Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $y_i$?

share|improve this question
    
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
1  
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
1  
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
show 2 more comments

3 Answers

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.

share|improve this answer
    
The equation comes from minimising a function with constraints, using Lagrange multipliers. –  user10389 Nov 24 '11 at 21:53
add comment

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:

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

share|improve this answer
add comment

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

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

share|improve this answer
    
Doesn't really help, but thanks anyay –  user10389 Nov 24 '11 at 3:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.