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I have been doing a few exercises involving implicit differentiation, and I encountered a contradiction(or that is at least what I think it is):

enter image description here

So here, in the beginning $\frac d {du}$ is done on both sides of equation.

However, in the book it says like this:

enter image description here

And an example from the book: enter image description here

So if I apply the logic from the book in my above problem the solution would say: The system of equations given in this problem corresponds to: $(\frac {du} {dx})_y$, and to calculate it we regard $u$ and $v$ as functions of $x$ and $y$ and differentiate the given equations with respect to x, holding y constant. Hence I would not differentiate this with $\frac d {du}$ like they did, but with $\frac d {dx}$. What am I missing?

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You are not missing anything.

You will get the same result anyway.

$ u= x^2+y$ therefore $1= (2x)x_u +y_u$ and $ v= x+y^2$ therefore $0 = x_u +(2y)y_u$

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  • $\begingroup$ This is the way the problem is solved in the solutions. But my question is: If I apply the logic from the book that I stated above, I would have to differentiate this with $\frac d {dx}$ and not $\frac d {du}$ as in the solutions? $\endgroup$ – Relax295 May 13 '18 at 13:34
  • $\begingroup$ $\frac d {dx}(x^2+y)=(2x)$ and you multiply by $x_u$ which gives you the same result. $\endgroup$ – Mohammad Riazi-Kermani May 13 '18 at 13:59
  • $\begingroup$ Is there a specific reason why do you multiply it by $x_u$, that is, why do we differentiate with respect to $u$, and not with respect to $x$? $\endgroup$ – Relax295 May 13 '18 at 14:11
  • $\begingroup$ we are applying implicit differentiation assuming that $x$ is a function of $u$ $\endgroup$ – Mohammad Riazi-Kermani May 13 '18 at 14:26

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