# Paritial derivative of function of dependent variables

 Math Emeritus Sci Advisor Thanks PF Gold P: 39,533 If F(x,y) is a function of the two variables, x and y, and it is true that y= g(x), then we can write f(x)= F(x, g(x)) and find the derivative of f with the chain rule: $$\frac{df}{dx}= \frac{\partial F}{\partial x}+ \frac{\partial F}{\partial y}\frac{dg}{dx}$$
P: 810
 Quote by Fredrik This sounds a bit odd actually. It's not that I have anything against calling free variables "constants". It's just that if I see the statement "There exists a real number x such that x2=e.", I'm not thinking that e is a free variable. I'm thinking that this statement is just a lazy way of writing this: Let e=exp(1). There exists a real number x such that x2=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.
The expression you're devoting your attention is important. You can also use this to define the notion of constancy of one variable with respect to another. In the above case, e is constant with respect to x. The reasoning would be something like "restricting your attention to the scope of the binder for x, e is free".
P: 30
lets consider a textbook question
 if ##f(x,y,z) = x^{2} + y^{2} + z^{2}##, ##x=t## , ##y=t^2##, ##z=2t## find out ##\frac{\partial f(x,y,z)}{\partial x}## .
here is my solution :
1. ##f(x,y,z) = x^{2} + y^{2} + z^{2}=x^{2}+x^{4}+4x^{2}=x^{4}+5x^{2}##
2. ##\Rightarrow f(x,y,z)=x^{4}+5x^{2}##
3. ##\Rightarrow\frac{\partial}{\partial x}f(x,y,z)=\frac{\partial}{\partial x}(x^{4}+5x^{2})##
4. ##\Rightarrow\frac{\partial}{\partial x}f(x,y,z)= 4x^{3} + 10x##

but the author says ##\frac{\partial}{\partial x}f(x,y,z)= 2x##

this might sound little annoying, but i still dont know how am i wrong ?

 Quote by Fredrik The derivative of f is another function, which I would denote by f' or maybe Df (but not ##\frac{d}{dx}f##). The value of that function at x is then denoted by f'(x), Df(x) or ##\frac{d}{dx}f(x)##. ##\frac{df}{dx}## is just a sloppy way of writing ##\frac{df(x)}{dx}##, which means the same thing as ##\frac{d}{dx}f(x)##.
"...sloppy...same thing as...". its little bit vague. so what are you saying,
1. ##\frac{df}{dx}=\frac{d}{dx}f(x)## and by extension ##f=f(x)## or
2. ##\frac{df}{dx}\neq\frac{d}{dx}f(x)##

 Quote by Fredrik The statement after the "and" is stated in a strange way (and you should use the \to arrow instead of the \mapsto arrow), but if you mean what I think you mean, then the answer is "sort of yes". You never have to worry about a possible relationship between the variables when you're asked to compute a partial derivative of a given function. However, if you're asked to compute a partial derivative of a function that is defined implicitly by a relationship between variables, then things are of course very different.
ofcourse i have to worry about that. i have to make sure (or at least assume) that the input variables are independent or not ? without that i cant move forward.

 Quote by Fredrik I'm not sure I understand what you're asking. If you're asking if e.g. the string of text "f(x,y)" represents a set, then the answer is yes, because we're working within the branch of mathematics defined by ZFC set theory, and in ZFC set theory, the members of sets are themselves sets. However, if e.g. S={1,2,3} then we prefer to call the members of S "numbers" instead of "sets". Similarly, if f:X→Y (i.e. if f is a function with domain X and codomain Y), and x is a member of X, then we prefer to describe f(x) as a member of Y instead of as a "set".
now ##f(x)## is a set too ? i thought it is not a set i.e. (a non-set). i dont know how to imagine this ##-A## if ##A## is a set. for example
what is ##-1## since ##1## is a set ?
what is ##-\{1,2,3\}## ?

why make everything set ?

p.s. sorry for the late reply.

