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Kainui Group TitleBest ResponseYou've already chosen the best response.1
This is what I did: I took the natural log to both sides xln(xy)=1 I divided x out of each side ln(xy)=1/x Used properties of logarithms to separate out ln(xy) lnx + lny=1/x lny=lnx+1/x then I took the derivative to both sides y'/y=[(1/x^2)+(1/x)] I then multiplied both sides by y, which I just found by using the earlier equation: lny=lnx+1/x y=e^(lnx+1/x) y'=[(1/x^2)+(1/x)][e^(lnx+1/x)] But I'm sure there's an easier or different way, I just went about it this route since this is where I felt most comfortable.
 2 years ago

keketsu Group TitleBest ResponseYou've already chosen the best response.2
Very interesting. I asked the question somewhere else and got a different approach but same response. This is actually very helpful to see it done a couple different ways. And thanks for the extra explanations!
 2 years ago
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