can someone explain this

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November 29th 2009, 08:01 AM
johnnyc418

can someone explain this

i have a problem that says find dy/dx: ln(xy)=x+y. am i trying to find the derv of both sides? dont they equal the same thing?[IMG]file:///C:/Users/Johnny/AppData/Local/Temp/moz-screenshot-1.png[/IMG][IMG]file:///C:/Users/Johnny/AppData/Local/Temp/moz-screenshot.png[/IMG]
November 29th 2009, 08:11 AM
skeeter

Quote:

Originally Posted by johnnyc418

i have a problem that says find dy/dx: ln(xy)=x+y. am i trying to find the derv of both sides? dont they equal the same thing?[IMG]file:///C:/Users/Johnny/AppData/Local/Temp/moz-screenshot-1.png[/IMG][IMG]file:///C:/Users/Johnny/AppData/Local/Temp/moz-screenshot.png[/IMG]

note that $\ln(xy) = \ln(x) + \ln(y)$

$\frac{d}{dx} \left[\ln(x) + \ln(y) = x+y\right]$

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

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

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

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

$\frac{y - xy}{xy - x} = y'$
November 29th 2009, 08:17 AM
artvandalay11

Quote:

Originally Posted by johnnyc418

i have a problem that says find dy/dx: ln(xy)=x+y. am i trying to find the derv of both sides? dont they equal the same thing?

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

So,

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

Let's differentiate both sides implicitly,

$\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1+\frac{dy}{d x}$

And now move stuff around and solve for $\frac{dy}{dx}$

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