thank you
Emeritus
PF Gold
P: 9,382
 Quote by ato lets consider a textbook question here is my solution : 1. ##f(x,y,z) = x^{2} + y^{2} + z^{2}=x^{2}+x^{4}+4x^{2}=x^{4}+5x^{2}## 2. ##\Rightarrow f(x,y,z)=x^{4}+5x^{2}## 3. ##\Rightarrow\frac{\partial}{\partial x}f(x,y,z)=\frac{\partial}{\partial x}(x^{4}+5x^{2})## 4. ##\Rightarrow\frac{\partial}{\partial x}f(x,y,z)= 4x^{3} + 10x## but the author says ##\frac{\partial}{\partial x}f(x,y,z)= 2x## this might sound little annoying, but i still dont know how am i wrong ?
You're supposed to find the partial derivative with respect to the first variable of f, evaluated at (x,y,z). f is defined by ##f(x,y,z)=x^2+y^2+z^2## for all ##x,y,z\in\mathbb R##. (It wouldn't make much sense to think of it as defined by ##f(x,y,z)=x^2+y^2+z^2## for all ##x,y,z\in\mathbb R## such that ##x=t, y=t^2, z=2t##, because then the domain of f is a single point, and f can't have any partial derivatives).
\begin{align}
\frac{\partial f(x,y,z)}{\partial x}&=D_1f(x,y,z)=(s\mapsto f(s,y,z))'(x)=\lim_{h\to 0}\frac{f(x+h,y,z)-f(x,y,z)}{h}\\
&=\lim_{h\to 0}\frac{(x+h)^2-x^2}{h}=\frac{d}{dx}x^2.\end{align} The specification ##x=t, y=t^2, z=2t## is a red herring. It's only relevant if you're asked to compute
$$\frac{d}{dt}f(x(t),y(t),z(t))=\frac{d}{dt}(5t^2+4t^4).$$

 Quote by ato "...sloppy...same thing as...". its little bit vague. so what are you saying, f 1. ##\frac{df}{dx}=\frac{d}{dx}f(x)## and by extension ##f=f(x)## or 2. ##\frac{df}{dx}\neq\frac{d}{dx}f(x)##
Same thing as = is equal to.
##\frac{df}{dx}## is a bad notation for f'(x). I don't want to write things like f=f(x) since f is a function and f(x) an element of its range. But people sometimes do use these notations interchangeably.

 Quote by ato ofcourse i have to worry about that. i have to make sure (or at least assume) that the input variables are independent or not ? without that i cant move forward.
I think you didn't understand the sentences you quoted there. I'm saying that if you know the function, no additional information can be relevant (since it's not part of the definition of the function), so you can move forward.

 Quote by ato now ##f(x)## is a set too ? i thought it is not a set i.e. (a non-set). i dont know how to imagine this ##-A## if ##A## is a set. for example what is ##-1## since ##1## is a set ? what is ##-\{1,2,3\}## ? why make everything set ? p.s. sorry for the late reply. thank you
-{1,2,3} is not defined. 1 and -1 typically denotes two members of a field (a set on which there's a multiplication operation and an addition operation). 1 denotes the multiplicative identity and -1 its additive inverse.

It's convenient to make everything sets, because then the theory only needs to leave two things undefined: what a set is, and what it means for a set to be a member of a set.
P: 30
 Quote by Fredrik You're supposed to find the partial derivative with respect to the first variable of f, evaluated at (x,y,z). f is defined by ##f(x,y,z)=x^2+y^2+z^2## for all ##x,y,z\in\mathbb R##. (It wouldn't make much sense to think of it as defined by ##f(x,y,z)=x^2+y^2+z^2## for all ##x,y,z\in\mathbb R## such that ##x=t, y=t^2, z=2t##, because then the domain of f is a single point, and f can't have any partial derivatives). \begin{align} \frac{\partial f(x,y,z)}{\partial x}&=D_1f(x,y,z)=(s\mapsto f(s,y,z))'(x)=\lim_{h\to 0}\frac{f(x+h,y,z)-f(x,y,z)}{h}\\ &=\lim_{h\to 0}\frac{(x+h)^2-x^2}{h}=\frac{d}{dx}x^2.\end{align} The specification ##x=t, y=t^2, z=2t## is a red herring. It's only relevant if you're asked to compute $$\frac{d}{dt}f(x(t),y(t),z(t))=\frac{d}{dt}(5t^2+4t^4).$$
yes the question did ask for ##\frac{d}{dt}f(x(t),y(t),z(t))## . the theoram used to solve is

##\frac{\partial f(x,y,z)}{\partial t}=\frac{\partial f(x,y,z)}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f(x,y,z)}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial f(x,y,z)}{\partial z}\frac{\partial z}{\partial t}##
and i understood everything except ##\frac{\partial f(x,y,z)}{\partial x},\frac{\partial f(x,y,z)}{\partial y},\frac{\partial f(x,y,z)}{\partial z}##.

why should not i take the partial derivative from this equation, ##f(x,y,z)=x^4+5x^2##. whats wrong with that ?

