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I have to solve this question with this solution way. But I made some mistakes while solving. I cannot see thesemistakes. And I cannot reach the wanted result properly. Please somebody helps me. Thank you so much. I Will be happy to help me.

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I added an example from my notebook. I need to solve the question 11.6.4(above picture) with this example's solution way. I tried to solve the question 11.6.4 with this solution way. But I failed.

The example's photo is folowing..

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  • $\begingroup$ in 2) why do you have $4\times 4$ matrix rather than $4\times 2$? $\endgroup$ – Ilya Jul 19 '13 at 12:56
  • $\begingroup$ In fact, I have just learnt this solution way. I could not solve totally knowingly. But I need to use this solution way. I have an exam tommorow. I have to learn this. Please can you solve this question below answer in detail? Thank you so much @Ilya $\endgroup$ – B11b Jul 19 '13 at 12:59
  • $\begingroup$ There I Try to find $DF(u,v,x,y)$. For the similar question on the notebook, I wrote so @Ilya $\endgroup$ – B11b Jul 19 '13 at 13:01
  • $\begingroup$ This is really the implicit function theorem, and in finding the solution this way, you are deriving the proof of the implicit function theorem from the inverse function theorem. As for your answer, I am having trouble understanding what you mean by "at $ux=1/2$, $vx=1/2$, $\nabla DF \neq 0$". First of all, $\nabla$ is the sign for gradient, and you should write $\Delta DF$, or better still, $|DF|$ for the determinant. Second of all, which values of $x,y,u,v$ are you plugging in here? Do they satisfy the equations at the top? $\endgroup$ – Eric Auld Jul 19 '13 at 13:51
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    $\begingroup$ I have tried to post a clearer version of the question here. Perhaps we will get an answer. math.stackexchange.com/questions/447475/… $\endgroup$ – Eric Auld Jul 19 '13 at 15:29
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The Jacobian is $4uv(x^2-y^2)$, so it is nonzero where $u,v\neq 0$ and $|x|\neq |y|$. To derive the desired relation, simply add the two equations, and divide by $x+y$.

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  • $\begingroup$ I understand thank you:)) $\endgroup$ – B11b Jul 19 '13 at 16:26

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