[SOLVED] Implicit Differentiation

**radioheadfan** · August 4th 2009, 01:02 PM

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
${1+y/-2(xy)^-3}={dy/dx}$

**Krizalid** · August 4th 2009, 01:08 PM

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

**skeeter** · August 4th 2009, 01:10 PM

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
${1+y/-2(xy)^-3}={dy/dx}$

$\ln(xy) = x+y$

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

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

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

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

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

**Opalg** · August 4th 2009, 01:11 PM

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.

**radioheadfan** · August 4th 2009, 01:15 PM

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

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