My Math Forum Need help differentiating
 User Name Remember Me? Password

 Calculus Calculus Math Forum

 June 30th, 2009, 12:23 PM #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, 12:57 PM #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, 01:57 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, 06:17 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?

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Lucida Calculus 1 November 21st, 2012 02:57 PM jskrzy Calculus 5 November 19th, 2012 10:05 AM pomazebog Calculus 4 March 18th, 2012 06:40 PM David_Lete Calculus 1 June 9th, 2010 06:49 AM 450081592 Calculus 2 January 30th, 2010 09:29 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top