# Partial derivative of $f(u,v)$

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

• 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 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 at 18:33
• @AlexOrtiz I added the original question that caused the confusion for me . – S.H.W Apr 17 at 18:46
• @Firepi Please see the edit . – S.H.W Apr 17 at 18:55

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 $$-z(x,y) = x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y},\quad\text{also holds.}$$
• 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 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 at 19:05
• @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 at 19:19