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):
So here, in the beginning $\frac d {du}$ is done on both sides of equation.
However, in the book it says like this:
And an example from the book:
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?
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 $0 = x_u +(2y)y_u$