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I'm new to this community, so please bear with me if I make a mistake following rules and conventions, and thanks in advance.

I'm currently learning - following along from this video ("Implicit differentiation, what's going on here? | Essence of calculus").

At 2:42 in the video timeline, the teacher mentions that the implicit curve equation $x^2+y^2=5^2$, which is then resolved into its derivative form: $2x\,\mathrm{d}x+2y\,\mathrm{d}y=0$ can then be rearranged to the form expression: $\frac{\mathrm dy}{\mathrm dx} = -\frac{x}{y}$. I wanted to understand how this rearrangement happened? I tried to explore this equation rearrangement: $2x\,\mathrm{d}x+2y\,\mathrm{d}y=0$ to $\frac{\mathrm dy}{\mathrm dx} = -\frac{x}{y}$ from online sources, but to no joy. By the way, I also tried to make the question more readable in terms of using the proper maths equation symbols syntax, using this link, but I got an error message from this website saying that I don't have enough reputation score to post images in my question. So I hope I made this question as readable and clear as possible, with the limits I'm given. Thanks in advance.

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    $\begingroup$ Note that the method of implicit differentiation that Sanderson presents is an abuse of notation used provide a more intuitive understanding of why it works. A more formal process would resemble the following $$ x^2 + y^2 = 5 \implies \frac{d}{dx}\left(x^2 + y^2\right) = \frac{d}{dx} 5^2 \\\implies 2x + 2y\frac{dy}{dx} = 0 \implies 2y\frac{dy}{dx} = -2x \\\implies \frac{dy}{dx} = \frac{-x}{y} $$ $\endgroup$
    – Brian
    Mar 28 '19 at 11:59
  • $\begingroup$ That makes total sense. Thanks Brian. What confused me first is the fact Grant Sanderson, plugged dx and dy to multiply by the derivatives 2x and 2y (but then he mentioned that he will explain more on this later in the video). Also the notation dy/dx is actually just to represent that we're trying to find out the derivative, but I confused it as being part of the equation workout. Also, thanks for editing my question in a better format. How do I mark your answer as "best answer" or give it a vote, since it's a comment? $\endgroup$
    – Hazzaldo
    Mar 28 '19 at 12:17
  • $\begingroup$ Since my comment suffices, I incorporated it into an answer below so that this question can be removed the the unanswered tab. $\endgroup$
    – Brian
    Mar 28 '19 at 14:48
  • $\begingroup$ Of course, naively one can just rearrange $2x\,\mathrm dx+2y\,\mathrm dy=0 \implies 2y\,\mathrm dy=-2x\,\mathrm dx \implies \mathrm dy=-x/y\,\mathrm dx \implies \mathrm dy/\mathrm dx = -x/y$. $\endgroup$
    – user856
    Mar 28 '19 at 14:53
  • $\begingroup$ $dy$ and $dx$ can be used as independent algebraic units when needed. Some special handling is needed for higher order differentials, but you can still do it just fine. $\endgroup$
    – johnnyb
    Apr 30 '19 at 18:48
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As noted in the comments, the method of implicit differentiation that Sanderson presents is an abuse of notation used provide a more intuitive understanding of why it works. A more formal process would resemble the following \begin{align} x^2+y^2=5&\implies \frac{\mathrm{d}}{\mathrm{d}x}(x^2+y^2)=\frac{\mathrm{d}}{\mathrm{d}x}5^2 \\ &\implies 2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}=0 \implies 2y\frac{\mathrm{d}y}{\mathrm{d}x}=−2x \\ &\implies \frac{\mathrm{d}y}{\mathrm{d}x}=−\frac{x}{y} \end{align}

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  • $\begingroup$ Thanks Brian. I tried to upvote, but apparently I need at least score of 15 to be able to do that. $\endgroup$
    – Hazzaldo
    Mar 28 '19 at 14:51

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