Thread: Differentiating natural log

Differentiate f and find the domain of f

$\displaystyle f(x) = \sqrt{1-ln x}$

$\displaystyle f'(x) = 1/2 (1 - ln x) (1 - ln x)'$

$\displaystyle = [(1-ln x)/2] [-1/x]$

$\displaystyle = - [1 - ln x/2x]$

I know I'm doing something wrong, I did a few of these before but I'm not getting the chain rule being in here and what not. The answer SHOULD be

$\displaystyle \frac{-1}{2x\sqrt{1-ln x}}$

Can anyone help here?

The derivative of the natural log is $\displaystyle \frac{1}{x}$

$\displaystyle f(x)=\sqrt{1-lnx}$

$\displaystyle f(x)=({1-lnx})^{\frac{1}{2}}$

$\displaystyle f(x)'=\frac{1}{2}({1-lnx})^{\frac{-1}{2}}*\frac{-1}{x}$

Now just do some simplication

$\displaystyle \frac{-1}{2x\sqrt{1-lnx}}$