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My question is what is $F_x$.

When you find $y'$ from the implicit function $$F(x,y)= x^3+y^3-6xy=0$$

you can obtain $y'$ by $$\frac{dy}{dx} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} = -\frac{F_x}{F_y}$$

The text book says $F_x$ is $$3x^2-6y$$. And $F_y$ is $$3y^2-6x$$ and obtain $$\frac{dy}{dx}=-\frac{3x^2-6y}{3y^2-6x}$$

Well, isn't $y$ a function of $x$?

Like F(x,y)=F(x,f(x)) and should it not go like $$ F_x=3x^2-6y-6x\frac{\partial y}{\partial x}$$ and not $$F_x=3x^2-6y$$?

although it would directly lead to the answer $\frac{dy}{dx}$.

What I am asking is why in implicit differentiation it is ignoring $y$ is a function of $x$ and treat $y$ as some constant?

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  • $\begingroup$ Note that given $F(x, y) =0$, it is not always possible to write in the form $y=f(x)$. $\endgroup$ – Eddy Sep 8 '18 at 14:04
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It’s important to distinguish the function $F:\mathbb R^2\to\mathbb R$ defined as $F: (x,y)\mapsto x^2+y^2-6xy$ from the equation $F(x,y)=0$. For this function, $x$ and $y$ are independent variables—they can take any values whatsoever—so it makes perfect sense to compute its partial derivatives with respect to each variable by treating the other variable as a constant. The graph of this function is some surface in $\mathbb R^3$ and at points at which it’s differentiable (everywhere for this function), this surface has a tangent plane with normal vector $(F_x,F_y,-1)$.

The equation $F(x,y)=0$, describes a level curve of the function $F$, specifically the intersection of its graph with the $x$-$y$ plane. The tangent line to this curve at a point is just the intersection of the tangent plane to the graph of $F$ with the $x$-$y$ plane. The projection of the tangent plane’s normal onto the $x$-$y$ plane, namely $(F_x,F_y)=\nabla F$ is orthogonal to this tangent line. The slope of this vector is $m_1 = F_y/F_x$, and using the relation $m_1m_2=-1$ for the slopes of perpendicular lines, we see that the slope of the tangent line is $m_2=-F_x/F_y$.

To put it another way, when you compute $dy/dx$ via partial derivatives of $F$, you are viewing the curve implicitly defined by the equation $F(x,y)=0$ as a horizontal slice through the 3-D graph of $F$ and then finding a tangent line to this curve by taking the same slice through the tangent plane to the graph of $F$ at that point. By adding an extra dimension and stepping out of the $x$-$y$ plane in this way, you can temporarily ignore the fact that $x$ and $y$ are not independent for points on the level curve. This dependency doesn’t disappear completely, though. It still plays a role in computing the coordinates of a point on the level curve, which you need in order to compute the partial derivatives of $F$ there.

An important thing to note about the above approach is that nowhere does it require that there exists some function $f:\mathbb R\to\mathbb R$ such that $F(x,y)=0$ implies that $f(x)=y$. Indeed, there may not even be such a single function. E.g., $x^2+y^2=1$ can’t be the graph of any function since it fails the vertical line test. (It might still be possible to find such a function in the neighborhood of a point, though—that’s what the implicit function theorem is about.) When you perform implicit differentiation, you start off by assuming that there is such a function and then differentiate both sides of the equation $F(x,y)=0$ taking that into account.

These two approaches can be reconciled via the chain rule. Suppose that $y=f(x)$. To help keep things straight, we introduce the function $\phi:\mathbb R\to\mathbb R^2$ with $\phi:x\mapsto(x,f(x))$, and also $g=F\circ\phi$. Then $$g' = \nabla F\cdot\phi' = (F_x,F_y)\cdot(1,f') = F_x+F_yf'.$$ This expression takes into account the assumed dependence of $y$ on $x$, but it nevertheless involves partial derivatives of $F$. Since we have $F(x,y)=g(x)$, implicit differentiation is equivalent to differentiating both sides of the equation $g(x)=0$ and then solving for $f'$, but this gives $F_x+F_yf'=0$, from which $f'=-F_x/F_y$ as before.

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Since the function $x \mapsto F \bigl(x,f(x) \bigr)$ is constant, its derivative is zero. Compute this derivative using the multivariate chain rule: $$ 0 = \frac{d}{dx} \Bigl( F \bigl(x,f(x) \bigr) \Bigr) = F_x \bigl(x,f(x) \bigr) + F_y \bigl(x,f(x) \bigr) \, f'(x) . $$ From this it follows that $$ f'(x) = - \frac{F_x \bigl((x,f(x) \bigr)}{F_y \bigl((x,f(x) \bigr)} . $$ Here the notation $F_x \bigl((x,f(x) \bigr)$ means that you compute $F_x(x,y)$ considering $F$ as a function of two independent variables $x$ and $y$, and then you substitute $y=f(x)$. And similarly for $F_y$, of course.

The fact that $y$ is a function of $f$ is taken into account by the second term in the chain rule, the one involving $F_y$ and $f'$.

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We have that

$$F(x,y)=x^3+y^3-6xy=0\implies F_xdx+F_ydy=(3x^2-6y)dx+(3y^2-6x)dy=0$$

and therefore

$$\frac{dy}{dx}=-\frac{3x^2-6y}{3y^2-6x}$$

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If we assume that $y=y(x)$ is given, then we get $$3x^2+3y^2y'-6y-6xy'=0$$

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  • $\begingroup$ yes that is what I am saying. But in implicit differentiation it is ignoring the fact that y is a function of x and just goes $F_x=3x^2-6y$. $\endgroup$ – 강승태 Sep 8 '18 at 12:40

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