Are all notation equal by derivatives?When is the derivative of an inverse function equal to the reciprocal of the derivative?Can a function and its derivative have different input variables?What are some good examples for suggestive notation?Notation: Set of all functions with finite number of variables from $A$ to $B$?Is $d^2y/dx$ a valid mathematical notation?Notation of Hessian?Notation for Derivatives with Respect to TimeStrange notation for a decimal expansion of a transcendental numberA pedantic question about notation — “such that” symbolsArithmetic rules for big O notation, little o notation and so on…Quantifier notation: $forall n implies cdot$ versus $forall n, cdot$In Leibniz notation, how do you write the second derivative of y with respect to the square of x?
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Are all notation equal by derivatives?
When is the derivative of an inverse function equal to the reciprocal of the derivative?Can a function and its derivative have different input variables?What are some good examples for suggestive notation?Notation: Set of all functions with finite number of variables from $A$ to $B$?Is $d^2y/dx$ a valid mathematical notation?Notation of Hessian?Notation for Derivatives with Respect to TimeStrange notation for a decimal expansion of a transcendental numberA pedantic question about notation — “such that” symbolsArithmetic rules for big O notation, little o notation and so on…Quantifier notation: $forall n implies cdot$ versus $forall n, cdot$In Leibniz notation, how do you write the second derivative of y with respect to the square of x?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Here I read that
Suppose you have a general function: y = f(x). All of the following
notations can be read as "the derivative of y with respect to x" or
less formally, "the derivative of the function":f'(x),f',y',
df/dx,dy/dx,d/dx[f(x)].
Still if I want to describe the value of the derivative at a certain point I write f'(3) for example and not df(3)/d3. I am not sure if the latter is syntactically valid, because I have never seen it. Is it possible with all of these notations to describe the value of the derivative for a certain x? If so, then how? If not, then why? Are there abstract and concrete notations in math?
notation
asked Jun 30 at 6:11
inf3rnoinf3rno
1588 bronze badges
$endgroup$
add a comment |
$begingroup$
Here I read that
Suppose you have a general function: y = f(x). All of the following
notations can be read as "the derivative of y with respect to x" or
less formally, "the derivative of the function":f'(x),f',y',
df/dx,dy/dx,d/dx[f(x)].
Still if I want to describe the value of the derivative at a certain point I write f'(3) for example and not df(3)/d3. I am not sure if the latter is syntactically valid, because I have never seen it. Is it possible with all of these notations to describe the value of the derivative for a certain x? If so, then how? If not, then why? Are there abstract and concrete notations in math?
notation
asked Jun 30 at 6:11
inf3rnoinf3rno
1588 bronze badges
$endgroup$
3
$begingroup$
The correct answer has already been given but note that $fracdf(3)dx=0$ since you're differentiating a constant. And $fracdf(3)d3$ is meaningless.
$endgroup$
– Deepak
Jun 30 at 6:41
4
$begingroup$
The last two examples are really one. It's notd/dxand[f(x)]butd/dx [f(x)]- I don't know whether that was a transcription error (in which case please edit) or you are genuinely thinking square bracket notation indicates differentiation.
$endgroup$
– abligh
Jun 30 at 14:42
1
$begingroup$
Referring to the linked page, the exampled/dx [f(x)]really is one example, not two, and it suffers from the fact that the writer of that page did not know how to typeset the mathematics in the usual way (or did not think it worth the trouble). I don't think I have ever seen $d/dx, [f(x)]$ in a regular textbook, but I have seen $fracddx[f(x)].$
$endgroup$
– David K
Jun 30 at 15:53
1
$begingroup$
For a function of only one variable, right? Because once you get into a function of several variables, you can't use $f'$ notation, you either have to use subscript $f_x, f_y, f_xx, f_xy,...$ or partial operators $partial/partial x, partial/partial y, partial^2/partial x^2 ...$ or come up with a vector formulation, or something else.
$endgroup$
– smci
Jun 30 at 19:22
$begingroup$
@abligh Thanks, it is just a typo. I did not think much about what I am copy pasting. :-)
$endgroup$
– inf3rno
Jul 1 at 6:24
add a comment |
$begingroup$
Here I read that
Suppose you have a general function: y = f(x). All of the following
notations can be read as "the derivative of y with respect to x" or
less formally, "the derivative of the function":f'(x),f',y',
df/dx,dy/dx,d/dx[f(x)].
Still if I want to describe the value of the derivative at a certain point I write f'(3) for example and not df(3)/d3. I am not sure if the latter is syntactically valid, because I have never seen it. Is it possible with all of these notations to describe the value of the derivative for a certain x? If so, then how? If not, then why? Are there abstract and concrete notations in math?
notation
asked Jun 30 at 6:11
inf3rnoinf3rno
1588 bronze badges
$endgroup$
Here I read that
Suppose you have a general function: y = f(x). All of the following
notations can be read as "the derivative of y with respect to x" or
less formally, "the derivative of the function":f'(x),f',y',
df/dx,dy/dx,d/dx[f(x)].
Still if I want to describe the value of the derivative at a certain point I write f'(3) for example and not df(3)/d3. I am not sure if the latter is syntactically valid, because I have never seen it. Is it possible with all of these notations to describe the value of the derivative for a certain x? If so, then how? If not, then why? Are there abstract and concrete notations in math?
notation
notation
asked Jun 30 at 6:11
inf3rnoinf3rno
1588 bronze badges
asked Jun 30 at 6:11
inf3rnoinf3rno
1588 bronze badges
edited Jul 1 at 6:23
inf3rno
asked Jun 30 at 6:11
inf3rnoinf3rno
1588 bronze badges
asked Jun 30 at 6:11
inf3rnoinf3rno
1588 bronze badges
asked Jun 30 at 6:11
inf3rnoinf3rno
1588 bronze badges
1588 bronze badges
3
$begingroup$
The correct answer has already been given but note that $fracdf(3)dx=0$ since you're differentiating a constant. And $fracdf(3)d3$ is meaningless.
