# Show that $\partial ^2 y / \partial x \partial y = 0$ for implicit function $f(x,y(x))$?

I find Implicit function theorem (limited to scalar function) both fascinating and a bit puzzling. Simply said, total derivative of some $$f(x,y(x))$$ according to $$x$$ is:

$$\frac{d f(x,y(x)) }{ dx } = f_{x} \frac{\partial x}{\partial x} + f_{y} \frac{\partial y}{\partial x} = 0$$

It is straight forward to see, that such total derivative of $$f(x,y(x))$$ can be expressed through partial derivative:

$$\frac{\partial y}{\partial x} = - \frac{f_{x}(x, y(x))}{f_{y}(x,y(x))} \tag {1}$$

where for brevity $$f_{x} = \frac{\partial f(x,y(y))}{\partial x}$$ and $$f_{y} = \frac{\partial f(x,y(y))}{\partial y}$$

So far, so good. Now lets try to continue by taking partial derivative with respect to $$y$$. This can naturally happen when one needs to evaluate implicit function $$f(x,y(x))$$ for variable $$y$$ at variable $$x$$, it's first derivative $$\frac{\partial y}{\partial x}$$ and naturally one would expect second derivative $$\frac{\partial ^2 y}{\partial ^2 x}$$ and $$\frac{\partial ^2 y}{\partial x \partial y}$$ to exist, e.g. for calculating Hessian matrix.

Case 1: apply quotient rule on {1}

One could blindly do the following in order to obtain $$\frac{\partial}{\partial y}\frac{\partial y}{\partial x}$$ to get $$\frac{\partial ^2 y}{\partial x \partial y}$$:

$$\frac{\partial ^2 y}{\partial x \partial y} = - \frac{\partial (\frac{f_{x}}{f_{y}})}{\partial y}$$ which holds, if higher derivatives of $$f_x$$ and $$f_y$$ exists.

Case 2: analyze where the expression becomes constant

Obviously $$\frac{\partial y}{\partial y} = 1$$. This happens within $$\frac{\partial ^2 y}{\partial y \partial x}$$ and since $$\frac{\partial ^2 y}{\partial x \partial y} = \frac{\partial ^2 y}{\partial y \partial x}$$, then it is expected to happen for any mixed second-order partial derivative.

This way, one should have always $$\frac{\partial ^2 y}{\partial y \partial x} = 0$$ for an implicit function, since deriving $$\frac{\partial y}{\partial y} = 1$$ is the derivation of a constant, thus 0.

## Question

Which of the 2 cases is wrong and why?

## More details

To provide more details, but still keep the question general, lets assume the form of $$f(x,y(x))$$, to provide some visual clues.

$$f(x,y(x)) = e^{(x+y)} + x + y = 0$$

## Background (to give the equations intuitive motivation related to experimental observations)

Variables $$x$$ and $$y$$ are physical quantities (current, voltage). $$f(x,y(x)) = 0$$, thus this function is used to evaluate the quantities $$x$$ and $$y$$ only. For instance, I need the value of $$y$$ for $$x=c$$, then I evaluate $$f(c,y) = 0$$ to get $$y$$. This also means, that when one variable changes (e.g. $$x$$) then $$y$$ changes as well. Therefore, when I calculate derivatives, then naturally I take $$\frac{\partial x}{\partial y}$$, because that is what I observe as a derivative at the level of the physical system (e.g. lowering current, rises the voltage). This means, that in the Jacobian and Hessian I expect to see $$\frac{\partial x}{\partial y}$$, rather than $$\frac{\partial f(x,y(x))}{\partial y}$$.

• You are abbreviating far too much. Eqn (1) should read $\frac{\partial y}{\partial x}=-\frac{f_x(x,y(x))}{f_y(x,y(x)}$, it only holds when we substitute in for $x$ the value $y(x)$ of the implicitly defined function. So both sides are univariate functions of $x$, no $y$ occurs in either. Sep 6, 2021 at 16:07
• This makes no sense. Instead of $y(x)$, write the implicitly defined function as $y=g(x)$, and I hope you'll see that it's quite meaningless to ask about the derivative of $g(x)$ or $g'(x)$ with respect to $y$... Sep 6, 2021 at 17:20
• @ancientmathematician I have corrected the eq. 1 Sep 7, 2021 at 6:02
• @HansLundmark "Makes no sense" is the puzzling part, since these expressions still appear in Hessian. Does "meaningless" means the $\partial y / \partial y = 1$ or $0$ since $g(x)$ is not a function of $y$? There is this excellent answer math.stackexchange.com/questions/3600407/…, which I think hints the same thing as you, but I am getting lost in notation, when the expression in square brackets apperar Sep 7, 2021 at 6:08
• The Hessian of what, exactly? The Hessian of a one-variable function $g(x)$ is just the $1 \times 1$ matrix containing the single entry $g''(x)$. The Hessian of a two-variable function $f(x,y)$ is a $2 \times 2$ matrix containing the second partial derivatives of $f$. I think your confusion comes from writing the implicitly defined function as $y(x)$, which causes you to mix up this one-variable function with the independent variable $y$ appearing as an argument to the two-variable function $f(x,y)$. So, as I already said, don't write the function like that. Call it $g(x)$ instead. Sep 7, 2021 at 7:43

