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I have the following:

$\frac{dy}{dx}=-\frac{x^2}{y^2}$

I would like to find $\frac{d^2y}{dx^2}$. One solution suggested was to take the partial derivative of the above function with respect to x, treating y as a constant. This gives the following solution:

$\frac{\partial \:}{\partial \:x}\left(-\frac{x^2}{y^2}\right)=-\frac{2x}{y^2}$

Which is correct. Why can y be treated as a constant here; why does taking this partial derivative give the second derivative of this implicit function?

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  • $\begingroup$ I'm thinking about writing the whole equation as $F(x,y)=0$, thus defining the function $F$ with $x$ and $y$ as its two independent variables. Then you could take the partial derivative $\partial F/\partial x$ and essentially treat $y$ as a constant. But I'm not sure as to how then you'd calculate $d^2y/dx^2$; you should differentiate that with respect to $x$ as well. Sorry for asking more questions, I just thought I should comment my rationale in case it helps; it seems to me that using the implicit function theorem leads somehow to what you're looking for. $\endgroup$
    – Chris
    Sep 18, 2018 at 19:49
  • $\begingroup$ You can get displayed equations by enclosing them in double instead of single dollar signs. That makes them a lot easier to read, especially when you're mixing fractions and exponents. Please see this tutorial and reference on how to typeset math on this site. $\endgroup$
    – joriki
    Sep 18, 2018 at 20:04
  • $\begingroup$ I don't think it's correct. What are you basing that statement on? For instance, $y=-x$ satisfies the first equation, but the second derivative is zero and isn't given by second equation. $\endgroup$
    – joriki
    Sep 18, 2018 at 20:07
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    $\begingroup$ @joriki Thank you for your comment. It would appear you are correct; I misinterpreted my source. Thanks for the display equations tip as well. $\endgroup$
    – rodit
    Sep 18, 2018 at 20:28
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    $\begingroup$ You're welcome. I've added my comment as an answer so that you can mark the question as answered. $\endgroup$
    – joriki
    Sep 18, 2018 at 20:36

1 Answer 1

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This is not the second total derivative of $y$ with respect to $x$. For instance, $y=−x$ satisfies the first equation, but the second derivative is zero and isn't given by the second equation.

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