$endgroup$
– Deepak
Jun 30 at 6:41
4
$begingroup$
The last two examples are really one. It's notd/dxand[f(x)]butd/dx [f(x)]- I don't know whether that was a transcription error (in which case please edit) or you are genuinely thinking square bracket notation indicates differentiation.
$endgroup$
– abligh
Jun 30 at 14:42
1
$begingroup$
Referring to the linked page, the exampled/dx [f(x)]really is one example, not two, and it suffers from the fact that the writer of that page did not know how to typeset the mathematics in the usual way (or did not think it worth the trouble). I don't think I have ever seen $d/dx, [f(x)]$ in a regular textbook, but I have seen $fracddx[f(x)].$
$endgroup$
– David K
Jun 30 at 15:53
1
$begingroup$
For a function of only one variable, right? Because once you get into a function of several variables, you can't use $f'$ notation, you either have to use subscript $f_x, f_y, f_xx, f_xy,...$ or partial operators $partial/partial x, partial/partial y, partial^2/partial x^2 ...$ or come up with a vector formulation, or something else.
$endgroup$
– smci
Jun 30 at 19:22
$begingroup$
@abligh Thanks, it is just a typo. I did not think much about what I am copy pasting. :-)
$endgroup$
– inf3rno
Jul 1 at 6:24
add a comment |
3
$begingroup$
The correct answer has already been given but note that $fracdf(3)dx=0$ since you're differentiating a constant. And $fracdf(3)d3$ is meaningless.
$endgroup$
– Deepak
Jun 30 at 6:41
4
$begingroup$
The last two examples are really one. It's notd/dxand[f(x)]butd/dx [f(x)]- I don't know whether that was a transcription error (in which case please edit) or you are genuinely thinking square bracket notation indicates differentiation.
$endgroup$
– abligh
Jun 30 at 14:42
1
$begingroup$
Referring to the linked page, the exampled/dx [f(x)]really is one example, not two, and it suffers from the fact that the writer of that page did not know how to typeset the mathematics in the usual way (or did not think it worth the trouble). I don't think I have ever seen $d/dx, [f(x)]$ in a regular textbook, but I have seen $fracddx[f(x)].$
$endgroup$
– David K
Jun 30 at 15:53
1
$begingroup$
For a function of only one variable, right? Because once you get into a function of several variables, you can't use $f'$ notation, you either have to use subscript $f_x, f_y, f_xx, f_xy,...$ or partial operators $partial/partial x, partial/partial y, partial^2/partial x^2 ...$ or come up with a vector formulation, or something else.
$endgroup$
– smci
Jun 30 at 19:22
$begingroup$
@abligh Thanks, it is just a typo. I did not think much about what I am copy pasting. :-)
$endgroup$
– inf3rno
Jul 1 at 6:24
3
3
$begingroup$
The correct answer has already been given but note that $fracdf(3)dx=0$ since you're differentiating a constant. And $fracdf(3)d3$ is meaningless.
$endgroup$
– Deepak
Jun 30 at 6:41
$begingroup$
The correct answer has already been given but note that $fracdf(3)dx=0$ since you're differentiating a constant. And $fracdf(3)d3$ is meaningless.
$endgroup$
– Deepak
Jun 30 at 6:41
4
4
$begingroup$
The last two examples are really one. It's not
d/dx and [f(x)]but d/dx [f(x)] - I don't know whether that was a transcription error (in which case please edit) or you are genuinely thinking square bracket notation indicates differentiation.$endgroup$
– abligh
Jun 30 at 14:42
$begingroup$
The last two examples are really one. It's not
d/dx and [f(x)]but d/dx [f(x)] - I don't know whether that was a transcription error (in which case please edit) or you are genuinely thinking square bracket notation indicates differentiation.$endgroup$
– abligh
Jun 30 at 14:42
1
1
$begingroup$
Referring to the linked page, the example
d/dx [f(x)] really is one example, not two, and it suffers from the fact that the writer of that page did not know how to typeset the mathematics in the usual way (or did not think it worth the trouble). I don't think I have ever seen $d/dx, [f(x)]$ in a regular textbook, but I have seen $fracddx[f(x)].$$endgroup$
– David K
Jun 30 at 15:53
$begingroup$
Referring to the linked page, the example
d/dx [f(x)] really is one example, not two, and it suffers from the fact that the writer of that page did not know how to typeset the mathematics in the usual way (or did not think it worth the trouble). I don't think I have ever seen $d/dx, [f(x)]$ in a regular textbook, but I have seen $fracddx[f(x)].$$endgroup$
– David K
Jun 30 at 15:53
1
1
$begingroup$
For a function of only one variable, right? Because once you get into a function of several variables, you can't use $f'$ notation, you either have to use subscript $f_x, f_y, f_xx, f_xy,...$ or partial operators $partial/partial x, partial/partial y, partial^2/partial x^2 ...$ or come up with a vector formulation, or something else.
$endgroup$
– smci
Jun 30 at 19:22
$begingroup$
For a function of only one variable, right? Because once you get into a function of several variables, you can't use $f'$ notation, you either have to use subscript $f_x, f_y, f_xx, f_xy,...$ or partial operators $partial/partial x, partial/partial y, partial^2/partial x^2 ...$ or come up with a vector formulation, or something else.
