Posted by
**Jill** on
.

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)]?