Find the derivative of ln (-8 + 3z^2).

We use  d/dx {ln{fx} = (1/f(x))*(d/dx)
f(x).

Or

{lnf(x)}' =
(1/f(x)f'(x).

Here we apply the above method to  f(z) = (-8
+ 3z^2).

The differentiation is with respect to
z.

f'(z) = (-8+3z^2)' = (-8)'+(3z^2)' = 0+3*2z =
6z.

Therefore {lnf(z)} = {ln (-8 +
3z^2)}'

{ln (-8 + 3z^2)}' = {1/ (-8 + 3z^2)}* (-8 +
3z^2)'

{ln (-8 + 3z^2)}' =
6z/(-8+3z^2).

Therefore {ln (-8 + 3z^2)}' =
6z/(3z^2-8).