$endgroup$
– smci
Jun 30 at 19:22
$begingroup$
@abligh Thanks, it is just a typo. I did not think much about what I am copy pasting. :-)
$endgroup$
– inf3rno
Jul 1 at 6:24
$begingroup$
@abligh Thanks, it is just a typo. I did not think much about what I am copy pasting. :-)
$endgroup$
– inf3rno
Jul 1 at 6:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
All of those notations for the derivative:
$$f'(x),quadfracdfdx,quadfracddxf(x),quadfracdfdx(x),quadfracdf(x)dx$$
are equivalent. However, if we replace $x$ by a constant, some of these notations no longer make much sense. In Leibniz notation (all of the latter four, i.e. those with a $fracddx$), we are thinking of the derivative $fracddx$ as an operator acting on the function $f(x)$ (or $f$). So if we were to write
$$fracddxf(3),$$
this seems to imply that the differentiation operator is acting on the constant value $f(3)$. But then we are just differentiating a constant, which gives zero. On the other hand, the notation $f'(3)$ is interpreted as plugging in the value of $x=3$ into the function $f'(x)$, and the differentiation operator has already acted on $f(x)$ before plugging in the value. In this case $f'(3)$ is not zero (in general). If you want to achieve the effect of plugging in a constant value with Leibniz notation, we usually use
$$left.fracdfdxrightvert_x=3quadtextorquadleft(fracdfdxright)(3),$$
though I should warn you that the latter is rather uncommon and nonstandard. In either case, notation makes it clear that the differentiation is done first, and the plugging in of the constant value done second.
answered Jun 30 at 6:58
YiFanYiFan
6,4132 gold badges10 silver badges32 bronze badges
$endgroup$
1
$begingroup$
@inf3rno Indeed, that is the notation which is implemented in computer programs like Mathematica, where the notation isD[f,x]. But we are not doing programming here, we're doing math, and when people do math they like to use notation which provides conceptual clarity. In this case the $dx$ represents an "infinitesimal", in the sense that $df/dx$ denotes the limit $lim_Delta xto0(Delta f/Delta x)$. It is very helpful to think of differentials (the $dx$s and $df$s) in this way, having a life of their own, as opposed to always being a part of a derivative or an integral.
$endgroup$
– YiFan
Jun 30 at 8:31
1
$begingroup$
@inf3rno Although, you might be pleasantly surprised to learn that there is a notation that uses $Df$ to represent the derivative of $f$. In this way then $D^nf$ denotes the $n$th derivative of $f$. The notation $(Df)(3)$, or sometimes even $Df(3)$, denotes the function $Df(x)$ evaluated at $x=3$. But this is not as common as the other two notations for the reasons outlined above.
$endgroup$
– YiFan
Jun 30 at 8:36
1
$begingroup$
I would solve the nth derivative with recursion instead of having a separate denotation.r(d,n)(f,x)would return the nth derivative of f. Ofc. that is programming for math a separate notation works too. :-)
$endgroup$
– inf3rno
Jun 30 at 9:10
2
$begingroup$
I feel like the notation $fracdfdx(3)$, or at least its partial derivative equivalent $fracpartial fpartial x(3,3)$, is fairly common. It's used in this textbook, for instance.
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– Misha Lavrov
Jun 30 at 18:28
1
$begingroup$
$fracddxf(xleftarrow3)$ Problem solved. It's just a syntactic sugar; the d/dx notation depends on a relationship to the parameter symbols, so it becomes unclear when a symbol disappears via substitution; but if we notate the substation in some other way, the symbol doesn't disappear.
$endgroup$
– Kaz
Jun 30 at 22:33
|
show 8 more comments
$begingroup$
All of those not on the ends are equal, as they all denote the same thing. The last one, I've never seen. The first one is "close" - I'm going to get to that.
However, there is actually a very good question you ask here - namely, if we have
$$fracdfdx$$
how can we write, with this notation, what the value at $x = 3$ is? And the "best" answer to this is that we have to make a clear distinction between three different things:
- the meaning of the symbol "$f(x)$",
- the meaning of the symbol "$f$", and
- what kind of "thing", "differentiation", is.
And this stuff typically isn't touched on as well as it could be, and when it needs to be. The symbol $f(x)$ is not a function, even though it often gets abused by calling it such. Rather, what it is is the value of the function at $x$, and if $x$ is a variable that hasn't been assigned a value yet, this value is also unspecified (except that it must be within the range of the function). That is, $f(x)$ is a symbol for a (possibly unspecified) number.
What $f$ is, on the other hand, is the function itself: the mathematical object that you can plug numbers into to get other numbers out. Thus, you should, ideally, never say "the function $f(x)$", when what you really mean is "the function $f$", according to the above understanding.
Now, for the last part. What "differentation" is, is what is called an operator, or a higher-order function: it is a function that takes in other functions as inputs and pops out functions. Such things are typically denoted by prefixing their symbol against the function symbol: hence here, you should have $fracddx f$, which is shortened to $fracdfdx$, and this forms a combined function symbol.
Hence, if you want the value at $x = 3$, say, I'd argue you should "best" write
$$left[fracdfdxright](3)$$
which means to evaluate the function denoted by the symbol $fracdfdx$ at 3. And this, thus, generalizes likewise to higher derivatives, e.g.
$$left[fracd^2 fdx^2right](3)$$
and so forth.
Of course, if you think about this even more, technically we should not ever write something like
$$fracdfdx + x$$
even though this is very common, e.g. in writing a differential equation, and you can "get away with it" in that people will generally know what you mean. This is because here, $x$ is a number, and $fracdfdx$ is a function. To be "absolutely" correct, you will have two choices: either demote $fracdfdx$ to a number, which will make it
$$left[fracdfdxright](x) + x$$
or, to promote $x$ to a function, in which case, if you don't want to give another function symbol, you have to use the anonymous function notation
$$fracdfdx + left(x mapsto xright)$$
Either of these would be a fully-correct mathematical expression (though keep in mind, they actually have a different, but related, meaning each! What is it?) But since they are a bit ungainly, we accept the "wrong" notation as a shorthand: it's called "abuse of notation". And there isn't anything wrong with that, as long as you clearly understand what is going on, which you can check by trying to mentally translate it into the strictly proper, if not typically used, notation. Just like learning a foreign human natural language, where that in typical speech we'll never always follow the rules of grammar perfectly, but we always try to learn the ideal forms first, when learning the mathematical language we should keep in mind the same principle.
answered Jun 30 at 23:53
The_SympathizerThe_Sympathizer
10k2 gold badges29 silver badges51 bronze badges
$endgroup$
$begingroup$
You could also write $$fracmathrmdfmathrmdx + I qquadtextorqquad fracmathrmdfmathrmdx + operatornameid, $$ if you don't want to deal with the notation $xmapsto x$ in the middle of an expression.
