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

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

Solution:

f(x)  =
u(x)*v(x),  f'(x) = u(x)v'(x) + u'(x)v(x)

Here u(x) = x-2 
and v(x) = 1/(x^2+3)

u'(x) = (x-2) =1, v'(x) {1/(x^2+3)^2}
= -(1/(x^2+3))(x^2)' = -2x/(x^2+3),  {u(v(x))' =
u'(x)*v'(x)

So f(x) = (x-2){-2x/(x^2+3)^2
+1/(x^2+3)

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

={-2x^2+4x+x^2+3}/(x^2+3)

=
(-x^2+4x+3)/(x^2+3)^2

=-(x^2-4x-3)/(x^2+3)