okay im really struggling with this. how would i differentiate the function
y=1
---
ln (x)
ive tried it a couple times but cant quite get it...can someone please help?!?
I'm guessing you mean
$\displaystyle \frac{d}{dx} \lparen\frac{1}{ln(x)}\rparen $
Do you know how do differentiate $\displaystyle \frac{1}{x} $?
Use the chain rule
$\displaystyle \frac{d}{dx} \lparen\frac{1}{u}\rparen = -\frac{1}{u^2} \frac{du}{dx}$
In this case
$\displaystyle \frac{d}{dx} \lparen\frac{1}{ln(x)}\rparen = -\frac{1}{(ln(x))^2} \frac{d(ln(x))}{dx}$
I'm assuming you mean $\displaystyle y = \frac{1}{\ln(x)}$.
You can rewrite this as $\displaystyle y = (\ln(x))^{-1}$ and apply the chain rule
$\displaystyle \frac{dy}{dx} = -(\ln(x))^{-2}\frac{1}{x} = -\frac{(\ln(x))^{-2}}{x} = \frac{1}{x(\ln(x))^{2}}$