# If f(x) = (3x-1)/(x^2 -2) find f ' (1)

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

Let us differentiate f:

f(x) = u/v such that:

u= 3x-1 ==> u' = 3

v= x^2 -2 ==> v'= 2x

==> f'(x) = (u'v- uv')/v^2

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

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

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

Now let us substitute with x= 1

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

= -7/1 = -7

**==> f'(1) = -7**

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

To find f'(1).

We first find f'(x).

f(x) is in u(x)/v(x) form

{u(x)/v(x)}' = {u'(x)v(x) -u(x)v'(x)}/{v(x)}^2....(1)

Here u(x) = (3x-1).

Differentiating both sides wrt x we get u'(x) = (3x-1)' =3.

v(x) = x^2-2

Differenting we get v'(x) = (x^2-2)' = 2x.

So substituting inthe formula (1) we get:

{(3x-1)/(x^2-2)}' = {(3x-1)'(x^2-2)-(3x-1)(x^2-2)}/(x^2-2)^2

{(3x-1)/(x^2-2)}'={3(x^2-2)-(3x-1)*2x}/(x^2-2)^2

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

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

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