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I am working on this textbook problem but I am not sure how to go about this because of all the variable names confusing me...I know that if you have a function u= x+y and you want the partial derivative of u with respect to x then you just differentiate x and keep y as a constant..But in the below equation it is somewhat different.

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Normally they want you to find du/dx or dv/dy but here it is the opposite. Also, y here is used in both u and v so how can you differentiate with respect to u.

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Hint: We have: $$ \frac{d}{du} u=1= 2x \frac{\partial x}{\partial u} -2y \frac{\partial y}{\partial u} \\ \frac{d}{du} v=0= 2x \frac{\partial x}{\partial u} -\frac{\partial y}{\partial u} $$ Then $$ \left(\begin{array}{cc}2x&-2y\\2x&-1\end{array}\right)\left(\begin{array}{c}x_u\\ y_u\end{array}\right)=\left(\begin{array}{c}1\\ 0\end{array}\right) $$ and we can find $x_u$ and $y_u$. Same idea to find $x_v$ and $y_v$.

I recommend you look at a book of multivariable calculus for full details.

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  • $\begingroup$ Why are you taking the total derivative w.rt. $u$ instead of the partial? $\endgroup$ – Ataraxia Sep 27 '13 at 16:24
  • $\begingroup$ Formally, $u$ is a function of $x$ and $y$. $\endgroup$ – Pocho la pantera Sep 27 '13 at 17:22

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