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)]?

To find the n-th derivative of f(x) = ln(1-x), we can continue applying the chain rule multiple times:

Start with the first derivative we found: f'(x) = -1/(1-x)

For the second derivative, we take the derivative of (-1/(1-x)) with respect to x using the quotient rule:

f''(x) = [(-1)'(1-x) - (-1)(1-x)'] / (1-x)^2
= (0 - (-1)) / (1-x)^2
= 1 / (1-x)^2

For the third derivative, we take the derivative of (1 / (1-x)^2) using the quotient rule:

f'''(x) = [(1)'(1-x)^2 - 1(2(1-x))(1-x)'] / (1-x)^4
= (0 - 2(1-x)(-1)) / (1-x)^4
= 2(1-x) / (1-x)^4
= 2 / (1-x)^3

By observing the pattern, we can derive an expression for the n-th derivative of f(x):

f^(n)(x) = n! / (1-x)^(n+1)

where n! represents the factorial of n.