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– Xander Henderson
Jul 1 at 0:03
$begingroup$
For me the $fracd^2dx^2f(3)$ is just as clear as the $left[fracd^2dx^2fright](3)$ because the order of the operators can matter in math. What I find the real problem here, that people can confuse $fracd^2 f(3)dx^2$ even to $fracd^2 cdot f(3)d cdot x^2$if we use sugar syntax like $fracd^2 fdx^2$ instead of $fracd^2dx^2f$. Even the $fracd^2dx^2f$ is not the best, but at least it gives a hint of what we are talking about.
$endgroup$
– inf3rno
Jul 1 at 6:11
$begingroup$
Maybe my question is not perfect, sometimes I just feel that something is wrong, but I cannot explain why until I think about it more. It's hard to write a good question. I like how you explained the difference between $f$ and $f(x)$ in your answer. I think being strict with notations is important, because it makes easier to understand what we are doing. It is just like coding in team, you write quality code, because others need to understand it and because if you read your own code again after a few months you might end up scratching your head too. It actually happened to me a few times. :D
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– inf3rno
Jul 1 at 6:13
1
$begingroup$
@inf3rno : Yes, you can also use $fracddx f(x)$ to reference the value of the derivative at the point $x$, as the operator binds more tightly to the function symbol, than to the evaluation. Also, I would not suggest to write $fracd^2 f(3)dx^2$ - either $fracd^2 fdx(3)$, or $fracd^2dx f(3)$, or as given with the brackets.
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– The_Sympathizer
Jul 1 at 7:46
1
$begingroup$
And I find it interesting you mention code. I also think that Maths has a fair bit it can learn from computer programming. One of my professors said that programming ideally tries to be like Maths, but I've also come to think that there's a lot of merit in things going the other way, too.
$endgroup$
– The_Sympathizer
Jul 1 at 7:47
add a comment |
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2 Answers
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$begingroup$
All of those notations for the derivative:
$$f'(x),quadfracdfdx,quadfracddxf(x),quadfracdfdx(x),quadfracdf(x)dx$$
are equivalent. However, if we replace $x$ by a constant, some of these notations no longer make much sense. In Leibniz notation (all of the latter four, i.e. those with a $fracddx$), we are thinking of the derivative $fracddx$ as an operator acting on the function $f(x)$ (or $f$). So if we were to write
$$fracddxf(3),$$
this seems to imply that the differentiation operator is acting on the constant value $f(3)$. But then we are just differentiating a constant, which gives zero. On the other hand, the notation $f'(3)$ is interpreted as plugging in the value of $x=3$ into the function $f'(x)$, and the differentiation operator has already acted on $f(x)$ before plugging in the value. In this case $f'(3)$ is not zero (in general). If you want to achieve the effect of plugging in a constant value with Leibniz notation, we usually use
$$left.fracdfdxrightvert_x=3quadtextorquadleft(fracdfdxright)(3),$$
though I should warn you that the latter is rather uncommon and nonstandard. In either case, notation makes it clear that the differentiation is done first, and the plugging in of the constant value done second.
answered Jun 30 at 6:58
YiFanYiFan
6,4132 gold badges10 silver badges32 bronze badges
$endgroup$
1
$begingroup$
@inf3rno Indeed, that is the notation which is implemented in computer programs like Mathematica, where the notation isD[f,x]. But we are not doing programming here, we're doing math, and when people do math they like to use notation which provides conceptual clarity. In this case the $dx$ represents an "infinitesimal", in the sense that $df/dx$ denotes the limit $lim_Delta xto0(Delta f/Delta x)$. It is very helpful to think of differentials (the $dx$s and $df$s) in this way, having a life of their own, as opposed to always being a part of a derivative or an integral.
$endgroup$
– YiFan
Jun 30 at 8:31
1
$begingroup$
@inf3rno Although, you might be pleasantly surprised to learn that there is a notation that uses $Df$ to represent the derivative of $f$. In this way then $D^nf$ denotes the $n$th derivative of $f$. The notation $(Df)(3)$, or sometimes even $Df(3)$, denotes the function $Df(x)$ evaluated at $x=3$. But this is not as common as the other two notations for the reasons outlined above.
$endgroup$
– YiFan
Jun 30 at 8:36
1
$begingroup$
I would solve the nth derivative with recursion instead of having a separate denotation.r(d,n)(f,x)would return the nth derivative of f. Ofc. that is programming for math a separate notation works too. :-)
$endgroup$
– inf3rno
Jun 30 at 9:10
2
$begingroup$
I feel like the notation $fracdfdx(3)$, or at least its partial derivative equivalent $fracpartial fpartial x(3,3)$, is fairly common. It's used in this textbook, for instance.
$endgroup$
– Misha Lavrov
Jun 30 at 18:28
1
$begingroup$
$fracddxf(xleftarrow3)$ Problem solved. It's just a syntactic sugar; the d/dx notation depends on a relationship to the parameter symbols, so it becomes unclear when a symbol disappears via substitution; but if we notate the substation in some other way, the symbol doesn't disappear.
