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.


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 $\frac{\partial f(x,y(x))}{\partial y}$.

  • 2
    $\begingroup$ 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. $\endgroup$ Sep 6, 2021 at 16:07
  • 1
    $\begingroup$ 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 $y$... $\endgroup$ Sep 6, 2021 at 17:20
  • $\begingroup$ @ancientmathematician I have corrected the eq. 1 $\endgroup$
    – Martin G
    Sep 7, 2021 at 6:02
  • $\begingroup$ @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 $\endgroup$
    – Martin G
    Sep 7, 2021 at 6:08
  • $\begingroup$ 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. $\endgroup$ Sep 7, 2021 at 7:43

2 Answers 2


Be careful about the order.

$\frac{\partial^2 y}{\partial x\partial y}$ 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 $$ \frac{\mathrm{d}}{\mathrm{d}x}\frac{\mathrm{d}}{\mathrm{d}y}f(x)-\frac{\mathrm{d}}{\mathrm{d}y}\frac{\mathrm{d}}{\mathrm{d}x}f(x)=-\frac1{2x^2}f'(x)\neq 0. $$

  • $\begingroup$ 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. $\endgroup$
    – Martin G
    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, $\frac{\partial y}{\partial x} = 0$.

  • $\begingroup$ 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. $\endgroup$
    – Martin G
    Sep 7, 2021 at 8:58

You must log in to answer this question.

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