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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. $\displaystyle df=(\frac{2}{1+y^2}-3(x+y)^2)dx-(\frac{4xy}{(1+y^2)^2}-3(x+y)^2)dy$

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