$endgroup$
– Kaz
Jun 30 at 22:33
|
show 8 more comments
$begingroup$
All of those notations for the derivative:
$$f'(x),quadfracdfdx,quadfracddxf(x),quadfracdfdx(x),quadfracdf(x)dx$$
are equivalent. However, if we replace $x$ by a constant, some of these notations no longer make much sense. In Leibniz notation (all of the latter four, i.e. those with a $fracddx$), we are thinking of the derivative $fracddx$ as an operator acting on the function $f(x)$ (or $f$). So if we were to write
$$fracddxf(3),$$
this seems to imply that the differentiation operator is acting on the constant value $f(3)$. But then we are just differentiating a constant, which gives zero. On the other hand, the notation $f'(3)$ is interpreted as plugging in the value of $x=3$ into the function $f'(x)$, and the differentiation operator has already acted on $f(x)$ before plugging in the value. In this case $f'(3)$ is not zero (in general). If you want to achieve the effect of plugging in a constant value with Leibniz notation, we usually use
$$left.fracdfdxrightvert_x=3quadtextorquadleft(fracdfdxright)(3),$$
though I should warn you that the latter is rather uncommon and nonstandard. In either case, notation makes it clear that the differentiation is done first, and the plugging in of the constant value done second.
answered Jun 30 at 6:58
YiFanYiFan
6,4132 gold badges10 silver badges32 bronze badges
$endgroup$
1
$begingroup$
@inf3rno Indeed, that is the notation which is implemented in computer programs like Mathematica, where the notation isD[f,x]. But we are not doing programming here, we're doing math, and when people do math they like to use notation which provides conceptual clarity. In this case the $dx$ represents an "infinitesimal", in the sense that $df/dx$ denotes the limit $lim_Delta xto0(Delta f/Delta x)$. It is very helpful to think of differentials (the $dx$s and $df$s) in this way, having a life of their own, as opposed to always being a part of a derivative or an integral.
$endgroup$
– YiFan
Jun 30 at 8:31
1
$begingroup$
@inf3rno Although, you might be pleasantly surprised to learn that there is a notation that uses $Df$ to represent the derivative of $f$. In this way then $D^nf$ denotes the $n$th derivative of $f$. The notation $(Df)(3)$, or sometimes even $Df(3)$, denotes the function $Df(x)$ evaluated at $x=3$. But this is not as common as the other two notations for the reasons outlined above.
$endgroup$
– YiFan
Jun 30 at 8:36
1
$begingroup$
I would solve the nth derivative with recursion instead of having a separate denotation.r(d,n)(f,x)would return the nth derivative of f. Ofc. that is programming for math a separate notation works too. :-)
$endgroup$
– inf3rno
Jun 30 at 9:10
2
$begingroup$
I feel like the notation $fracdfdx(3)$, or at least its partial derivative equivalent $fracpartial fpartial x(3,3)$, is fairly common. It's used in this textbook, for instance.
$endgroup$
– Misha Lavrov
Jun 30 at 18:28
1
$begingroup$
$fracddxf(xleftarrow3)$ Problem solved. It's just a syntactic sugar; the d/dx notation depends on a relationship to the parameter symbols, so it becomes unclear when a symbol disappears via substitution; but if we notate the substation in some other way, the symbol doesn't disappear.
$endgroup$
– Kaz
Jun 30 at 22:33
|
show 8 more comments
$begingroup$
All of those notations for the derivative:
$$f'(x),quadfracdfdx,quadfracddxf(x),quadfracdfdx(x),quadfracdf(x)dx$$
are equivalent. However, if we replace $x$ by a constant, some of these notations no longer make much sense. In Leibniz notation (all of the latter four, i.e. those with a $fracddx$), we are thinking of the derivative $fracddx$ as an operator acting on the function $f(x)$ (or $f$). So if we were to write
$$fracddxf(3),$$
this seems to imply that the differentiation operator is acting on the constant value $f(3)$. But then we are just differentiating a constant, which gives zero. On the other hand, the notation $f'(3)$ is interpreted as plugging in the value of $x=3$ into the function $f'(x)$, and the differentiation operator has already acted on $f(x)$ before plugging in the value. In this case $f'(3)$ is not zero (in general). If you want to achieve the effect of plugging in a constant value with Leibniz notation, we usually use
$$left.fracdfdxrightvert_x=3quadtextorquadleft(fracdfdxright)(3),$$
though I should warn you that the latter is rather uncommon and nonstandard. In either case, notation makes it clear that the differentiation is done first, and the plugging in of the constant value done second.
answered Jun 30 at 6:58
YiFanYiFan
6,4132 gold badges10 silver badges32 bronze badges
$endgroup$
All of those notations for the derivative:
$$f'(x),quadfracdfdx,quadfracddxf(x),quadfracdfdx(x),quadfracdf(x)dx$$
are equivalent. However, if we replace $x$ by a constant, some of these notations no longer make much sense. In Leibniz notation (all of the latter four, i.e. those with a $fracddx$), we are thinking of the derivative $fracddx$ as an operator acting on the function $f(x)$ (or $f$). So if we were to write
$$fracddxf(3),$$
this seems to imply that the differentiation operator is acting on the constant value $f(3)$. But then we are just differentiating a constant, which gives zero. On the other hand, the notation $f'(3)$ is interpreted as plugging in the value of $x=3$ into the function $f'(x)$, and the differentiation operator has already acted on $f(x)$ before plugging in the value. In this case $f'(3)$ is not zero (in general). If you want to achieve the effect of plugging in a constant value with Leibniz notation, we usually use
$$left.fracdfdxrightvert_x=3quadtextorquadleft(fracdfdxright)(3),$$
though I should warn you that the latter is rather uncommon and nonstandard. In either case, notation makes it clear that the differentiation is done first, and the plugging in of the constant value done second.
answered Jun 30 at 6:58
YiFanYiFan
6,4132 gold badges10 silver badges32 bronze badges
answered Jun 30 at 6:58
YiFanYiFan
6,4132 gold badges10 silver badges32 bronze badges
answered Jun 30 at 6:58
YiFanYiFan
6,4132 gold badges10 silver badges32 bronze badges
answered Jun 30 at 6:58
YiFanYiFan
6,4132 gold badges10 silver badges32 bronze badges
6,4132 gold badges10 silver badges32 bronze badges
1
$begingroup$
@inf3rno Indeed, that is the notation which is implemented in computer programs like Mathematica, where the notation isD[f,x]. But we are not doing programming here, we're doing math, and when people do math they like to use notation which provides conceptual clarity. In this case the $dx$ represents an "infinitesimal", in the sense that $df/dx$ denotes the limit $lim_Delta xto0(Delta f/Delta x)$. It is very helpful to think of differentials (the $dx$s and $df$s) in this way, having a life of their own, as opposed to always being a part of a derivative or an integral.
