Differentiating natural log

**arkhampatient** · Nov 22nd 2009, 10:03 AM

Differentiate f and find the domain of f

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

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

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

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

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

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

Can anyone help here?

**RockHard** · Nov 22nd 2009, 10:05 AM

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

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

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

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

Now just do some simplication

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

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