Can someone please help with this homework problem: Given the equations \begin{align} x^2 - y^2 - u^3 + v^2 + 4 &= 0 \\ 2xy + y^2 - 2u^2 + 3v^4 + 8 &= 0 \end{align} find $\frac{\partial u}{\partial x}$ at $(x,y) = (2,-1)$.
In case it may be helpful, we know from part (a) of this question (this is actually part (b)) that these equations determine functions $u(x,y)$ and $v(x,y)$ near the point $(x,y,u,v) = (2, -1, 2, 1)$.
While I am taking a high-level multivariable calc class, I have not yet taken linear algebra or diffeq yet, so please refrain from using such techniques in your solutions.
Thanks a lot!
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$\begingroup$Consider $u(x,y), v(x,y)$ as functions and $x,y$ variables. Take the $x$ derivative of both equations ($y$ is a constant). Use the chain rule to derivate $u$ and $v$. Then set $x=2, y = -1$ and solve.You have 4 equations and 4 unknowns ($u, v, \frac{\partial u}{\partial x}, \frac{\partial v}{\partial x}$) , so you should be able to do that.
If you have seen it, depending on the level of rigor of your course, you should use the implicit function theorem to show than you can solve for $u,v$ and everything is well-defined.
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