# Thread: [SOLVED] Implicit Differentiation

1. ## [SOLVED] Implicit Differentiation

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

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

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

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

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

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

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

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

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

Edit. Just beaten to it by skeeter.

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