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Paritial derivative of function of dependent variables 
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#1
Dec3112, 09:01 AM

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i am having a hard time understanding partial derivative for function of dependent variables.
for example lets consider $$z=x+y$$ so by usual steps that are mentioned on e.g wikipedia etc. $$\frac{\partial}{\partial x}z=1$$ but what if its also true that $$y=x$$ (or in other words why the steps does not take into account that the input variables might depend on each other) $$\frac{\partial}{\partial x}z=2$$ so the question is how paritial derivative is defined for function of dependent variables ? 


#2
Dec3112, 09:45 AM

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On what do x and y depend.
If y = x, then they are the same variable, no? I believe it is customary to have a function depending on an independent variable, e.g., a set of coordinates and/or time. 


#3
Dec3112, 10:36 AM

P: 30

if x and y are truly independent would not $$\frac{\partial f(x,y)}{\partial x}=\frac{df(x,y)}{dx}$$ thank you 


#4
Jan213, 11:40 AM

P: 30

Paritial derivative of function of dependent variables
still waiting for a reply, guys !
so, I am gonna bump the thread, just this once . 


#5
Jan213, 01:27 PM

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Righthand side: "The value at x, of the derivative of the function ##t\mapsto f(t,y)##". Yes, those two are the same number, regardless of the values of x and y. 


#6
Jan213, 01:46 PM

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$$\frac{\partial }{\partial x}z$$ that informs us what function we are to take a partial derivative of. What you are describing is the convention to use this notation even though it's ambiguous, and to specify what function we're dealing with in a separate statement. To say that z=x+y is to suggest that the function is the f such that f(s,t)=s+t for all s,t, and that we're supposed to take the partial derivative with respect to the first variable. To supplement this by saying that y=x, is to suggest that the function is the g such that g(t)=f(t,t)=2t for all t. I'm using the word "suggest" because these specifications of the functions are still ambiguous. For example, in the first case, the function could be the h defined by h(r,s,t)=s+t for all r,s,t. Then the partial derivative would be with respect to the second variable. 


#7
Jan213, 02:56 PM

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f(x,y=2) = 2x. f(x,y=4) = 4x. 


#8
Jan213, 03:28 PM

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I'm also not sure what you're saying here. At first I thought you meant this: We can use the function ##f:\mathbb R^2\to\mathbb R## to define a function ##f_y:\mathbb R\to\mathbb R## for each ##y\in\mathbb R## by ##f_y(x)=f(x,y)## for all ##x\in\mathbb R##. For all ##x\in\mathbb R##, we have \begin{align} f_2(x)=x+2\\ f_4(x)=x+4. \end{align} But you wrote 2x and 4x. 


#9
Jan213, 03:42 PM

P: 810

The problem you're having here is that physics (and traditional calculus notation) tend to conflate variables and functions.
The function here is z. We write z(x, y) = x + y. It is a function of two parameters. The nice thing about functions is that their parameters are *always* independent. We don't even need to use the word "independent". The choice of the parameters is always at the discretion of the person using the function. Being a smooth function of two parameters, f has two partial derivatives: z_1(x, y) = 1 and z_2(x, y) = 1. Notational aside: I'm using an alternate notation here. The Leibniz notation (the one you used, and the standard one in calculus) mentions the parameter name instead of its position. However, one valid "move" in math is renaming of bound variables. Even though I wrote z(x, y) = x + y, I could just as easily say z(p, q) = p + q. They denote the exact same function, however, once I write $$\frac{\partial}{\partial x}z$$, I have "locked myself into" a particular choice of variable, and that valid move of mine suddenly stops working. Leibniz notation is convenient for backofthenapkin calculations, but it's kinda gross. Now your question is that, "what if $$x = y$$"? Well, that changes things a little bit. As I said above, function parameters are independent always. In order to capture their dependence on each other I need.... another function! Let w(x) = z(x, x). Now we have a new function w. It is a function of only ONE parameter. Computing w(x) gives us z(x, y) under the restriction that x = y. Since w is a function of one variable, it doesn't have partial derivatives. It just has the regular 1variable derivative from Calc I. My (somewhat controversial) advice is, whenever you can, try to formulate your problems in terms of functions instead of just equations. It might end up looking a bit different from what's in your textbook, and it might be a little bit longer to write out by hand, but it's the correct formalization. The standard notation becomes a useful shorthand later once you understand the underlying concept. 


