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 June 30th, 2009, 11:23 AM #1 Newbie   Joined: Jun 2009 Posts: 1 Thanks: 0 Need help differentiating Find dy/dx for f(x,y)=3xe^(x/y^2). Simplify Completely. I believe you have to differentiate Fx(x,y) and Fy(x,y) then use the formula dy/dx=-Fx(x,y)/Fy(x,y)
 June 30th, 2009, 11:57 AM #2 Senior Member   Joined: Jun 2009 Posts: 150 Thanks: 0 Re: Need help differentiating Looks senseless to me. To compute dy/dx we need y to be a function of x. But we are merely given a function of both variables.
 June 30th, 2009, 12:57 PM #3 Senior Member   Joined: May 2008 From: Sacramento, California Posts: 299 Thanks: 0 Re: Need help differentiating I'm guessing we solve for y and differentiate. $f(x,y)=3xe^{\frac{x}{y^2}}\\\frac{f(x,y)}{3x}=e^{\ frac{x}{y^2}}\\\ln{\frac{f(x,y)}{3x}}=\frac{x}{y^2 }\\y=\sqrt{\frac{x}{\ln{\frac{f(x,y)}{3x}}}}\\y=\s qrt{\frac{x}{\ln{f(x,y)}-\ln{3}-\ln{x}}}$ Differentiate: 1. Apply the chain rule to $y=\sqrt{\frac{x}{\ln{f(x,y)}-\ln{3}-\ln{x}}}$. 2. Apply the quotient rule to $\frac{x}{\ln{f(x,y)}-\ln{3}-\ln{x}}$. 3. Apply the chain rule to $\ln{f(x,y)}$. You should get: $\frac{dy}{dx}=\frac{1}{2}\left(\frac{\ln{f(x,y)}-\ln{3}-\ln{x}-x(\frac{f#39;(x,y)}{f(x,y)}-\frac{1}{x})}{(\ln{f(x,y)}-\ln{3}-\ln{x})^2}\right)\left(\frac{x}{\ln{f(x,y)}-\ln{3}-\ln{x}}\right)^{-\frac{1}{2}}$ Is that the way to do it?
 July 1st, 2009, 05:17 AM #4 Senior Member   Joined: Jun 2009 Posts: 150 Thanks: 0 Re: Need help differentiating That's one guess. Another guess would be that the problem really was: Find $\partial/\partial x$ of ... But until the OP clarifies, who knows?

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