or

f'(x)+6x/(x^2-1) = 0