$endgroup$
– YiFan
Jun 30 at 8:31
1
$begingroup$
@inf3rno Although, you might be pleasantly surprised to learn that there is a notation that uses $Df$ to represent the derivative of $f$. In this way then $D^nf$ denotes the $n$th derivative of $f$. The notation $(Df)(3)$, or sometimes even $Df(3)$, denotes the function $Df(x)$ evaluated at $x=3$. But this is not as common as the other two notations for the reasons outlined above.
$endgroup$
– YiFan
Jun 30 at 8:36
1
$begingroup$
I would solve the nth derivative with recursion instead of having a separate denotation.r(d,n)(f,x)would return the nth derivative of f. Ofc. that is programming for math a separate notation works too. :-)
$endgroup$
– inf3rno
Jun 30 at 9:10
2
$begingroup$
I feel like the notation $fracdfdx(3)$, or at least its partial derivative equivalent $fracpartial fpartial x(3,3)$, is fairly common. It's used in this textbook, for instance.
$endgroup$
– Misha Lavrov
Jun 30 at 18:28
1
$begingroup$
$fracddxf(xleftarrow3)$ Problem solved. It's just a syntactic sugar; the d/dx notation depends on a relationship to the parameter symbols, so it becomes unclear when a symbol disappears via substitution; but if we notate the substation in some other way, the symbol doesn't disappear.
$endgroup$
– Kaz
Jun 30 at 22:33
|
show 8 more comments
1
$begingroup$
@inf3rno Indeed, that is the notation which is implemented in computer programs like Mathematica, where the notation isD[f,x]. But we are not doing programming here, we're doing math, and when people do math they like to use notation which provides conceptual clarity. In this case the $dx$ represents an "infinitesimal", in the sense that $df/dx$ denotes the limit $lim_Delta xto0(Delta f/Delta x)$. It is very helpful to think of differentials (the $dx$s and $df$s) in this way, having a life of their own, as opposed to always being a part of a derivative or an integral.
$endgroup$
– YiFan
Jun 30 at 8:31
1
$begingroup$
@inf3rno Although, you might be pleasantly surprised to learn that there is a notation that uses $Df$ to represent the derivative of $f$. In this way then $D^nf$ denotes the $n$th derivative of $f$. The notation $(Df)(3)$, or sometimes even $Df(3)$, denotes the function $Df(x)$ evaluated at $x=3$. But this is not as common as the other two notations for the reasons outlined above.
$endgroup$
– YiFan
Jun 30 at 8:36
1
$begingroup$
I would solve the nth derivative with recursion instead of having a separate denotation.r(d,n)(f,x)would return the nth derivative of f. Ofc. that is programming for math a separate notation works too. :-)
$endgroup$
– inf3rno
Jun 30 at 9:10
2
$begingroup$
I feel like the notation $fracdfdx(3)$, or at least its partial derivative equivalent $fracpartial fpartial x(3,3)$, is fairly common. It's used in this textbook, for instance.
$endgroup$
– Misha Lavrov
Jun 30 at 18:28
1
$begingroup$
$fracddxf(xleftarrow3)$ Problem solved. It's just a syntactic sugar; the d/dx notation depends on a relationship to the parameter symbols, so it becomes unclear when a symbol disappears via substitution; but if we notate the substation in some other way, the symbol doesn't disappear.
$endgroup$
– Kaz
Jun 30 at 22:33
1
1
$begingroup$
@inf3rno Indeed, that is the notation which is implemented in computer programs like Mathematica, where the notation is
D[f,x]. But we are not doing programming here, we're doing math, and when people do math they like to use notation which provides conceptual clarity. In this case the $dx$ represents an "infinitesimal", in the sense that $df/dx$ denotes the limit $lim_Delta xto0(Delta f/Delta x)$. It is very helpful to think of differentials (the $dx$s and $df$s) in this way, having a life of their own, as opposed to always being a part of a derivative or an integral.$endgroup$
– YiFan
Jun 30 at 8:31
$begingroup$
@inf3rno Indeed, that is the notation which is implemented in computer programs like Mathematica, where the notation is
D[f,x]. But we are not doing programming here, we're doing math, and when people do math they like to use notation which provides conceptual clarity. In this case the $dx$ represents an "infinitesimal", in the sense that $df/dx$ denotes the limit $lim_Delta xto0(Delta f/Delta x)$. It is very helpful to think of differentials (the $dx$s and $df$s) in this way, having a life of their own, as opposed to always being a part of a derivative or an integral.$endgroup$
– YiFan
Jun 30 at 8:31
1
1
$begingroup$
@inf3rno Although, you might be pleasantly surprised to learn that there is a notation that uses $Df$ to represent the derivative of $f$. In this way then $D^nf$ denotes the $n$th derivative of $f$. The notation $(Df)(3)$, or sometimes even $Df(3)$, denotes the function $Df(x)$ evaluated at $x=3$. But this is not as common as the other two notations for the reasons outlined above.
$endgroup$
– YiFan
Jun 30 at 8:36
$begingroup$
@inf3rno Although, you might be pleasantly surprised to learn that there is a notation that uses $Df$ to represent the derivative of $f$. In this way then $D^nf$ denotes the $n$th derivative of $f$. The notation $(Df)(3)$, or sometimes even $Df(3)$, denotes the function $Df(x)$ evaluated at $x=3$. But this is not as common as the other two notations for the reasons outlined above.
