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Let $f(u,v) = c$ where $u(x,y) , v(x,y)$ are functions and $c$ is constant. Can we conclude $\frac{\partial f}{\partial v} = \frac{\partial f}{\partial u} = 0$ ? It really sounds confusing to me but I've tried many examples and also the definition of partial derivative , and it was true ! What's the problem here ?

Main question : Suppose $f$ is a differentiable function . If $z$ is a differentiable function with respect to $x$ and $y$ and defined in $f(xz,yz) = 1$ prove that : $x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = -z$

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  • $\begingroup$ Are you defining $f$ to be a constant function (in which case it is unclear what the role of $x,y$ is), or are you parametrizing the level set $f(u,v) = c$ with parameters $x,y$ and parametric functions $u=u(x,y),v=v(x,y)$? In the latter case, it may well be the case that the partial derivatives $\partial_uf, \partial_vf$ are nonzero. $\endgroup$ – Alex Ortiz Apr 17 at 18:25
  • $\begingroup$ Maybe the confusion lies in the following: if $f\colon D\rightarrow\mathbb{R}$ is the constant function $f(x,y)=c$ and $g\colon A\rightarrow \mathbb{R}^2$ is given by the correspondence $(s,t)\mapsto (u(s,t),v(s,t))$ (so that $g(A)\subset D$), then what you're really trying to compute is $\partial_1 (f\circ g)$ which is equal to $(\partial_1 f)(u,v)\partial_1(u)+(\partial_2 f)(u,v)\partial_1(v)=0$. $\endgroup$ – Firepi Apr 17 at 18:33
  • $\begingroup$ @AlexOrtiz I added the original question that caused the confusion for me . $\endgroup$ – S.H.W Apr 17 at 18:46
  • $\begingroup$ @Firepi Please see the edit . $\endgroup$ – S.H.W Apr 17 at 18:55
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There is some confusion being caused by the employment of dummy variables. Strictly speaking, if we have a differentiable function $f\colon \mathbf R^2\to\mathbf R$, then we can write it as $f = f(x,y) = f(u,v) = f(\uparrow,\downarrow), \dots$. The partial derivatives of $f$ with respect to the dummy variables we are using for $\mathbf R^2$ then use the same dummy variables by convention.

Now, in the question you are trying to solve, you are asked to show that for all $x,y$ such that $$ f\big(x\cdot z(x,y), y\cdot z(x,y)\big) = 1, $$ the equation $$ -z(x,y) = x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} $$ holds. We are not saying that $f = f(\text{dummy variable 1},\text{dummy variable 2})$ is the constant function $1$, in which case we of course would have $\partial_1f = \partial_2 f = 0$. Instead, we are saying, look at all the points $(x,y)\in\mathbf R^2$ at which $$ f\big(x\cdot z(x,y), y\cdot z(x,y)\big) = 1\quad\text{holds}. $$ For any such point $(x,y)$, show that $$ -z(x,y) = x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y},\quad\text{also holds.} $$

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  • $\begingroup$ I'm sorry but I'm still confused . If we let $u = xz$ and $v = yz$ obviously we have $\frac{\partial f}{\partial v} = \frac{\partial f}{\partial u} = 0$ . $\endgroup$ – S.H.W Apr 17 at 19:02
  • $\begingroup$ @S.H.W: The problem is that in the two statements $u$ and $v$ have different meanings. In "let $u = xz$" or "let $v = yz$," $u$ and $v$ are functions of $(x,y)$. As you are using the letters $u,v$ in the statement $\frac{\partial f}{\partial u} = \frac{\partial f}{\partial v} = 0$, $u$ and $v$ are the dummy variables of $\mathbf R^2$. The two meanings are incompatible! $\endgroup$ – Alex Ortiz Apr 17 at 19:05
  • $\begingroup$ As you said they are dummy variables . So we can use anything instead of them and this means that the two meanings are the same! Sorry , I'm really baffled by this problem . $\endgroup$ – S.H.W Apr 17 at 19:14
  • $\begingroup$ @S.H.W: If we say $f = f(u,v)$, then we are specifying which "global" variables we will use for the input to the function $f$. Saying, let $u = xz$ and $v = yz$ is now saying, "let us specialize the input to $f$ to be only inputs $u,v$ of the special form where $u = xz$ and $v = yz$." If we now decide we want to write $\partial_uf$ or $\partial_vf$, we have to be careful about what we mean, and that is the two partial derivatives of $f$ with respect to its "global" variables. $\endgroup$ – Alex Ortiz Apr 17 at 19:19

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