$$\frac{\partial^2 y}{\partial x\partial y}$$ means $$\frac{\partial}{\partial x}\left(\frac{\partial y}{\partial y}\right)$$. If you want to calculate it the other order (differentiate with respect to $$x$$ first), it is $$\frac{\partial^2 y}{\partial y\partial x}$$.

Recall in the proof of symmetry of partials that we need to be able to change $$x$$ and $$y$$ independently (and the twice-differentiability of $$f\colon U\to E$$). Now symmetry of partial derivatives $$\frac{\partial^2 f(x,y,\dots)}{\partial x\partial y}=\frac{\partial^2 f(x,y,\dots)}{\partial y\partial x}$$ does not hold if $$x,y$$ are dependent in general. For example, on $$\mathbb{R}^+$$ with local coordinate $$x$$, consider also another local coordinate $$y=x^2$$. The two differential operators $$\frac{\mathrm{d}}{\mathrm{d}x}$$ and $$\frac{\mathrm{d}}{\mathrm{d}y}$$ do not commute: We have $$\frac{\mathrm{d}}{\mathrm{d}y}=\frac1{2x}\frac{\mathrm{d}}{\mathrm{d}x}$$, so

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}y}\frac{\mathrm{d}}{\mathrm{d}x} &=\frac1{2x}\cdot\frac{\mathrm{d}^2}{\mathrm{d}x^2} \\ \frac{\mathrm{d}}{\mathrm{d}x}\frac{\mathrm{d}}{\mathrm{d}y} &=\frac{\mathrm{d}}{\mathrm{d}x}\frac1{2x}\frac{\mathrm{d}}{\mathrm{d}x}\\ &=\frac1{2x}\frac{\mathrm{d}}{\mathrm{d}x}\frac{\mathrm{d}}{\mathrm{d}x} +\left(\frac{\mathrm{d}}{\mathrm{d}x}\frac1{2x}\right)\frac{\mathrm{d}}{\mathrm{d}x}\\ &= \frac1{2x}\frac{\mathrm{d}^2}{\mathrm{d}x^2} -\frac1{2x^2}\frac{\mathrm{d}}{\mathrm{d}x}\\ \end{align*} In other words, for any nonconstant twice-differentiable function $$f\colon\mathbb{R}^+\to\mathbb{R}$$ we have $$\frac{\mathrm{d}}{\mathrm{d}x}\frac{\mathrm{d}}{\mathrm{d}y}f(x)-\frac{\mathrm{d}}{\mathrm{d}y}\frac{\mathrm{d}}{\mathrm{d}x}f(x)=-\frac1{2x^2}f'(x)\neq 0.$$

• I believe you are onto something here, I dont know who downvoted the answer. But excuse my lack of knowledge, but $\frac{\partial}{\partial x}$ and $\frac{d}{d x}$ are technically operators. Dont you then mean $\frac{d}{d x} \frac{d}{d y} x^2$ ? If I am wrong or I miss read your notation, please rewrite your answer, such that differentiation operators are separated from operands. Sep 7, 2021 at 8:51

Most confusion about partial derivatives can be cured by being explicit about which function you are taking partial derivative of. So, what is the function $$y$$? What are its parameters?

Perhaps you are talking about the function $$y(x,f)$$?

Then, indeed $$\frac{\partial y}{\partial x} = -\frac{f_x}{f_y}$$. However, for this function, $$\frac{\partial y}{\partial y}$$ is a meaningless combination of symbols since $$y$$ is not one of its parameters.

Or, maybe, you mean function $$y(x,y)$$?

Now you can have a partial derivative with respect to $$y$$: $$\frac{\partial y}{\partial y} = 1$$. But, for this function, $$\frac{\partial y}{\partial x} = 0$$.

• I have added More details and also in question comments I have asked @HansLundmark to elaborate on $y(x,f)$ or $g(x)$ and such. I think you and Hans try to guide me the right way, but I somewhat fail to recognize the general case. May I ask to edit your answer and clearly state, what $y(x,f)$ or $y(y,x)$ would be in case of the provided $f(x,y(x))$? If it seems that I am asking for too specific answer (e.g. solving for specific $f(x,y(x))$) , then after I solve this issue, I will edit your answer (if necessary) to apply to general case. And of course accept it as correct answer. Sep 7, 2021 at 8:58