This is from Tenenbaum & Pollard's "Ordinary Differential Equations" book, chapter 1, Exercise 2, problem # 16.

There are two relations introduced in problems 14 and 15:

(1) $\sqrt{x^2 -y^2} + \arccos{(x/y)} = 0, y\neq0$

(2) $\sqrt{x^2 -y^2} + \arcsin{(x/y)} = 0$

Problem 16 asks:

Can you apply the method of differentiation as taught in the calculus to the function of problem 14, of problem 15?

The book says the answer for both is "no".

I think I Understand why this is for (2), because (2) doesn't implicitly define y as a function of x (domain of 1st and 2nd terms are contradictory).

But for (1), I'm not sure why you can't implicitly differentiate. I think I came up with an explanation but want to verify if this is the correct reasoning:

So I start by finding $\frac{d}{dx}$ of both sides:

$\sqrt{x^2 -y^2} + \arccos{(x/y)} = 0, y\neq0$

$\frac{d}{dx}(\sqrt{x^2 -y^2} + \arccos{(x/y)}) = \frac{d}{dx}(0)$

$\frac{d}{dx}(\sqrt{x^2 -y^2}) + \frac{d}{dx}(\arccos{(x/y)}) = 0$

$\frac{1}{2}(x^2 -y^2)^{\,(-\frac{1}{2})}(2x-2y\frac{dy}{dx}) + \Big(\frac{-1}{\sqrt{1-(x/y)^2}}\cdot\frac{y-x\frac{dy}{dx}}{y^2}\Big) = 0$

$\frac{(x-y\frac{dy}{dx})}{\sqrt{x^2 -y^2}} - \Big(\frac{y-x\frac{dy}{dx}}{y^2\sqrt{1-(x/y)^2}}\Big) = 0$

$\frac{(x-y\frac{dy}{dx})}{\sqrt{x^2 -y^2}} = \Big(\frac{y-x\frac{dy}{dx}}{y^2\sqrt{1-(x/y)^2}}\Big)$

So everything seems fine so far, but if we look at the domain of these two terms, they are contradictory.

The left term implies:

$x^2-y^2>0 \Rightarrow x^2>y^2 \Rightarrow |x|>|y|$

And the right hand side implies:

$1-\frac{x^2}{y^2}>0 \Rightarrow 1>\frac{x^2}{y^2} \; \Rightarrow \; y^2>x^2 \; \Rightarrow |y|>|x|$

...which contradict each other, and therefore the equation we arrived at by implicitly differentiating (1) is meaningless, and therefore we cannot implicitly derive equation (1). Is this the correct reason for why the book says that we cannot apply implicit differentiation to the function of problem 14 (equation (1))?

Note: the question is asking to differentiate "the function of problem 14." Problem 14 gives equation (1) and asks you to find a function g(x) that is implicitly defined by equation (1). The answer for #14 was: $y = g(x) = x$.

The wording of this was unclear to me, I wasn't sure if I was supposed to be implicitly differentiating $y=g(x)=x$, or the original equation, eq (1). If I'm supposed to do $y = g(x) = x$, then I'm really not sure at all why the answer is "No".

If we implicitly differentiate $y = g(x) = x$ wouldn't we get $\frac{dy}{dx}=1$ ?


Note that $\arccos$ and $\arcsin$ are both defined on $[-1,1]$, so you have to have $\left| \dfrac x y \right| \le 1$. On the other hand, existence of the square root requires that $x^2 - y^2 \ge 0$ or, since $y \ne 0$, that $\left( \dfrac x y \right) ^2 \ge 1$, so the only possibility is $y = \pm x$, but this equality does not define on open set, only a finite set of points, and the concept of derivability makes no sense on such sets.

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  • $\begingroup$ $y = -x$ doesn't satisfy the original relation (eq. 1), so it would be $y=x$. So why can't we differentiate? And just to be clear, when you say $y=\pm x$ here, you are saying that the relation (1) implicitly defines the function $y=g(x)=\pm x$, correct? I'm not sure what you mean by "this equality does not define on open set". Do you mean that $y=\pm x$ isn't a valid function, therefore can't be differentiated? $\endgroup$ – mathiness Dec 28 '15 at 3:46
  • $\begingroup$ @mathiness, so, did you finally understand why it is impossible? I am stuck at this point too. $\endgroup$ – Turkhan Badalov Jun 24 '18 at 23:44
  • $\begingroup$ Why doesn't the differentiation make sense on finite set of points? I can't even understand why it makes sense only on an open set $\endgroup$ – Turkhan Badalov Jun 24 '18 at 23:56

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