differentiate F(z)=1/Ln(z)?

Question

Can I see the steps to get to the answer

Answer 1

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

by chain rule

Answer 2

F(z) = 1/Ln(z) = [Ln(z)]^-1

f(z) = d( [Ln(z)]^-1 )/dx

d( [Ln(z)]^-1 )/dx = -1 (1/x)*[Ln(z)]^-2

d( [Ln(z)]^-1 )/dx = -1/[x*(Ln z)²]

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