Not sure how to interpret this question or where to start.
$\text{Assuming that the equation}$ $$F(x,y,z) = 0$$ $\text{defines} z \text{implicitly as a differentiable function of} \: x \: \text{and} \: y \: \text{and that}$ $$F_{zx} = F_{xz}$$ $\text{show that}$ $$\frac{\partial ^2 z}{\partial x^2} = \frac{-(F_{z})^2 F_{xx} + 2F_{z}F_{x}F_{xz} - (F_{x})^2 F_{zz}}{(F_{z})^3}.$$

I have no idea how to use the given equation to imply that. I know how to implicity differentiate functions but when they actually give a function...
Also, from their definition, it means that this is true right?
$z \equiv z(x,y)$
(We haven't learn the implicit function theorem btw).


A heuristic approach: from $F(x,y,z)=0$, you differentiate in terms of $x$ and $y$ repsectively (just treat $z=z(x,y)$) to get \begin{align} F_x+F_zz_x &= 0 \\ F_y+F_zz_y &=0 \end{align} and you solve $(z_x,z_y)=-\frac1F_z(F_x,F_y)$.

Then you differentiate $F_x+F_zz_x=0$ wrt $x$ again to obtain $$F_{xx} + F_{zx}z_x + z_x(F_{xz}+F_{zz}z_x)+F_zz_{xx}=0$$ and you get $z_{xx}$. It turns out $z_y$ is unused, but in general even if you only need the second partial derivative in a particular defining variable, you will still have to differentiate in all the defining variables as all first order partial derivatives can arise later on.

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  • $\begingroup$ Thanks. How did you get your two results in the beginning? I used chain rule on $F$ and obtained $$F_{x} = F_{z} z_{x} + F_{y} y_{x}$$. Is it true that $y_{x} = 0$ (as is with any other combination of x,y,z)? Since we can think of it as partial derivative of y w.r.t x. $\endgroup$ – Twenty-six colours Apr 28 '17 at 8:47
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    $\begingroup$ since you let $(x,y)$ define $z$, then $(x,y)$ themselves should be independent variables and $y_x=0$. $\endgroup$ – Vim Apr 28 '17 at 8:48
  • $\begingroup$ I see, the sign should be the opposite then right? (Might've just been a typo in your paragraph) $\endgroup$ – Twenty-six colours Apr 28 '17 at 8:49
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    $\begingroup$ which sign? I don't think anything is wrong in my first order derivatives. It's just the first entry of $$[F_x,F_y,F_z]\begin{bmatrix}\frac{\partial x}{\partial x} & \frac{\partial x}{\partial y}\\ \frac{\partial y}{\partial x}& \frac{\partial y}{\partial y}\\ \frac{\partial z}{\partial x}& \frac{\partial z}{\partial y}\end{bmatrix}$$ Actually I wonder how you got yours. Also, it seems that the denominator in the answer should be $F_z^3$ instead of $F_z^2$, could you confirm it again? $\endgroup$ – Vim Apr 28 '17 at 8:53
  • $\begingroup$ My apologies, you are correct about the denominator.. My first order derivative for x is found by chain rule. Am I using it incorrectly? $\frac{\partial F}{\partial x} = \frac{\partial F}{\partial x}\frac{\partial z}{\partial x} + \frac{\partial F}{\partial y}\frac{\partial y}{\partial x}$. $\endgroup$ – Twenty-six colours Apr 28 '17 at 8:55

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