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If I have the equation and function $$f_1(x_1,x_2,x_3,...,x_n) = 0,\qquad x_1 = g_1(y_1, y_2, y_3,...,y_m)$$ then what is $\frac {\partial f_1}{\partial x_1}$ in terms of $g_1$ and $y_i$?

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The first thing that you list is not a function; it is an equation. Implicitly, any of the $x_i$ can almost always be locally viewed as functions of the other $x_i$, but still, what you have written is not a function. – alex.jordan Nov 23 '11 at 23:18
@alex yes, I've edited it now. – user10389 Nov 23 '11 at 23:31
I guess my point is, why include the "$=0$"? You are asking for a derivative of $f_1$, so that means you want to think of $f_1$ as having output that changes as input changes. Yet if you have set $f_1(\ldots)$ equal to zero, then $\frac{\partial f_1}{\partial x_1}$ is trivially equal to $0$. I still don't know what you mean to ask, but I'm pretty sure that this is not it. – alex.jordan Nov 24 '11 at 0:37
@alex $df_1=0$ doesn't imply $\frac{\partial f_1}{\partial x_1} = 0$ – user10389 Nov 24 '11 at 3:04
I'm not sure what "$df_1$" means in this context , but put that aside. If I defined $f(x)=0$ and asked you to compute $\frac{\partial f}{\partial x}$, what would you tell me? The derivative a constant function is $0$. That is exactly what is going on with your question the way it is currently worded. – alex.jordan Nov 24 '11 at 3:30

Suppose that you have a function of more than one variable. Let's take the example of $f$, where: $$f(x,y)=x^2-2xy+y^2-1$$ This function can take differing values as $x$ and $y$ vary. And that means we have "interesting" derivatives we can study. \begin{align*} \frac{\partial f}{\partial x} & =2x-2y\\ \frac{\partial f}{\partial y} & =-2x+2y \end{align*}

Sometimes, we will become interested in studying one of $f$'s level sets. That means the set of all $(x,y)$ such that the output of $f$ is some constant level $k$. In our example, we might consider $$f(x,y) = 0$$ and what we mean is the set of all $(x,y)$ where $$x^2-2xy+y^2-1=0$$ It doesn't really make sense to speak of $\frac{\partial f}{\partial x}$ anymore. Our attention is focused on the curve with the equation $x^2-2xy+y^2-1=0$, and on that curve $f$ is identically $0$. We might consider a small change in $x$, but the constraint will force a small change in $y$ so that together, the corresponding change in $f$ will be $0$. So if $\frac{\partial f}{\partial x}$ means anything at all, it means the derivative of the zero function: $0$.

Of course nothing is stopping us from going back and working with the partial derivative computed earlier: $\frac{\partial f}{\partial x} =2x-2y$. And maybe that is what you intend in your question. But then I ask why include the "$=0$" in your question at all?

Assuming that you would be interested in the partial derivative $2x-2y$ for this example, then the answer is just as Abramodj has said. To keep this example going, suppose $x=g(s,t)=\sin(s)\cos(t)$. Then \begin{align*} \frac{\partial f}{\partial x} & = 2x-2y\\ &=2g(s,t)-2y\\ &=2\sin(s)\cos(t)-2y \end{align*}

May I ask how this question arose? It looks like a familiar issue that arises in a vector calculus course when you learn that on surfaces $\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}$. But I might be mistaken.

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The equation comes from minimising a function with constraints, using Lagrange multipliers. – user10389 Nov 24 '11 at 21:53

Not sure if that is what you need, but if you have a function of $n$ variables $f_{1}(x_{1}, x_{2}, x_{3},...,x_{n})$, where $x_{1} = g_{1}(y_{1}, y_{2}, y_{3},...,y_{m})$, $x_{2} = g_{2}(y_{1}, y_{2}, y_{3},...,y_{m}),\dots, x_{n} = g_{n}(y_{1}, y_{2}, y_{3},...,y_{m})$ are also some functions of $m$ variables, then you can find partial derivatives of $f_{1}$ with respect to $y_{i}, i=\overline{1,n} $ using the formula:

$$\frac{\partial f_{1}}{\partial y_{i}} = \frac{\partial f_{1}}{\partial x_{1}}\frac{\partial x_{1}}{\partial y_{i}} + \frac{\partial f_{1}}{\partial x_{2}}\frac{\partial x_{2}}{\partial y_{i}} + \ldots + \frac{\partial f_{1}}{\partial x_{n}}\frac{\partial x_{n}}{\partial y_{i}} = \sum_{k=1}^{n} \frac{\partial f_{1}}{\partial x_{k}}\frac{\partial x_{k}}{\partial y_{i}}$$

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In my opinion it is just

$$ \frac {\partial f_1}{\partial x_1} $$

and the relation you ask is given by substituting $x_1$ in this expression with

$$ g_1(y_1, y_2, y_3,...,y_m) $$

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Doesn't really help, but thanks anyay – user10389 Nov 24 '11 at 3:06

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