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I am solving past calculus exams, and I came across the following question.

Does the equation: $$ F(x,y,z) = 2\sin(x^2yz) - 3x + 5y^2 - 2e^{yz} = 0 $$ define a differentiable function $z = f(x,y)$ in a neighborhood of $p = (1, 1, 0)$?

At first, I thought this was a natural candidate for the implicit function theorem, but: $$ \left.\begin{matrix} \frac{\partial }{\partial z}F \end{matrix}\right|_{(1,1,0)} = \begin{matrix} 2x^2y\cos(x^2yz) -2ye^{yz} \end{matrix}_{(1,1,0)} = 0 $$ hence, the theorem doesn't hold in this case.

Ideas? Thanks!

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1 Answer 1

up vote 2 down vote accepted

Following Fleming, Functions of several variables, page 150, if $F(x,y,z(x,y))=0$, then $F_1+F_3 z_1=0$. But in our case $F_3=0$ and $F_1=-3$ at $(1,1,0)$. Hence no differentiable function $z=z(x,y)$ can exists such that $F(x,y,z(x,y))=0$ identically in a neighborhood of $(1,1,0)$.

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