$endgroup$
– YiFan
Jun 30 at 8:36
1
1
$begingroup$
I would solve the nth derivative with recursion instead of having a separate denotation.
r(d,n)(f,x) would return the nth derivative of f. Ofc. that is programming for math a separate notation works too. :-)$endgroup$
– inf3rno
Jun 30 at 9:10
$begingroup$
I would solve the nth derivative with recursion instead of having a separate denotation.
r(d,n)(f,x) would return the nth derivative of f. Ofc. that is programming for math a separate notation works too. :-)$endgroup$
– inf3rno
Jun 30 at 9:10
2
2
$begingroup$
I feel like the notation $fracdfdx(3)$, or at least its partial derivative equivalent $fracpartial fpartial x(3,3)$, is fairly common. It's used in this textbook, for instance.
$endgroup$
– Misha Lavrov
Jun 30 at 18:28
$begingroup$
I feel like the notation $fracdfdx(3)$, or at least its partial derivative equivalent $fracpartial fpartial x(3,3)$, is fairly common. It's used in this textbook, for instance.
$endgroup$
– Misha Lavrov
Jun 30 at 18:28
1
1
$begingroup$
$fracddxf(xleftarrow3)$ Problem solved. It's just a syntactic sugar; the d/dx notation depends on a relationship to the parameter symbols, so it becomes unclear when a symbol disappears via substitution; but if we notate the substation in some other way, the symbol doesn't disappear.
$endgroup$
– Kaz
Jun 30 at 22:33
$begingroup$
$fracddxf(xleftarrow3)$ Problem solved. It's just a syntactic sugar; the d/dx notation depends on a relationship to the parameter symbols, so it becomes unclear when a symbol disappears via substitution; but if we notate the substation in some other way, the symbol doesn't disappear.
$endgroup$
– Kaz
Jun 30 at 22:33
|
show 8 more comments
$begingroup$
All of those not on the ends are equal, as they all denote the same thing. The last one, I've never seen. The first one is "close" - I'm going to get to that.
However, there is actually a very good question you ask here - namely, if we have
$$fracdfdx$$
how can we write, with this notation, what the value at $x = 3$ is? And the "best" answer to this is that we have to make a clear distinction between three different things:
- the meaning of the symbol "$f(x)$",
- the meaning of the symbol "$f$", and
- what kind of "thing", "differentiation", is.
And this stuff typically isn't touched on as well as it could be, and when it needs to be. The symbol $f(x)$ is not a function, even though it often gets abused by calling it such. Rather, what it is is the value of the function at $x$, and if $x$ is a variable that hasn't been assigned a value yet, this value is also unspecified (except that it must be within the range of the function). That is, $f(x)$ is a symbol for a (possibly unspecified) number.
What $f$ is, on the other hand, is the function itself: the mathematical object that you can plug numbers into to get other numbers out. Thus, you should, ideally, never say "the function $f(x)$", when what you really mean is "the function $f$", according to the above understanding.
Now, for the last part. What "differentation" is, is what is called an operator, or a higher-order function: it is a function that takes in other functions as inputs and pops out functions. Such things are typically denoted by prefixing their symbol against the function symbol: hence here, you should have $fracddx f$, which is shortened to $fracdfdx$, and this forms a combined function symbol.
Hence, if you want the value at $x = 3$, say, I'd argue you should "best" write
$$left[fracdfdxright](3)$$
which means to evaluate the function denoted by the symbol $fracdfdx$ at 3. And this, thus, generalizes likewise to higher derivatives, e.g.
$$left[fracd^2 fdx^2right](3)$$
and so forth.
Of course, if you think about this even more, technically we should not ever write something like
$$fracdfdx + x$$
even though this is very common, e.g. in writing a differential equation, and you can "get away with it" in that people will generally know what you mean. This is because here, $x$ is a number, and $fracdfdx$ is a function. To be "absolutely" correct, you will have two choices: either demote $fracdfdx$ to a number, which will make it
$$left[fracdfdxright](x) + x$$
or, to promote $x$ to a function, in which case, if you don't want to give another function symbol, you have to use the anonymous function notation
$$fracdfdx + left(x mapsto xright)$$
Either of these would be a fully-correct mathematical expression (though keep in mind, they actually have a different, but related, meaning each! What is it?) But since they are a bit ungainly, we accept the "wrong" notation as a shorthand: it's called "abuse of notation". And there isn't anything wrong with that, as long as you clearly understand what is going on, which you can check by trying to mentally translate it into the strictly proper, if not typically used, notation. Just like learning a foreign human natural language, where that in typical speech we'll never always follow the rules of grammar perfectly, but we always try to learn the ideal forms first, when learning the mathematical language we should keep in mind the same principle.
answered Jun 30 at 23:53
The_SympathizerThe_Sympathizer
10k2 gold badges29 silver badges51 bronze badges
$endgroup$
$begingroup$
You could also write $$fracmathrmdfmathrmdx + I qquadtextorqquad fracmathrmdfmathrmdx + operatornameid, $$ if you don't want to deal with the notation $xmapsto x$ in the middle of an expression.
$endgroup$
– Xander Henderson
Jul 1 at 0:03
$begingroup$
For me the $fracd^2dx^2f(3)$ is just as clear as the $left[fracd^2dx^2fright](3)$ because the order of the operators can matter in math. What I find the real problem here, that people can confuse $fracd^2 f(3)dx^2$ even to $fracd^2 cdot f(3)d cdot x^2$if we use sugar syntax like $fracd^2 fdx^2$ instead of $fracd^2dx^2f$. Even the $fracd^2dx^2f$ is not the best, but at least it gives a hint of what we are talking about.