#11
Jan313, 06:28 AM

P: 30

assuming for a function,all parameters are independent. then you cant say and if you are saying a function contains only those variables that are independent of each other. then z(x,y) is undefined for x=y. so i dont understand why its false $$\frac{\partial}{\partial t}f(x_{i},x_{j},x_{k},...) = \frac{\partial}{\partial t}f(x_{i})=\frac{d}{dt}f(x_{i})$$ for example total derivative does not assume any relation between x_{1},x_{2},x_{3}... $$\frac{d}{dx}z(x,y)=\frac{d}{dx}(x+y)=1+\frac{dy}{dx}$$ so if x and y are indendent z' = 1 and if x and y are dependent then z' != 1 and y' != 0 as TacTics suggested, i AM confused about function and variable. first let me give all relevent "i think" points, so you can tell me where i am wrong 1. in an equation, if a side is variable then the other side is also a variable. e.g., $$\left(z=x+y\; and\; z\: is\: variable\right)\Rightarrow x+y\: is\: variable$$ $$\left(x+y=f(x,y)\; and\; x+y\: is\: variable\right)\Rightarrow f(x,y)\: is\: variable$$ 2. f(x_{1},x_{2},x_{3}...) is a short notation which says f is equal to some expression involving x_{1},x_{2},x_{3}... . eg, $$x+y=f(x,y)=f(x)=f(y)$$ more importantly f(x_{1},x_{2},x_{3}...) does not assumes no relation between x_{1},x_{2},x_{3}... . moer than one function notation can be formed out of a single expression. thank you 


#12
Jan313, 07:53 AM

P: 810

Again, suppose f(x, y) = x + y. And let's take the partial derivative with respect to x on both sides. The ultimate problem is that "taking a partial derivative of an equation" (partial or otherwise) doesn't actually make any sense. Derivatives (again, partial or not) are something you do to a function. NOT to one or both sides of an equation. This is stated clearly in the early chapters of every calculus text book: Let f be a function from R > R. It's derivative is defined as f'(x) = lim h→0 of (f(x + h)  f(x)) / h, or Let f be a function from RxR > R. It's first partial derivative is defined as f_0(x, y) = lim h→0 of (f(x+h, y)  f(x, y)) / h, and it's second partial derivative is f_1(x, y) = lim h→0 of (f(x, y+h)  f(x, y)) / h. Both of these definitions assume we start with a function and we get another function for free (and we call it its derivative). What is a function? It's a rule for assigning an output to each input. In our f example, our output is x + y, and our input is a pair (x, y). Standard mathematical notation for defining the function is as an equation f(x, y) = x + y. But what that means is "f applied to the arguments x and y equals x + y". It tells us what f(x, y) equals.... but what does f itself equal? Borrowing some notation, we might write f = (x, y) ↦ x + y. The meaning here is that f is a function that maps pairs of numbers to their sum. This notation makes the input and output of our function f explicit: everything to the left of the ↦ is input, while everything to the right is output. So back to derivatives (including partial derivatives). We know in order to take a derivative, we need a function. Where is our function in the equation f(x, y) = x + y? What are we even taking the derivative of? This is where textbooks and teachers will quickly start abusing notation. (And most of the time, they won't even realize they're doing it). As long as you are lucky, this abuse is harmless, and "taking the derivative of an equation" works out just the same. However, when you start throwing questions like "what is x = y?", you run into trouble. The best bet is to always try to be as explicit as possible with what functions you are deriving, and keep it in the back of your head that you cannot "take derivatives of both sides of an equation". in fact i added x = y so its false that x and y are parameters. may be x or y is a parameter but both cant be . so for a general function f(x_{1},x_{2},x_{3}...) before saying x_{i},x_{j},x_{k}... are parameters of f, first it has to be proved that they are independent variables. and if you are saying a function contains only those variables that are independent of each other. then z(x,y) is undefined for x=y. so i dont understand why its false $$\frac{\partial}{\partial t}f(x_{i},x_{j},x_{k},...) = \frac{\partial}{\partial t}f(x_{i})=\frac{d}{dt}f(x_{i})$$ for example total derivative does not assume any relation between x_{1},x_{2},x_{3}... $$\frac{d}{dx}z(x,y)=\frac{d}{dx}(x+y)=1+\frac{dy}{dx}$$ so if x and y are indendent z' = 1 and if x and y are dependent then z' != 1 and y' != 0 how do you mean ? i dont see the difference . as TacTics suggested, i AM confused about function and variable. first let me give all relevent "i think" points, so you can tell me where i am wrong 1. in an equation, if a side is variable then the other side is also a variable. e.g., $$\left(z=x+y\; and\; z\: is\: variable\right)\Rightarrow x+y\: is\: variable$$ $$\left(x+y=f(x,y)\; and\; x+y\: is\: variable\right)\Rightarrow f(x,y)\: is\: variable$$ 2. f(x_{1},x_{2},x_{3}...) is a short notation which says f is equal to some expression involving x_{1},x_{2},x_{3}... . eg, $$x+y=f(x,y)=f(x)=f(y)$$ more importantly f(x_{1},x_{2},x_{3}...) does not assumes no relation between x_{1},x_{2},x_{3}... . moer than one function notation can be formed out of a single expression. thank you[/QUOTE] 


