# Thread: can someone explain this

1. ## 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]

2. 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 $\displaystyle \ln(xy) = \ln(x) + \ln(y)$

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

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

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

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

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

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

3. 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?

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

So,

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

Let's differentiate both sides implicitly,

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

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