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If the equation $F(x,y,z)=0$ defines $z$ implicitly as a differentiable function of x and y, then by taking a partial derivative with respect to one of the independent variables (in this case x), you get

$\large F_x(x,y,z)\frac{\partial x}{\partial x}+F_y(x,y,z)\frac{\partial y}{\partial x}+F_z(x,y,z)\frac{\partial z}{\partial x}=0.$

Because dx/dx = 1 and dy/dx = 0, you can solve for the desired partial derivative:

$\large \frac{\partial z}{\partial x}=-\frac{F_x(x,y,z)}{F_z(x,y,z)} $

The bolded dy/dx = 0 is what I don't get. I mean, it makes sense that an independent variable doesn't change in response to another, but it doesn't seem very formal and I feel like there's more to it than that. So basically, is there a more formal or detailed explanation or is that all there is to it?

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  • $\begingroup$ I asked my calculus teacher the exact same question and he said that because both x and y are independent variables, dy/dx therefore has no relationship and is 0. $\endgroup$ – user208818 Jan 18 '15 at 22:26
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    $\begingroup$ In the described context we have $z=z(x,y)$. So you are taking partially differentiating the (2-variable!!) identity $$F(x,y,z(x,y))=0.$$ Does that make it clear what happens? The good ole chain rule in action. Implicit function theorem gives other results too. If you had two equations $F(x,y,z)=0=G(x,y,z)$, then you would think $y=y(x)$, $z=z(x)$, differentiate both equations w.r.t. $x$ and solve for $dy/dx$ and $dz/dx$. $\endgroup$ – Jyrki Lahtonen Jan 18 '15 at 22:49
  • $\begingroup$ @JyrkiLahtonen: Since you gave the answer in a comment the OP could not accept it. As a consequence this question will be bumped back to the main board every month or so for years to come. $\endgroup$ – Christian Blatter Jul 7 '17 at 12:56
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Going from $F(x,y,z)=0$ to $$F_x(x,y,z)\frac{\partial x}{\partial x}+F_y(x,y,z)\frac{\partial y}{\partial x}+F_y(x,y,z)\frac{\partial z}{\partial x}=0\tag{A}$$ is a formal manipulation, we are just applying $\frac{\partial}{\partial x}$ to both sides of $F(x,y,z)=0$. Then we are assuming that $z$, at least locally, can be written as a (smooth) function of $x$ and $y$. If we wish to consider $\frac{\partial z}{\partial x}$, we may recall what a partial/directional derivative stands for: the rate of change of a smooth function along a direction. In particular $\frac{\partial z}{\partial x}$ accounts for what happens to $z=z(x,y)$ when we move from $(x,y)$ to $(x+\varepsilon,y)$. What happens to $y$ during this travel? Absolutely nothing, it does not change. So by assuming $z=z(x,y)$ and recalling the geometric meaning of $\frac{\partial}{\partial x}$ we get that $(A)$ implies $$ F_x(x,y,z)+F_z(x,y,z)\frac{\partial z}{\partial x}=0 \tag{B} $$ as wanted. Remark: we would have got $(B)$ also by assuming that $F$ is locally linear, and not by chance. Indeed, the differentiable functions are the functions which are well-approximated by their tangent plane at any point.

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