1. [SOLVED] Implicit Differentiation

My teacher didn't really teach this section out of our book that well in class.
$\displaystyle ln(xy) = x + y$
Especially problems with ln(s) and log(s) I tried the problem out and got
$\displaystyle {1+y/-2(xy)^-3}={dy/dx}$

2. $\displaystyle \ln(xy)=\ln x+\ln y,$ now $\displaystyle \frac1x+\frac1y\cdot y'=1+y',$ and now just isolate $\displaystyle y'.$

My teacher didn't really teach this section out of our book that well in class.
$\displaystyle ln(xy) = x + y$
Especially problems with ln(s) and log(s) I tried the problem out and got
$\displaystyle {1+y/-2(xy)^-3}={dy/dx}$
$\displaystyle \ln(xy) = x+y$

$\displaystyle \ln(x) + \ln(y) = x+y$

$\displaystyle \frac{1}{x} + \frac{y'}{y} = 1 + y'$

$\displaystyle \frac{y'}{y} - y' = 1 - \frac{1}{x}$

$\displaystyle y'\left(\frac{1}{y} - 1\right) = 1 - \frac{1}{x}$

$\displaystyle y' = \frac{1 - \frac{1}{x}}{\frac{1}{y} - 1}$

My teacher didn't really teach this section out of our book that well in class.
$\displaystyle ln(xy) = x + y$
Especially problems with ln(s) and log(s) I tried the problem out and got
$\displaystyle {1+y/-2(xy)^-3}={dy/dx}$
This becomes a lot easier if you first use the fact that $\displaystyle \ln(xy) = \ln x+\ln y$. Then when you differentiate $\displaystyle \ln x+\ln y = x + y$ implicitly, you get $\displaystyle \frac1x + \frac1y\frac{dy}{dx} = 1 + \frac{dy}{dx}$. Finally, rearrange that to get an expression for $\displaystyle \frac{dy}{dx}$.

Edit. Just beaten to it by skeeter.

5. You guys are great, I see exactly how this makes sense! Thanks!