#13
Jan313, 04:30 PM

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Most functions can be defined by specifying a relationship between variables. For example, the specification xy=1 implicitly defines two functions ##f,g:\mathbb R\to\mathbb R## that can also be defined by f(t)=1+t for all ##t\in\mathbb R##, and g(t)=1t for all ##t\in\mathbb R##. (Note that it never matters what variable is used in a "for all" statement). These functions are differentiable. For all ##t\in\mathbb R##, we have f'(t)=1 and ##g'(t)=1##. Note that this makes f' and g' constant functions, not to be confused with constants as defined above. The notation dy/dx doesn't refer to a derivative of y. It can't, because y isn't a function. The y in the numerator and the x in the denominator lets us know that we're supposed to compute the derivative of the function that "takes x to y" and then plug x into the result (which is another function) to get the final result. If xy=1, then the function that "takes x to y" is the one I called f. There's no such thing as a partial derivative of a variable. Typically, a math book will say something like this: Let n be a positive integer. Let E be a subset of ##\mathbb R^n##. Let x be an interior point of ##E##. Let ##\{e_1,e_2,\dots,e_n\}## be the standard basis for ##\mathbb R^n##. Let ##k\in\{1,2,\dots,n\}## be arbitrary. If there's a number A such that the limit $$\lim_{t\to 0}\frac{f(x+te_k)f(x)}{t}$$ exists, then this number is called the kth partial derivative of f at x, or the "partial derivative of f at x, with respect to the kth variable". There are many different notations for it, for example ##D_kf(x)##, ##\partial_kf(x)##, ##\partial f(x)/\partial x_k##, ##f^{(k)}(x)## and ##f_{,\,k}(x)##. Now, if n=2, we will usually write f(x,y) instead of f(x) (with ##x\in\mathbb R^2##) or ##f(x_1,x_2)##. Because x is traditionally put into the first variable "slot" of f, the notation ##\partial f(x,y)/\partial x## is an alternative to the five notations mentioned above (with k=1 and (x,y) replacing x). On the other hand, if you say that x,y,z are variables representing real numbers, and then say that z=x+y, then this (i.e. the string of text "z=x+y") is a constraint that prevents us from assigning arbitrary values to all three variables, and also ensures that if we assign values to any two of them, the value of the third is fixed. This means that the constraint implicitly defines at least three functions, and we can compute the partial derivatives of those. If you also specify that x=y, then this further reduces our ability to assign values to the three variables. Now if we assign a value to any one of them, the values of the other two are fixed. So the pair of constraints (z=x+y, x=y) defines at least six functions implicitly, and we can compute the derivatives of those. For example, if ##f:\mathbb R^2\to\mathbb R## is defined by ##f(x,y)=5x+y^2## for all ##x,y\in\mathbb R##, then f is a function, x and y are real numbers, (x,y) is an ordered pair of real numbers, i.e. a member of ##\mathbb R^2##, f(x,y) is a real number (but we only know which one if we know the values of x and y), f(7,3) is a real number (and we know that it's 44). It is never OK to write f(x,y)=f(x), because the lefthand side only makes sense if the domain of f is a subset of ##\mathbb R^2## and the righthand side only makes sense if the domain of f is a subset of ##\mathbb R##. Note that the definition of a function is always a "for all" statement, even if the words "for all" have been omitted. For example, the words "the function ##x^2##" is strictly speaking nonsense. The correct way to say it is "the function ##f:\mathbb R\to\mathbb R## defined by ##f(x)=x^2## for all ##x\in\mathbb R##". Even the phrase "the function ##f(x)=x^2##" is flawed in four(!) different ways: 1. Neither of the strings of text "f(x)=x^{2}" or "f(x)" represents a function. The function is denoted by f. 2. There's no specification of the domain. 3. There's no specification of the codomain. 4. The absence of the words "for all" hides the fact that x is a dummy variable (one that can be replaced by any other symbol without changing the meaning of the statement). 


