# How to show that $\frac{dy}{dx}=-\frac{F_x(x,y)}{F_y(x,y)}$

Suppose that $y$ is defined implicitly as a function $y(x)$ by an equation on the form $F(x,y)=0$. I'm trying to show that $$\frac{dy}{dx}=-\frac{F_x(x,y)}{F_y(x,y)},$$ but I don't know where to start. Can someone please give me a hint?

Both $y(x)$ and $F(x,y)$ are differentiable and $F_y(x,y)\neq 0$.

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 Use the partial derivatives notation, then repleace them with their equivalant limit form, the answer would be looking back at you! Alternatively look at the definition of Total Derivative. – Arjang Apr 19 '11 at 11:51 @Arjang: Thanks. What do you mean with their equivalent limit form? – please delete me Apr 19 '11 at 12:41 @Evind : using the definition of derivative in terms of limits. But from the lhf's answer I see there is no need for that, unless one wants to show it directly, but I prefer lhf's answer. – Arjang Apr 27 '11 at 0:04

The key point is to use the chain rule. From $F(x,y(x))=0$ get $F_x(x,y)\cdot1+F_y(x,y)y'(x)=0$.

The derivation above has a geometric interpretation: The gradient of $F$ is orthogonal to the level curve $F=0$. Hence, it's orthogonal to the tangent vector.

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Thanks. I understand it now. – please delete me Apr 19 '11 at 12:45
Can you please expand the step using chain rule to get from $F(x,y(x))=0$ to $F_x(x,y)\cdot1+F_y(x,y)y'(x)=0$ , for the life of me I can't! Please help – Arjang Apr 27 '11 at 0:07
@Arjang, if $g(x)=F(a(x),b(x))$ then $g'(x)=F_x(a(x),b(x))a'(x)+F_y(a(x),b(x))b'(x)$. See en.wikipedia.org/wiki/… – lhf Apr 27 '11 at 0:40
aha, it was the Jakobian after all, I asuumed it was just the usual one dimentional version of chain rule ( wrongly of course!) +1 for your time , thank you – Arjang Apr 27 '11 at 22:49

What happens when you take a small example, say you want to find $\mathrm{d}y/\mathrm{d}x$ of $x^2+y^2=9$, you have: \begin{align} 2x+\frac{\mathrm{d}y}{\mathrm{d}x}2y&=0\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{x}{y} \end{align} Now you have $F(x,y)=0$, differentiation gives: \begin{align} F_x(x,y)+F_y(x,y)\frac{\mathrm{d}y}{\mathrm{d}x}&=0\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{F_x(x,y)}{F_y(x,y)} \end{align}

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 Thanks. That helped a lot. – please delete me Apr 19 '11 at 12:44