What is the derivative y' of x + y + ln(xy) = 0

Friday, December 26, 2014

Differentiate x + y + ln(xy) =
0

$\displaystyle{\left.{d}{x}\right.}+{\left.{d}{y}\right.}+\frac{{1}}{{{x}{y}}}{d}{\left({x}{y}\right)}={0}$ Using the chain
rule

d(xy) = xdy+ydx using the product
rule

$\displaystyle{\left.{d}{x}\right.}+{\left.{d}{y}\right.}+\frac{{1}}{{{x}{y}}}{\left({x}{\left.{d}{y}\right.}+{y}{\left.{d}{x}\right.}\right)}={0}$

Distributing
$\displaystyle\frac{{1}}{{{x}{y}}}$ we get

$\displaystyle{\left.{d}{x}\right.}+{\left.{d}{y}\right.}+\frac{{\left.{d}{y}}}{{y}}+\frac{{\left.{d}{x}}}{{x}}=$
0

Moving the dx terms to the right
side

$\displaystyle{\left.{d}{y}\right.}+\frac{{1}}{{y}}{\left.{d}{y}\right.}=-{\left({\left.{d}{x}\right.}+\frac{{1}}{{x}}\right.}$
dx)

Factoring dx and dy we
get

$\displaystyle{\left.{d}{y}\right.}{\left({1}+\frac{{1}}{{y}}\right)}=$
-(1+1/x)dx

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=-\frac{{{1}+\frac{{1}}{{x}}}}{{{1}+\frac{{1}}{{y}}}}=$
-((x+1)/x)/((y+1)/y)=-(y(x+1))/(x(y+1))

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