 Quote by Fredrik It's only relevant if you're asked to compute $$\frac{d}{dt}f(x(t),y(t),z(t))=\frac{d}{dt}(5t^2+4t^4).$$
so i am supposed to ignore ##x=t, y=t^2, z=2t## .

you cant ignore anything like that.

 Quote by Fredrik ##\frac{df}{dx}## is a bad notation for f'(x). I don't want to write things like f=f(x) since f is a function and f(x) an element of its range.
so f(x) IS a variable . so what was the point of "we dont take total derivative of variable but function" as now you can easily define total derivative (probably the whole calculus) without bringing the concept of function,
let ##y=f(x)## so ##\frac{df(x)}{dx}=\frac{dy}{dx}=lim_{h\rightarrow0}\frac{\left.y\right |_{x=x}^{x=x+h}}{h}##

(what good this (weird) definition of function do anyway ?)

 Quote by Fredrik -{1,2,3} is not defined.
since 1,2,3... all are sets, -1 is defined means - operator for a set is defined so i wanted to know how it is defined for a generel set.

 Quote by Fredrik 1 and -1 typically denotes two members of a field (a set on which there's a multiplication operation and an addition operation). 1 denotes the multiplicative identity and -1 its additive i inverse.
no i asked could you give me something resembling '1 = {...}'

 Quote by Fredrik It's convenient to make everything sets, because then the theory only needs to leave two things undefined: what a set is, and what it means for a set to be a member of a set.
could give me something like a paradox or a problem that it solves ?

thank you
Emeritus
PF Gold
P: 9,382
 Quote by ato yes the question did ask for ##\frac{d}{dt}f(x(t),y(t),z(t))## .
The part of it that you posted didn't. The simplest way to find ##\frac{d}{dt}f(x(t),y(t),z(t))## is
$$\frac{d}{dt}f(x(t),y(t),z(t))=\frac{d}{dt}(t^4+5t^2)=4t^3+10t.$$
 Quote by ato the theoram used to solve is ##\frac{\partial f(x,y,z)}{\partial t}=\frac{\partial f(x,y,z)}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f(x,y,z)}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial f(x,y,z)}{\partial z}\frac{\partial z}{\partial t}##
This is another way.

 Quote by ato why should not i take the partial derivative from this equation, ##f(x,y,z)=x^4+5x^2##. whats wrong with that ?
The problem is that there's no open set ##E\subset\mathbb R^3## such that ##f(x,y,z)=x^4+5x^2## for all ##(x,y,z)\in E##. If there had been such a set (this is only possible with a different definition of f), and (x,y,z) is a point in that set, then we would have had ##D_1f(x,y,z)=4x^3+10x##.

 Quote by ato so i am supposed to ignore ##x=t, y=t^2, z=2t## . you cant ignore anything like that.
Yes you can, because a definition of a function is always a "for all" statement, and you can always replace the variable in a "for all" statement without changing the meaning of the statement.
$$f(x,y,z)=x^2+y^2+z^2$$ really means
$$\forall x,y,z\in\mathbb R,\quad f(x,y,z)=x^2+y^2+z^2.$$ And this statement is equivalent to
$$\forall p,q,r\in\mathbb R,\quad f(p,q,r)=p^2+q^2+r^2.$$ If this is followed by a separate statement x=t, then this can't be the same x, since the first one was a dummy (replaceable) variable and this one isn't.

The statements
$$x=t, y=t^2, z=2t$$ are intended as definitions of three more functions x,y,z. So what they really mean is this:
\begin{align}
\end{align} To find ##D_1f##, you only need the definition of f. (Just look at the definition. It doesn't involve any other functions). So these three functions are irrelevant when you compute ##D_1f##.

