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 May 6th, 2016, 02:08 AM #1 Newbie   Joined: Mar 2016 From: Australia Posts: 21 Thanks: 1 Total differential of function $\displaystyle z(x,y)=\frac{2x}{1+y^2}-e^{3\ln(x+y)}$ $\displaystyle df=\frac{2x}{1+y^2}-3(x+y)^2dx-\frac{4xy}{(1+y^2)^2}-3(x+y)^2dy$ $\displaystyle =-9dx-11dy$ Does my total differential to the above function at point (3, -1) look correct? Last edited by skipjack; May 6th, 2016 at 04:00 AM. May 6th, 2016, 04:08 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2203 No. May 6th, 2016, 04:15 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 One minor problem is that you have "df" when "f" itself has not been defined. If you mean "f(x,y)= z(x, y)", you need to say that. Best would have been to write "dz= ". Another minor point is that you should have parentheses: "(g(x,y)+ h(x,y))dx", not "g(x,y)+ h(x,y)dx" The important thing is that your derivatives are simply wrong. How did you differentiate $\displaystyle \frac{2x}{1+ y^3}- e^{3 ln(x+y)}$ with respect to x and y? May 6th, 2016, 02:56 PM #4 Global Moderator   Joined: May 2007 Posts: 6,805 Thanks: 716 Taking into account all the caveats, the only term in df that looks wrong is the first. The numerator is 2 not 2x. Also the final arithmetic is wrong altogether. May 6th, 2016, 03:51 PM #5 Newbie   Joined: Mar 2016 From: Australia Posts: 21 Thanks: 1 I used the total differential general formula $\displaystyle df=f'(x,y)dx+f'y(x,y)dy$ And yes there was a typo in df, should be: $\displaystyle df=\frac{2}{1+y^2}-3(x+y)^2dx-\frac{4xy}{(1+y^2)^2}-3(x+y)^2dy$ And the final arithmetic should be: $\displaystyle -11dx-9dy$ May 6th, 2016, 05:09 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2203 You've again omitted the parentheses that Country Boy pointed out are needed. May 6th, 2016, 08:21 PM #7 Newbie   Joined: Mar 2016 From: Australia Posts: 21 Thanks: 1 Indeed I did. $\displaystyle df=(\frac{2}{1+y^2}-3(x+y)^2)dx-(\frac{4xy}{(1+y^2)^2}-3(x+y)^2)dy$ Tags differential, function, total Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post golomorf Differential Equations 0 December 1st, 2014 03:34 AM adagus Differential Equations 1 November 19th, 2014 06:27 AM piotrek Differential Equations 2 May 23rd, 2013 07:22 AM Azhar Economics 1 March 11th, 2013 10:56 AM Taurai Mabhena Differential Equations 11 January 30th, 2012 02:55 PM

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