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a) Show that the equations \begin{align*} x^3-y^2+2u-v&=1,\\ x^2+y^3+u-v&=2 \end{align*} can be solved simultaneously for functions $u=g(x,y),\ v=h(x,y)$ in a neighbourhood of $(1,1)$ such that $g(1,1) = h(1,1) = 1$.

b) Calculate $\frac{dg}{dx}$ and $\frac{dh}{dy}$ at the point $(1,1)$.

For part (a), I got the answer $-1$. For part (b), my $\frac{dg}{dx}$ is $2x - 3x^2$ and $\frac{dh}{dy}$ is $6y^2 + 2y$.

Please let me know if I am wrong and why; and if i'm correct, please let me know too! Much appreciated!

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  • $\begingroup$ How do you get the answer $-1$ at all? How do you get a number ? ! The exercise cries loud "Implicit Function Theorem". And your Question here looks a bit like homework. $\endgroup$ – Dominic Michaelis May 30 '13 at 15:16
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    $\begingroup$ >How do you get the answer −1 at all? Apparently, she computed the determinant. The second part is also correct, though doesn't answer the question as posed. If this is a homework question from a textbook or a lecture on the implicit function theorem, the author (or the professor) should be reminded that solving an explicit 2 by 2 linear system symbolically is not quite what all that stuff is about. $\endgroup$ – fedja May 30 '13 at 15:29
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For part (a) you should be finding the functions $g(x,y)$ and $h(x,y)$ by solving for $u$ and $v$ (do this either by the substitution or elimination methods you would have learnt). The answers should involve $x$ and $y$.

For part (b) you then differentiate the above functions with respect to the appropriate variables. Your current answer is close, but not quite right.

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