This question is giving me a hard time conceptually. I know with respect to u the derivative should be Fx Xu + Fy Yu and then substitute v instead of u for respect with v. Out of the information given, I'm having trouble finding the Xu, Yu, Xv, Yv. Any help would be greatly appreciated.
Multi-function partial derivatives confusion.
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multivariable-calculus
1 Answers
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By the multi-variable chain rule:
$$z_u=f_xx_u+f_yy_u$$
It's easy to see that
$$\frac{\partial x}{\partial u}=\frac\partial{\partial u}u^2+v^2=2u$$
$$\frac{\partial y}{\partial u}=\frac\partial{\partial u}\frac uv=\frac1v$$
Plugging in known values, we have
$$f_u(-18,-6)=f_xx_u+f_yy_u=-14\cdot2\cdot18+\frac16=-\frac{3023}6$$
Can you find $f_v(-18,-6)$ like this?
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0by Fx (360, 3)=-11 and Fy(360,3) = 18, the shouldn't Zu = (-11)(-36)+(18)(-1/6)=393? I still don't understand how you chose the plug in values. – 2017-02-22
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0@Megannn I calculated $x_u$ and $y_u$. $f_x$ and $f_y$ are already given, and they are not the coordinate points. – 2017-02-22
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0Xu = 2u, so should it not be 14(-36)+(1/6) ?? I'm sorry I still can't see where the -14 and 18 is from. – 2017-02-22
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0the answer was 393 – 2017-02-22
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0@Megannn Yes, $x_y=2u$, that was my mistake. And that's odd, I'm not sure it should be 393... Hm... – 2017-02-23