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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).

Twenty-six colours
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1 Answers1

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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.

Vim
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  • 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.     – Twenty-six colours Apr 28 '17 at 08:47
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    since you let $(x,y)$ define $z$, then $(x,y)$ themselves should be independent variables and $y_x=0$. – Vim Apr 28 '17 at 08:48
  • I see, the sign should be the opposite then right? (Might've just been a typo in your paragraph) – Twenty-six colours Apr 28 '17 at 08:49
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    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? – Vim Apr 28 '17 at 08:53
  • My apologies, you are correct about the denominator.. My first order derivative for x is found by chain rule. Am I using it incorrectly?
    $\frac{\partial F}{\partial x} = \frac{\partial F}{\partial x}\frac{\partial z}{\partial x} + \frac{\partial F}{\partial y}\frac{\partial y}{\partial x}$.     – Twenty-six colours Apr 28 '17 at 08:55
  • @PaulWoch the chain rule in the multivariate world is in matrix form as given in my last comment. Google Jacobian matrix for details. – Vim Apr 28 '17 at 08:58
  • I don't think I'm interpreting/understanding correctly. Won't the matrix on the right have just zeroes everywhere besides the entry in row1,column1 and row 2, column 2? – Twenty-six colours Apr 28 '17 at 09:00
  • @PaulWoch yes you are right. The whole point is, if I treat $F(x,y,z)$ as a function of $(x,y)$, then its Jacobian matrix is just $DF \cdot D_{(x,y)}\left(\begin{bmatrix} x \ y \z\end{bmatrix}\right)$ according to the chain rule. You just have to compute each entry accordingly. My solution actually uses only the given fact that $F(x,y,z)=0$ for each $(x,y)$ in some neighbourhood on which $z$ is implicitly defined by $(x,y)$, and then uses the chain rule to differentiate the equation $F(x,y,z(x,y))=0$. What's hard is the existence of such an implicit function $z=z(x,y)$ on a nbhd. – Vim Apr 28 '17 at 09:16
  • @PaulWoch in fact, since the problem already asserted the existence of this implicit function which is the implicit function's job, the only tool we need here is multivariate differentiation and we don't need to invoke the implicit function theorem at all. If you know this big theorem, then you'll see that we only need $F_z\ne 0$ and $F\in C^1$ to assert an implicit map $(x,y)\mapsto z$ locally exists (meaning on a nbhd) where the above two conditions hold. – Vim Apr 28 '17 at 09:26
  • when you say I have to compute each entry accordingly, I'm still quite unsure how it gives me the partial derivative of $F$ w.r.t. to $x$.
    Why am I not able to use this result?:
    $\frac{\partial F}{\partial x} = \frac{\partial F}{\partial z}\frac{\partial z}{\partial x} + \frac{\partial F}{\partial y}\frac{\partial y}{\partial x}$     – Twenty-six colours Apr 28 '17 at 09:32
  • @PaulWoch what's the rationale behind your result? – Vim Apr 28 '17 at 09:41
  • Hmm, I'm not sure. I've used that result (and have been advised that it's the multivaraite chain rule) in other questions and it seems to be working fine. My understanding is that since $F$ is a function of $x,y,z$ and $z$ is a function of $x,y$, then I'm basically finding the partial of $F$ w.r.t. $x$ through finding the other partials as well... – Twenty-six colours Apr 28 '17 at 09:50
  • @PaulWoch the one term you are missing on the rhs is $F_x\partial x/\partial x$. I don't understand why you would leave this one out, isn't $x$ also a function of $x$ (itself)? – Vim Apr 28 '17 at 09:53
  • Won't this then cancel with the $F_x$ term on the LHS? since $x_x = 1$? – Twenty-six colours Apr 28 '17 at 10:01
  • @PaulWoch ok there seems to be a notation conflict. Your LHS doesn't mean the partial derivative in $x$ in its usual sense, but the partial derivative in $x$ of the function $F(x,y,z(x,y))$ in which $z$ is no longer independent of $x$. In the RHS, I meant the usual partial derivative, or, for clarity, the partial derivative of $F$ in its first variable (when its three variables are independent). – Vim Apr 28 '17 at 10:05
  • Thus, sometimes people use $F_x,F_y$ or even $F_1,F_2$ etc to mean the usual partial derivatives when $F$ is deemed as a function of multiple free variables without intervariable dependence like the implicit map. – Vim Apr 28 '17 at 10:09
  • Thanks for your patience. This is quite confusing. What would the RHS of my equation be written as? Or is it correct right now? – Twenty-six colours Apr 28 '17 at 10:12
  • @PaulWoch rhs as I said should include the missing term $F_x\partial x/\partial x$. I advise your LHS be written more explicitly as $\partial F(x,y,z(x,y))/\partial x$. A quick note: $F(x,y,z(x,y))$ is a function of $(x,y)$, only two variables whereas $F(x,y,z)$ has three. Does it clarify? – Vim Apr 28 '17 at 10:15
  • Ah I see. Thanks! – Twenty-six colours Apr 28 '17 at 10:38
  • A small technical question - if z is an implicit function of x and y, wouldn't this mean x is an implicit function of z and y as well? Something like $xyz^2 + yxz = 0$ is an implicit function of z and x isn't it? – Twenty-six colours Apr 28 '17 at 11:01
  • @PaulWoch of course not. For example x^2+y^2+z^2-1=0 determines an implicit function $z=\sqrt{1-x^2-y^2}$ locally at $(0,0,1)$, but at this point there doesn't exist implicit function to express $x$ or $y$ as their partial derivatives vanish here. – Vim Apr 28 '17 at 11:09