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 $y_i$?
Changing variables in a partial derivative
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multivariable-calculus
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0The 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. – 2011-11-23
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0@alex yes, I've edited it now. – 2011-11-23
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2I 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. – 2011-11-24
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0@alex $df_1=0$ doesn't imply $\frac{\partial f_1}{\partial x_1} = 0$ – 2011-11-24
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1I'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. – 2011-11-24
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0@alex $f_1$ in my question is a function of more than one variable, so it's not the same as your example. $df_1$ is the total differential of $f_1$ from basic calculus – 2011-11-24
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0There's too much confusion to continue this way. For example, if the total differential of $f$ is zero, it most certainly does imply that all of $f$'s partial derivatives are zero. But if you are convinced otherwise, I'm not sure how I can change your mind. I would suggest taking your question to a multivariable calculus instructor in person. – 2011-11-25