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How do you find f'(x), f''(x) of f(x) =ln (1-x)

IF f(x) = ln (1-x)
Then use the chain rule with u(x) = 1-x
f(u) = ln u
df/dx = df/du du/dx
f'(x) = -1/(1-x)
Use the chain rule again:
f"(x) = -[-1/(1-x)^2]*(-1) = -1/(1-x)^2

What is the pattern there how can you give an expression for f ^(n) (x) [the n-th derivative of f(x)]?

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