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Let $F_x$ and $F_y$ denote partial derivatives with respect to $x$ and $y$ respectively. I want to prove that given a function $F(x, y)$ and the equation $$F(x,y) = 0$$ implicitly defines a function $y(x)$ satisfying the equation, then the following holds: $$\frac{dy}{dx} = - \frac{-F_x}{F_y}$$ Therefore, we can start from this equation: $$F(x,y(x))= 0$$ And now I think the best thing we can do is to differentiate both sides with respect to x: The only problem is that I am not entirely sure how to do this. On the one hand, I would like to write this: $$F_x = 0$$ However, there is also the $y$ part which depends on $x$ and cannot be ignored. Therefore, I should try using a chain rule: $$F_x\frac{\partial x}{\partial x} + F_y \frac{\partial y}{\partial x} = 0$$ Now, I can solve this and get: $$\frac{\partial y}{\partial x} = - \frac{F_x}{F_y}$$ Now, is it enough to simply replace $\partial$ with $d$? Also, is it the correct way of differetiating such functions and would work for, for example $\varphi(x,y,z,a(x,y,z)) = 0$?

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One problem here is that you're using $y$ in two different roles: that of a variable in $F$, and as a function of $x$. Instead, I'm going to write $$ F_1(x, y) $$ for the partial of $F$ with respect to the first argument, and similarly for $F_2$. Now things get easier. If we define $$ G(x) = F(x, y(x)) $$ then $$ \frac{dG}{dx} = F_1(x, y(x)) \frac{dx}{dx} + F_2(x, y(x)) \frac{dy(x)}{dx} $$ Now if we further assume that $G$ is the constant function $0$ (or any other constant!), then the expression on the right is zero, hence \begin{align} 0 &= F_1(x, y(x)) \frac{dx}{dx} + F_2(x, y(x)) \frac{dy(x)}{dx}\\ -F_1(x, y(x)) \frac{dx}{dx} &= F_2(x, y(x)) \frac{dy(x)}{dx}\\ -F_1(x, y(x)) &= F_2(x, y(x)) \frac{dy(x)}{dx}\\ \frac{-F_1(x, y(x))} {F_2(x, y(x))} &= \frac{dy(x)}{dx} \end{align} which is what you were hoping to prove.

As for the question about $$ \frac{\partial y}{\partial x} \text{ vs. } \frac{dy}{dx} $$ that's mostly a convention: when a function has more than one argument, we use $\partial$; when it has only one argument, we use $d$. [And then, later, people who study fluids go back to using $d$ to mean something quite different involving time, but you'll know that when you get to it.]

As for your more general question (does this work for any number of arguments?), the answer is "yes, but you should read up on the implicit function theorem to see the exact hypotheses needed to make it work." In particular, in the present example, it's possible for $dy/dx$ to exist even if $F_2(x, y) = 0$ (you have to set things up so that $F_1(x, y) = 0$ as well!), so the real rule you can derive from the chain rule is that $$ 0 = F_1(x, y(x)) + F_2(x, y(x)) \frac{dy(x)}{dx} $$ with the division to determine $dy/dx$ being valid only if $F_2$ is nonzero.

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