 Quote by ato so f(x) IS a variable .
I don't consider "f(x)" a variable, because it's a string of text that consists of four characters. I would only use the term "variable" about 1-character strings, not longer strings.

 Quote by ato so what was the point of "we dont take total derivative of variable but function" as now you can easily define total derivative (probably the whole calculus) without bringing the concept of function, let ##y=f(x)## so ## ##\frac{df(x)}{dx}=\frac{dy}{dx}=lim_{h\rightarrow0}\frac{\left.y\right |_{x=x}^{x=x+h}}{h}## (what good this (weird) definition of function do anyway ?)
The concept of "function" is so extremely important in all areas of mathematics that I wouldn't know where to begin to answer that. I would have had more sympathy for the complaint if you had found these calculations really easy to do without involving the concept of "function", but you are clearly still struggling, and I'm not the only one in this thread who has said that it's probably because you're thinking in terms of variables than instead of in terms of functions.

 Quote by ato since 1,2,3... all are sets, -1 is defined means - operator for a set is defined so i wanted to know how it is defined for a generel set.
It's not.

 Quote by ato no i asked could you give me something resembling '1 = {...}'
Ah. OK, then we first have to define the natural numbers as sets. A standard way to do this is
0=∅
1={0}
2={0,1}
3={0,1,2}
...

I haven't seen a similar definition of the integers in a textbook, but it's not hard to think of one. Let S be any set with two members. Denote those members by p,n. Now consider the set
$$Z=\{(n,1),(n,2),\dots\}\cup \{0\}\cup\{(p,1),(p,2),\dots\}.$$
With an appropriate definition of addition, multiplication, etc, the subset
$$\{0\}\cup\{(p,1),(p,2),\dots\}$$ will have all the same properties (as far as the standard operations are concerned) as the set {0,1,2,...}. So it has just as much right to be called "the set of positive integers" as {0,1,2,...}. And now we can define the term "integer" as a member of Z. At this point, it would be convenient to simplify the notation like this:

Write k instead of (p,k) for all k in {1,2,...}.
Write -k instead of (n,k) for all k in {1,2,...}.

These things are very far from the things you need to understand to compute partial derivatives.

 Quote by ato could give me something like a paradox or a problem that it solves ? thank you
The problem it solves is just that it's annoying and kind of dangerous to have more undefined concepts and more axioms than is absolutely necessary. Every new concept that requires its own set of assumptions increases the probability that we will screw up by making inconsistent assumptions.
P: 30
 Quote by Fredrik The problem is that there's no open set ##E\subset\mathbb R^3## such that ##f(x,y,z)=x^4+5x^2## for all ##(x,y,z)\in E##.
##E## does exist, ##\{(a,a^{2},2a)|a\in R\}\subset R^3##

 Quote by Fredrik It's not. Ah. OK, then we first have to define the natural numbers as sets. A standard way to do this is 0=∅ 1={0} 2={0,1} 3={0,1,2} ...
we konw -2 exist right. should not -{0,1} exists.

##3 - 2 = 1##

##\{ 0,1,2 \} - \{ 0,1 \} = \{ 2 \} \neq 1##

thank you
Emeritus
PF Gold
P: 9,382
 Quote by ato ##E## does exist, ##\{(a,a^{2},2a)|a\in R\}\subset R^3##
That's a curve, not an open set.

 Quote by ato we konw -2 exist right. should not -{0,1} exists.
Right, but {0,1} isn't an arbitrary set. It's a member of a set on which there's a unary operation for which we have chosen the notation ##x\mapsto -x##.

 Quote by ato ##3 - 2 = 1## ##\{ 0,1,2 \} - \{ 0,1 \} = \{ 2 \} \neq 1## is not that a paradox/contradiction.
The minus sign here isn't the one defined by ##A-B=\{x\in A|x\notin B\}##. It's a binary operation on the set ##\mathbb N##. I don't have time to think about its definition right now. I suggest a book on set theory. Hrbacek and Jech is good for set theory in general. Goldrei is good alternative if you're especially interested in the constructions of the number systems.

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