MHB Implicit Functions: How Can Derivatives Be Calculated?

[mathmari]

Hey!

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 $$\frac{dy}{dx}=-\frac{\frac{\partial{G}}{\partial{x}}}{\frac{\partial{G}}{\partial{y}}} , \text{ if } \frac{\partial{G}}{\partial{y}} \neq 0$$
First of all, what does it mean that $y(x)$ is defined implicitly by $G(x,y(x))=0$ ?? (Wondering) 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 ?? Is the substitution $f(x)=x$ necessary?? (Wondering)



Then 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??

How could I continue ?? (Wondering)

 

[mathmari]

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?? (Wondering)

 

[Opalg]

We have that $${\color{red}0=\frac{\partial{G_1}}{\partial{x}} +\frac{\partial{G_1}}{\partial{y_1}} \frac{dy_1}{dx}+\frac{\partial{G_1}}{\partial{y_2}} \frac{dy_2}{dx}} \Rightarrow \frac{\partial{G_1}}{\partial{y_1}} \frac{dy_1}{dx}= -\frac{\partial{G_1}}{\partial{x}} -\frac{\partial{G_1}}{\partial{y_2}} \frac{dy_2}{dx} \ \ \ \ (*) \\ {\color{green}0=\frac{\partial{G_2}}{\partial{x}} +\frac{\partial{G_2}}{\partial{y_1}} \frac{dy_1}{dx}+\frac{\partial{G_2}}{\partial{y_2}} \frac{dy_2}{dx}} \Rightarrow \frac{\partial{G_2}}{\partial{y_2}} \frac{dy_2}{dx}=-\frac{\partial{G_2}}{\partial{x}} -\frac{\partial{G_2}}{\partial{y_1}} \frac{dy_1}{dx} \\ \Rightarrow \frac{dy_2}{dx}=-\frac{\frac{\partial{G_2}}{\partial{x}} +\frac{\partial{G_2}}{\partial{y_1}} \frac{dy_1}{dx}}{\frac{\partial{G_2}}{\partial{y_2}}} \\ (*) \Rightarrow \frac{\partial{G_1}}{\partial{y_1}} \frac{dy_1}{dx}= -\frac{\partial{G_1}}{\partial{x}} +\frac{\partial{G_1}}{\partial{y_2}} \frac{\frac{\partial{G_2}}{\partial{x}} +\frac{\partial{G_2}}{\partial{y_1}} \frac{dy_1}{dx}}{\frac{\partial{G_2}}{\partial{y_2}}} \Rightarrow \frac{dy_1}{dx}=\frac{ -\frac{\partial{G_1}}{\partial{x}} +\frac{\partial{G_1}}{\partial{y_2}} \frac{\frac{\partial{G_2}}{\partial{x}} +\frac{\partial{G_2}}{\partial{y_1}} \frac{dy_1}{dx}}{\frac{\partial{G_2}}{\partial{y_2}}}}{\frac{\partial{G_1}}{\partial{y_1}}}$$

Is this correct?? Could I improve something?? Could we write the last equality in a simplified way?? (Wondering)

The equations are correct. The elaborate notation obscures the fact that these are simply a pair of simultaneous equations for $\frac{dy_1}{dx}$ and $\frac{dy_2}{dx}$. So multiply the red equation by $\frac{\partial{G_2}}{\partial{y_2}}$, multiply the green equation by $\frac{\partial{G_1}}{\partial{y_2}}$, then subtract. The terms involving $\frac{dy_2}{dx}$ will cancel, and you will be left with a formula for $\frac{dy_1}{dx}$ in terms of the partial derivatives of $G_1$ and $G_2$.