f(x) = (x-3)/(x^2+2) find f ' (1) .

We'll use the limit method to calculate the value of
derivative of a function in a given point.

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

lim
[(x-3)/(x^2+2) + 2/3]/(x-1) = lim (3x-9+2x^2+4)/(x-1)

lim
(3x-9+2x^2+4)/(x-1) = lim (3x-5+2x^2)/(x-1)

We'll
substitute x by 1:

lim (3x-5+2x^2)/(x-1) = (3-5+2)/(1-1) =
0/0

Since we've obtained an indeterminacy, we'll apply
L'Hospital rule:

lim (3x-5+2x^2)/(x-1) = lim
(3x-5+2x^2)'/(x-1)'

 lim (3x-5+2x^2)'/(x-1)' = lim
(3+4x)/1

We'll substitute x by
1:

lim (3+4x)/1 =
(3+4)/1

f'(1) = lim (3x-5+2x^2)/(x-1) =
7

f'(1) =
7