[tex]f(x)=\ln(1-5x)\\\\f'(x)=\left[\ln(1-5x)\right]'=\dfrac{1}{1-5x}\cdot(1-5x)'=\dfrac{1}{1-5x}\cdot(-5)=-\dfrac{5}{1-5x}\\\\Used:\\\\(\ln(x))'=\dfrac{1}{x}\\\\\left[g(f(x))]\right]'=g'(f(x))\cdot f'(x)[/tex]
The formula for the derivative of the ln function, with chain rule, is
(d/dx) ln u = (1/u)(du/dx).
Here, u = 1 - 5x, and du/dx = -5.
Then df/dx = [ 1 / 1-5x ]*(-5), or
df -5
---- = --------
dx 1-5x
