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May 6th, 2016, 02:08 AM   #1
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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.
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May 6th, 2016, 04:08 AM   #2
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No.
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May 6th, 2016, 04:15 AM   #3
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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?
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May 6th, 2016, 02:56 PM   #4
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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.
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May 6th, 2016, 03:51 PM   #5
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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:
$\displaystyle -11dx-9dy$
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May 6th, 2016, 05:09 PM   #6
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You've again omitted the parentheses that Country Boy pointed out are needed.
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May 6th, 2016, 08:21 PM   #7
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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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