#14
Jan313, 09:03 PM

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If y is independent of x then [itex]\partial[/itex]y/[itex]\partial[/itex]x = 0 


#15
Jan413, 01:36 PM

P: 30

1. a function is a set , where each element is an orderd pair and composed of two elements taken from two sets. the vice versa may or may not be true. (f is still a variable though). anyway if it is assumed that a constant is not a variable, then my 1st point is wrong (only for constant though, but still wrong). but i dont assume . i dont have any reason to. please mention if there are any. the reason i mentioned it here is that even though f is a set why it is false that 2. ##f(x_{1},x_{2},...)## is a nonset variable (i dont know what else to call, please tell me if there is a standard term). when ##f(x_{1},x_{2},...)=y## , ##y## is a nonset variable , it implies that ##f(x_{1},x_{2},...)## is nonset variable too. 3. it seems you and TacTicks are saying the correct total derivative notation is ##\frac{d}{dx}f## not ##\frac{d}{dx}f(x)## . (first let me add it to the set of bizzare notations that i dont understand ##\left\{ \frac{df}{dx},dx,df,...\right\}##) luckily i dont have to understand, because i know ##\frac{df}{dx}=lim_{h\rightarrow0}\frac{f(x+h)f(x)}{h}##, so whenever i need to deal with lhs i just calculate the rhs. lets assume ##f:A\mapsto R##, ##A\subset R^{n}## so it can be written that ##f(x_{1},x_{2},...,x_{n})=y## but later it is found that ##A=\{(x_{1},x_{1}+1,x_{1}+2,...,x_{1}+n1):x_{1}\in R\}## so for all following implication, mention if it is correct/incorrect and why ? ##f(x_{1},x_{2},...,x_{n})=f(x_{1},x_{1}+1,...,x_{1}+n1)## ##f(x_{1},x_{2},...,x_{n})=f(x_{1})## ##f(x_{1},x_{2},...,x_{n})=g(x_{1})## ##A=\{(x_{1},x_{1}+1,x_{1}+2,...,x_{1}+n1):x_{1}\in R\}## is contradictory statement does ##f:A\mapsto R##, ##A\subset R^{n}## and ##f(x_{1},x_{2},...,x_{n})=y\Rightarrow(x_{1},x_{2},...,x_{n})## are independent. my another question is how to make a function out of an expression ? in other words , how to make a function (a graph,a table or a mapping in your words) depicting the changes between the expression and the variables involved in that expression ? assuming all variables belongs to real set, will the domain be ##R^{n}## if n is the total number of variables involved ? what if the constants are assumed as variables ? is it neccerry to expresse the same expression in terms of independent variables only ? if not would not a expression led to more than one function mapping. would not it give different partial derivative of functions of the expression ? is it ok to give different results ? if yes why is it ok ?  its really annoying when oversimplification leads to more doubts. thank you 


