f(x) = ln x / (x + 2)

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

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

For x = 1

f'(1) = (3 - ln 1) / 3^2

=> 1/3

**The value of f'(1) = 1/3**

f(x) = lnx/(x+2)

We need to find the values of f'(1)

First we will need to find the first derivative f'(x).

Let f(x) = u/v such that:

u= ln x ==> u' = 1/x

v= x+2 ==> v' = 1

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

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

==> f'(x) = (x+2)/x - ln x] / (x+2)^2

==> Now we will substitute with x = 1

==> f'(1) = (1+2)/1 - ln 1] / (1+2)^2

= (3 - 0) / 9 = 3/9 = 1/3

**==> f'(1)= 1/3**