# Implicit function theorem exercise with higher derivatives

Consider the equation $e^{xz}+y-z=e$. Using the implicit function theorem shows $z$ is a smooth function of $x,y$ about $(1,1,1)$. I needed to calculate a directional derivative of $z$ at $(1,1)$ and managed that using the implicit function theorem to recover the gradient of $z$.

Now I'm asked whether the partial derivatives of $z$ are symmetric about $(1,1)$ and furthermore, I need to calculate them.

I think the partial derivatives are symmetric because the original function $e^{xz}+y-z=e$ is smooth, which means so is $z=z(x,y)$. I don't understand however how to find second order derivatives. The "formula" $$\frac{\partial z}{\partial x}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}$$ (read with matrix inverse instead of quotient in matrix case) does not really make sense before it's evaluated at a point, since the RHS has additional variables.

So how to find $\frac{\partial ^2z}{\partial x\partial y}(1,1)$?

Formulas for second partial derivatives of $z$ are given at this link: implicit surface derivatives. Some discussion of how to compute these formulas is here: curvature of implicit surface

• Wait, am I right about the symmetry? – eleventeen Feb 9 '17 at 1:35
• The mixed partials of $z$ are symmetric because the mixed partials of $f(x,y,z)=e^{xz}+y-z-e$ are symmetric. See the formula for $z_{xy}$ at the link. If you switch $x$ and $y$, you get the same thing except for an $f_{yx}$ term instead of $f_{xy}.$ For these formulas to make sense, $f_z$ must also be non-zero, which it is at $(1,1,1)$. – J. Heller Feb 13 '17 at 6:34