2
$\begingroup$

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

$\endgroup$

2 Answers 2

3
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ Wait, am I right about the symmetry? $\endgroup$
    – 224160
    Feb 9, 2017 at 1:35
  • $\begingroup$ 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 $(1,1,1)$. $\endgroup$
    – J. Heller
    Feb 13, 2017 at 6:34
3
$\begingroup$

You can try these formulas \begin{cases} \begin{split} \dfrac{\partial^2z}{\,\partial\,\!x^2}&=\dfrac{1}{{F_z}^3} \begin{vmatrix} F_{xx}&F_{xz}&F_{x}\\ F_{zx}&F_{zz}&F_{z}\\ F_{x}&F_{z}&0\\ \end{vmatrix}\\ \\ \dfrac{\partial^2z}{\partial\,\!x\partial\,\!y}&=\dfrac{1}{{F_z}^3} \begin{vmatrix} F_{xy}&F_{yz}&F_{y}\\ F_{zx}&F_{zz}&F_{z}\\ F_{x}&F_{z}&0\\ \end{vmatrix}\\ \\ \dfrac{\partial^2z}{\,\partial\,\!y^2}&=\dfrac{1}{{F_z}^3} \begin{vmatrix} F_{yy}&F_{yz}&F_{y}\\ F_{zy}&F_{zz}&F_{z}\\ F_{y}&F_{z}&0\\ \end{vmatrix}\\ \end{split} \end{cases}

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .