Given ln(-8 + 3z^2)

We need to find the derivative.

Let f(z) = ln (-8+3z^2).

We will use the chain rule to solve for f'(z).

Let u = -8+3z^2 ==> u' = 6z

==> f(z) = ln u

==> f'(z) = (ln u) ' * u'

= 1/u * u'

We will substitute.

==> f'(z) = 1/(-8+3x^2) * 6z

= 6z / (-8+3z^2)

**==> f'(z) = 6z/ (-8+3z^2)**

We have to fine the derivative of ln (-8 + 3z^2),

Now we use the chain rule here.

The derivative of ln x is 1/x and the derivative of -8 + 3z^2 is 6z.

Using the same we get the derivative of ln (-8 + 3z^2) as 6z/(-8 + 3z^2)

**Therefore the derivative of ln (-8 + 3z^2) is 6z/(-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).