For $F(x,y)=ye^{x^2-y}$ find $F_x, F_{xx}, F_y, F_{xy}$ (partial derivatives)
I'm not sure if these are correct, but this is what I got:
$F_x=2xye^{x^2-y}$
$F_{xx}=4xye^{x^2-y}$
$F_y=-e^{x^2-y}$
$F_{xy}=-2xe^{x^2-y}$
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2Only $F_x$ is computed correctly. – egreg Jun 03 '16 at 17:56
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You need to use product rule for $F_{xx}$. – Chee Han Jun 03 '16 at 17:56
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1The product rule will come into play for all but the first one. – MPW Jun 03 '16 at 17:57
2 Answers
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Only $F_x$ is correct.
For $F_y$ you get $$ F_y=e^{x^2-y}-ye^{x^2-y} $$ using the product rule and the chain rule.
Similarly, $$ F_{xx}=2ye^{x^2-y}+4x^2ye^{x^2-y} $$
egreg
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Only $F_x=2xye^{x^2-y}$ is correct. Using both the product rule and chain rule, we get that $F_{xx}=(4x^2+2)ye^{x^2-y}$. Proceeding as you did for $F_x$, but now treating $x$ as a constant and taking the derivative with respect to $y$, we obtain $F_y=(1-y)e^{x^2-y}$. Lastly, using the product and chain rule, we obtain that $F_{xy}=F_{yx}=2xe^{x^2-y}-2xye^{x^2-y}$. If there are any steps you are confused about, I can write out the work.
David_Shmij
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