Let $y(x)$ be defined implicitly by $G(x,y(x))=0$, where $G$ is a given two-variable function. Show that if $y(x)$ and $G$ are differentiable, then $$\frac{dy}{dx}=-\frac{\frac{\partial{G}}{\partial{x}}}{\frac{\partial{G}}{\partial{y}}} , \text{ if } \frac{\partial{G}}{\partial{y}} \neq 0$$
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First of all, what does it mean that $y(x)$ is defined implicitly by $G(x,y(x))=0$ ??
I have done the following:
Let $f(x)=x$. Then $G(x,y(x))=G(f(x), y(x))=H(x)$.
We have that $G(x,y(x))=0 \Rightarrow H(x)=0$.
$H(x)=0 \Rightarrow \frac{H(x)}{dx}=0$
From the chain rule we have the following:
$$\frac{H(x)}{dx}=\frac{\partial{G}}{\partial{f}}\frac{df}{dx}+\frac{\partial{G}}{\partial{y}}\frac{dy}{dx} \\ \Rightarrow 0=\frac{\partial{G}}{\partial{x}}\frac{dx}{dx}+\frac{\partial{G}}{\partial{y}}\frac{dy}{dx} \\ \Rightarrow 0=\frac{\partial{G}}{\partial{x}}+\frac{\partial{G}}{\partial{y}}\frac{dy}{dx} \\ \Rightarrow \frac{\partial{G}}{\partial{y}}\frac{dy}{dx}=-\frac{\partial{G}}{\partial{x}} \\ \overset{ \text{ if } \frac{\partial{G}}{\partial{y}}\ \neq 0 }{\Longrightarrow }\frac{dy}{dx}=-\frac{\frac{\partial{G}}{\partial{x}}}{\frac{\partial{G}}{\partial{y}}}$$
Is this correct ?? Could I improve something ??
EDIT1:
I have to find a similar formula if $y_1$, $y_2$ are defined implicitly by $$G_1(x, y_1(x), y_2(x))=0 \\ G_2(x, y_1(x), y_2(x))=0$$
I have the following:
From the chain we have the following:
$$\frac{dG_1}{dx}=\frac{\partial{G_1}}{\partial{x}} \frac{dx}{dx}+\frac{\partial{G_1}}{\partial{y_1}} \frac{dy_1}{dx}+\frac{\partial{G_1}}{\partial{y_2}} \frac{dy_2}{dx} \Rightarrow 0=\frac{\partial{G_1}}{\partial{x}} +\frac{\partial{G_1}}{\partial{y_1}} \frac{dy_1}{dx}+\frac{\partial{G_1}}{\partial{y_2}} \frac{dy_2}{dx}$$
Similar, we get $$0=\frac{\partial{G_2}}{\partial{x}} +\frac{\partial{G_2}}{\partial{y_1}} \frac{dy_1}{dx}+\frac{\partial{G_2}}{\partial{y_2}} \frac{dy_2}{dx}$$
Is it correct so far??
Hw could I continue??
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EDIT2:
We have that $$0=\frac{\partial{G_1}}{\partial{x}} +\frac{\partial{G_1}}{\partial{y_1}} \frac{dy_1}{dx}+\frac{\partial{G_1}}{\partial{y_2}} \frac{dy_2}{dx} \Rightarrow \frac{\partial{G_1}}{\partial{y_1}} \frac{dy_1}{dx}= -\frac{\partial{G_1}}{\partial{x}} -\frac{\partial{G_1}}{\partial{y_2}} \frac{dy_2}{dx} \ \ \ \ (*) \\ 0=\frac{\partial{G_2}}{\partial{x}} +\frac{\partial{G_2}}{\partial{y_1}} \frac{dy_1}{dx}+\frac{\partial{G_2}}{\partial{y_2}} \frac{dy_2}{dx} \Rightarrow \frac{\partial{G_2}}{\partial{y_2}} \frac{dy_2}{dx}=-\frac{\partial{G_2}}{\partial{x}} -\frac{\partial{G_2}}{\partial{y_1}} \frac{dy_1}{dx} \\ \Rightarrow \frac{dy_2}{dx}=-\frac{\frac{\partial{G_2}}{\partial{x}} +\frac{\partial{G_2}}{\partial{y_1}} \frac{dy_1}{dx}}{\frac{\partial{G_2}}{\partial{y_2}}} \\ (*) \Rightarrow \frac{\partial{G_1}}{\partial{y_1}} \frac{dy_1}{dx}= -\frac{\partial{G_1}}{\partial{x}} +\frac{\partial{G_1}}{\partial{y_2}} \frac{\frac{\partial{G_2}}{\partial{x}} +\frac{\partial{G_2}}{\partial{y_1}} \frac{dy_1}{dx}}{\frac{\partial{G_2}}{\partial{y_2}}} \Rightarrow \frac{dy_1}{dx}=\frac{ -\frac{\partial{G_1}}{\partial{x}} +\frac{\partial{G_1}}{\partial{y_2}} \frac{\frac{\partial{G_2}}{\partial{x}} +\frac{\partial{G_2}}{\partial{y_1}} \frac{dy_1}{dx}}{\frac{\partial{G_2}}{\partial{y_2}}}}{\frac{\partial{G_1}}{\partial{y_1}}}$$
Is this correct?? Could I improve something?? Could we write the last equality in a simplified way??
"$y(x)$ is defined implicitly by $G(x,y(x))=0$" means that for each $x$, $y(x)$ is defined to be "the" solution to $G(x,y(x))=0$. This is only an implicit definition because it doesn't give a recipe for calculating $y$. I put the word "the" in scare quotes because typically there are several solutions, and this $y$ is only defined on a neighborhood of a particular solution. ($G(x,y)=x^2+y^2-1$ is a great example for how to see this; the two functions you get come from the top and bottom semicircles.)
Your derivation of the derivative is correct (minus a minor typo). In the one variable situation ($x,y \in \mathbb{R}$), this is just implicit differentiation as you probably saw in Calculus I.