#16
Jan413, 05:09 PM

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I should probably have mentioned that those definitions of "constant" and "variable" are my own. I haven't seen those terms defined in textbooks. If ##f:\mathbb R\to\mathbb R##, then we can define a function ##df:\mathbb R^2\to\mathbb R## by ##df(x,h)=f'(x)h## for all ##x,h\in\mathbb R##. If we then use the notation "dx" for the number h, we have ##df(x,dx)=f'(x)dx##, and if we recklessly write the lefthand side as "df", we have df=f'(x)dx. Note that this does not provide any justification for tricks like $$\frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}.$$ This is the chain rule, which I prefer to write as ##(f\circ g)'(x)=f'(g(x))g'(x)##. It has a relatively complicated proof. *) The open ball of radius r around x is defined as the set ##B_r(x)=\{y\in\mathbb R^n:yx<r\}##. By the way, the \mapsto arrow is supposed to be used to specify what happens to the members of the domain. Example: The function ##f:\mathbb R\to\mathbb R## defined by ##f(x)=x^2## for all ##x\in\mathbb R## shouldn't be written as just ##x^2##, but it can be written as ##x\mapsto x^2##. Some people like to state the definition of this f like this: ##f:\mathbb R\to\mathbb R:x\mapsto x^2##. In the complicated cases, the answer is very nontrivial. You need a pretty hard theorem called the implict function theorem. $$\frac{\partial}{\partial x}(x+2y+z^3)=\frac{\partial}{\partial x}x+2\frac{\partial}{\partial x}y+\frac{\partial}{\partial x}z^3$$ and then think about how to evaluate each term. (Note that this involves using the constraints you've been given to reinterpret y and z as functions instead of as variables representing real numbers). 


#17
Jan413, 05:57 PM

P: 810

Variables don't exist on their own. Every variable you use in an expression must first be declared. We call expressions that declare variables "binding forms". Here are some examples of binding forms: f(x) = ... (a function definition declares x) Σi=0 to 100 ... (a summation declares i) ∫ .... dt (integration declares t) let x = 2 + 3 .... (a variable definition declares x... obviously :) Each binding form also has a scope for the variable. The variable "comes into existence" in the scope, and ceases to exist entirely outside of the scope. The scopes above are given by the ... part (which can be any expression). Looking at a particular expression and a variable x. We say x is bound if its binder appears in the expression. Otherwise, we say it is free. Some examples: let x = 1, x + x = 2. (x is bound by the "let") f(r) = π * r^2 (r is bound by the function definition. π is also a variable, but it's free!) Σi=1 to 5 = 10 (i is bound by the summation) ∫ ct^2 dt = c∫ t^2 dt (we have two variables: t is bound by the integral, c is free). Notice that whether a variable is bound or free depends on the expression we're looking at! If we restrict our attention to some subexpression, a variable's freedom may change: f(x) = x^2 (x is bound) but x^2 (x is free!) Now, here's the surprise twist ending: Free variables are also called constants. This explains why e and π are always called constants. We gave their definition a long time ago in high school, and no one bothers writing the definitions in their expressions. 


#18
Jan513, 06:05 AM

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Let e=exp(1). There exists a real number x such that x^{2}=e. And here e is not a free variable. I don't have any objections against using the lazy (and inaccurate) version of the statement, since everyone knows exactly what we're omitting. 


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