# 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 $F_y(x,y)\neq 0$.

• 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. Apr 19, 2011 at 11:51
• @Arjang: Thanks. What do you mean with their equivalent limit form? Apr 19, 2011 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. Apr 27, 2011 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.
• 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 Apr 27, 2011 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/…
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}