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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 : $x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = -z$

S.H.W
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  • 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. – Alex Ortiz Apr 17 '19 at 18:25
  • 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 $(\partial_1 f)(u,v)\partial_1(u)+(\partial_2 f)(u,v)\partial_1(v)=0$. – Firepi Apr 17 '19 at 18:33
  • @AlexOrtiz I added the original question that caused the confusion for me . – S.H.W Apr 17 '19 at 18:46
  • @Firepi Please see the edit . – S.H.W Apr 17 '19 at 18:55

2 Answers2

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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 $$ -z(x,y) = x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y},\quad\text{also holds.} $$

Alex Ortiz
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  • I'm sorry but I'm still confused . If we let $u = xz$ and $v = yz$ obviously we have $\frac{\partial f}{\partial v} = \frac{\partial f}{\partial u} = 0$ . – S.H.W Apr 17 '19 at 19:02
  • @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! – Alex Ortiz Apr 17 '19 at 19:05
  • 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 . – S.H.W Apr 17 '19 at 19:14
  • @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. – Alex Ortiz Apr 17 '19 at 19:19
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I think I've found a convincing answer. Also see this, this and this related posts. Define the function $G(x,y) :\mathbb{R}^2\to \mathbb{R}$ as $$G(x,y) = f(xz(x,y) , yz(x,y)).$$ We assume that there is an open set $A$ in $\mathbb{R}^2$ for which $$G(x,y) = 1,$$ holds for all $(x,y)\in A$. So for $(x,y) \in A$ we have $$\frac{\partial G}{\partial x} = 0, \frac{\partial G}{\partial y} = 0.$$Let $s(x,y) = xz(x,y)$ and $t(x,y) = yz(x,y)$. According to the chain rule, we have $$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial s}\frac{\partial s}{\partial x} + \frac{\partial f}{\partial t}\frac{\partial t}{\partial x}, \\ \frac{\partial f}{\partial y} = \frac{\partial f}{\partial s}\frac{\partial s}{\partial y} + \frac{\partial f}{\partial t}\frac{\partial t}{\partial y}.$$ Note that $$\frac{\partial s}{\partial x} = z + x\frac{\partial z}{\partial x} , \frac{\partial s}{\partial y} = x\frac{\partial z}{\partial y}, \\ \frac{\partial t}{\partial x} = y\frac{\partial z}{\partial x} , \frac{\partial t}{\partial y} = z + y\frac{\partial z}{\partial y}.$$Therefore we can conclude that $$0 = \frac{\partial G}{\partial x} = \frac{\partial f}{\partial x} = \frac{\partial f}{\partial s}(z + x\frac{\partial z}{\partial x}) + \frac{\partial f}{\partial t}(y\frac{\partial z}{\partial x}), \\ 0 = \frac{\partial G}{\partial y} =\frac{\partial f}{\partial y} = \frac{\partial f}{\partial s}(x\frac{\partial z}{\partial y}) + \frac{\partial f}{\partial t}(z + y\frac{\partial z}{\partial y}).$$ By finding the function$$\frac{\frac{\partial f}{\partial s}}{\frac{\partial f}{\partial t}}, $$ from the two relations and equating them we have, $$xy\frac{\partial z}{\partial x}\frac{\partial z}{\partial y} = z^2 + zy\frac{\partial z}{\partial y} + xz\frac{\partial z}{\partial x} + xy\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}.$$ This implies that $$-z = y\frac{\partial z}{\partial y} + x\frac{\partial z}{\partial x},$$ which is the desired conclusion.

Example: Take $A = \mathbb{R}^2 - \{(0,0)\}$ and $$z(x,y) = \frac{1}{x+y},f(s,t) = s+t.$$For this choice of $z(x,y)$ and $f(s,t)$ we have, $$G(x,y) = f(s(x,y),t(x,y)) = f(\frac{x}{x+y} , \frac{y}{x+y}) = \frac{x}{x+y} + \frac{y}{x+y} = 1, \\ \frac{\partial G}{\partial x} = \frac{\partial f}{\partial x} = 0, \frac{\partial G}{\partial y} = \frac{\partial f}{\partial y} = 0 , \frac{\partial f}{\partial s} = 1,\frac{\partial f}{\partial t} = 1.$$

S.H.W
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