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To differentiate an implicit function $y(x)$, defined by an equation $R(x, y) = 0$ one can totally differentiate $R(x, y) = 0$ with respect to $x$ and $y$ and then solve the resulting linear equation for $\frac{dy}{dx}$ to explicitly get the derivative in terms of $x$ and $y$.

Consider the following example: Let $y(x)$ be defined by the following relation:

$$(x^2-y^2)^{1/2}+\arccos\frac{x}{y}=0. \,(y\neq 0.)$$

Clearly, the equation defines $y$ as a function of $x$. In fact, it is easy to see that $y=x$. However, when I apply the method of implicit differentiation to $(x^2-y^2)^{1/2}+\arccos\frac{x}{y}=0$, I failed to get the desired result $\frac{dy}{dx}=1$ (since $y=x$). Why does implicit differentiation fail here?

Edit: I did not do the implicit differentiation by hand as it is too tedious; instead I trusted the result on WolframAlpha:

enter image description here

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  • $\begingroup$ What result did you get? $\endgroup$
    – JRN
    May 28, 2020 at 5:33
  • $\begingroup$ @JoelReyesNoche, I have edited my question; thank you for your comment! $\endgroup$
    – Zuriel
    May 28, 2020 at 5:44
  • $\begingroup$ It's too tedious to answer, why don't you ask Wolfram Alpha? $\endgroup$
    – user436658
    May 28, 2020 at 6:00
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    $\begingroup$ It means your R(x,y) does not satisfy the conditions of the implicit function theorem. Note for instance that R(x,y) is not defined for the half plane where $y>x$, therefore it can not be continuously differentiable on the line $y=x$. $\endgroup$ May 28, 2020 at 6:03
  • $\begingroup$ @ProfessorVector, WolframAlpha gives great answers to many questions but I am not sure how to ask this particular one. $\endgroup$
    – Zuriel
    May 28, 2020 at 6:04

3 Answers 3

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With a bit of work, you can show that $R(x,y)$ is only defined for $y=x$ and $y=-x$ excluding the point $(0,0)$. Hence your function does not satisfy the conditions of the implicit function theorem, which states that your function should be at least continuously differentable to apply the method of implicit differentiation.

Parts of the plane not in the domain of R

Note, you could extend your problem to complex numbers, in that case the domain of applicability of $R(x,y)$ would be extended. However, even in that case, it would still not be the entire plane because no definition of the square root or the inverse cosine is analytic on the entire complex plane. In particular, if you choose the most "natural" extension to the complex plane of the inverse cosine, it is not defined on the branch points $-1$ and $1$. Those correspond exactly to the cases $y=x$ and $y=-x$. In other words, naïve implicit differentiation does not work in that case either. You would therefore need analytic extensions of the square root and the inverse cosine such that they are valid in the case $y=x$. But, those extensions will not correspond to the definitions you adopted in the real case.

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Let's see. $$1/2(x^2-y^2)^{-1/2}(2x-2y\dfrac{\rm dy}{\rm dx})-1/\sqrt{1-(x/y)^2}(1/y-x/y^2\dfrac{\rm dy}{\rm dx})=0\implies \dfrac{\rm dy}{\rm dx}(y-x/y)/\sqrt{x^2-y^2}=-(x-1)/\sqrt{y^2-x^2}\implies\dfrac{\rm dy}{\rm dx}=y/x(x-1)$$ etc.

It looks like the chances of this agreeing with what "you" got are, at least, decent. Though I probably have a careless mistake (or two) "in there ".

That being said, and notwithstanding the comments in the other answer, and also noting that the equation is not differentiable, since you get division by zero or $i$ appears, I must say that I don't quite see why it is obvious that $y=x$. I mean if you plug in $y=x$ you do get a true statement. But so what. What about necessity?

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  • $\begingroup$ Here we are considering real functions. So both terms in the equation are non-negative. $\endgroup$
    – Zuriel
    May 28, 2020 at 12:53
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$ dR(x,y)=\dfrac{\partial R}{\partial y}dy+\dfrac{\partial R}{\partial x}dx=0 $

Since $y(x)$ it follows that $dy=\frac{dy}{dx}dx$

$dR(x,y)=\dfrac{\partial R}{\partial y}dy+\dfrac{\partial R}{\partial x}dx=\dfrac{\partial R}{\partial y}\dfrac{dy}{dx}dx+\dfrac{\partial R}{\partial x}dx=\big(\dfrac{\partial R}{\partial y}\dfrac{dy}{dx}+\dfrac{\partial R}{\partial x} \big)dx=0$

$\rightarrow \dfrac{dy}{dx}=-\dfrac{\dfrac{\partial R}{\partial x}}{\dfrac{\partial R}{\partial y}}$

If you try it on Wolfram Alpha with

-d/dx(√(x^2-y^2)+arccos(x/y))/d/dy(√(x^2-y^2)+arccos(x/y))

it provides the required answer

enter image description here

enter image description here

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