$endgroup$
– inf3rno
Jul 1 at 6:11
$begingroup$
Maybe my question is not perfect, sometimes I just feel that something is wrong, but I cannot explain why until I think about it more. It's hard to write a good question. I like how you explained the difference between $f$ and $f(x)$ in your answer. I think being strict with notations is important, because it makes easier to understand what we are doing. It is just like coding in team, you write quality code, because others need to understand it and because if you read your own code again after a few months you might end up scratching your head too. It actually happened to me a few times. :D
$endgroup$
– inf3rno
Jul 1 at 6:13
1
$begingroup$
@inf3rno : Yes, you can also use $fracddx f(x)$ to reference the value of the derivative at the point $x$, as the operator binds more tightly to the function symbol, than to the evaluation. Also, I would not suggest to write $fracd^2 f(3)dx^2$ - either $fracd^2 fdx(3)$, or $fracd^2dx f(3)$, or as given with the brackets.
$endgroup$
– The_Sympathizer
Jul 1 at 7:46
1
$begingroup$
And I find it interesting you mention code. I also think that Maths has a fair bit it can learn from computer programming. One of my professors said that programming ideally tries to be like Maths, but I've also come to think that there's a lot of merit in things going the other way, too.
$endgroup$
– The_Sympathizer
Jul 1 at 7:47
add a comment |
$begingroup$
All of those not on the ends are equal, as they all denote the same thing. The last one, I've never seen. The first one is "close" - I'm going to get to that.
However, there is actually a very good question you ask here - namely, if we have
$$fracdfdx$$
how can we write, with this notation, what the value at $x = 3$ is? And the "best" answer to this is that we have to make a clear distinction between three different things:
- the meaning of the symbol "$f(x)$",
- the meaning of the symbol "$f$", and
- what kind of "thing", "differentiation", is.
And this stuff typically isn't touched on as well as it could be, and when it needs to be. The symbol $f(x)$ is not a function, even though it often gets abused by calling it such. Rather, what it is is the value of the function at $x$, and if $x$ is a variable that hasn't been assigned a value yet, this value is also unspecified (except that it must be within the range of the function). That is, $f(x)$ is a symbol for a (possibly unspecified) number.
What $f$ is, on the other hand, is the function itself: the mathematical object that you can plug numbers into to get other numbers out. Thus, you should, ideally, never say "the function $f(x)$", when what you really mean is "the function $f$", according to the above understanding.
Now, for the last part. What "differentation" is, is what is called an operator, or a higher-order function: it is a function that takes in other functions as inputs and pops out functions. Such things are typically denoted by prefixing their symbol against the function symbol: hence here, you should have $fracddx f$, which is shortened to $fracdfdx$, and this forms a combined function symbol.
Hence, if you want the value at $x = 3$, say, I'd argue you should "best" write
$$left[fracdfdxright](3)$$
which means to evaluate the function denoted by the symbol $fracdfdx$ at 3. And this, thus, generalizes likewise to higher derivatives, e.g.
$$left[fracd^2 fdx^2right](3)$$
and so forth.
Of course, if you think about this even more, technically we should not ever write something like
$$fracdfdx + x$$
even though this is very common, e.g. in writing a differential equation, and you can "get away with it" in that people will generally know what you mean. This is because here, $x$ is a number, and $fracdfdx$ is a function. To be "absolutely" correct, you will have two choices: either demote $fracdfdx$ to a number, which will make it
$$left[fracdfdxright](x) + x$$
or, to promote $x$ to a function, in which case, if you don't want to give another function symbol, you have to use the anonymous function notation
$$fracdfdx + left(x mapsto xright)$$
Either of these would be a fully-correct mathematical expression (though keep in mind, they actually have a different, but related, meaning each! What is it?) But since they are a bit ungainly, we accept the "wrong" notation as a shorthand: it's called "abuse of notation". And there isn't anything wrong with that, as long as you clearly understand what is going on, which you can check by trying to mentally translate it into the strictly proper, if not typically used, notation. Just like learning a foreign human natural language, where that in typical speech we'll never always follow the rules of grammar perfectly, but we always try to learn the ideal forms first, when learning the mathematical language we should keep in mind the same principle.
answered Jun 30 at 23:53
The_SympathizerThe_Sympathizer
10k2 gold badges29 silver badges51 bronze badges
$endgroup$
$begingroup$
You could also write $$fracmathrmdfmathrmdx + I qquadtextorqquad fracmathrmdfmathrmdx + operatornameid, $$ if you don't want to deal with the notation $xmapsto x$ in the middle of an expression.
$endgroup$
– Xander Henderson
Jul 1 at 0:03
$begingroup$
For me the $fracd^2dx^2f(3)$ is just as clear as the $left[fracd^2dx^2fright](3)$ because the order of the operators can matter in math. What I find the real problem here, that people can confuse $fracd^2 f(3)dx^2$ even to $fracd^2 cdot f(3)d cdot x^2$if we use sugar syntax like $fracd^2 fdx^2$ instead of $fracd^2dx^2f$. Even the $fracd^2dx^2f$ is not the best, but at least it gives a hint of what we are talking about.
$endgroup$
– inf3rno
Jul 1 at 6:11
$begingroup$
Maybe my question is not perfect, sometimes I just feel that something is wrong, but I cannot explain why until I think about it more. It's hard to write a good question. I like how you explained the difference between $f$ and $f(x)$ in your answer. I think being strict with notations is important, because it makes easier to understand what we are doing. It is just like coding in team, you write quality code, because others need to understand it and because if you read your own code again after a few months you might end up scratching your head too. It actually happened to me a few times. :D
$endgroup$
– inf3rno
Jul 1 at 6:13
1
$begingroup$
@inf3rno : Yes, you can also use $fracddx f(x)$ to reference the value of the derivative at the point $x$, as the operator binds more tightly to the function symbol, than to the evaluation. Also, I would not suggest to write $fracd^2 f(3)dx^2$ - either $fracd^2 fdx(3)$, or $fracd^2dx f(3)$, or as given with the brackets.
$endgroup$
– The_Sympathizer
Jul 1 at 7:46
1
$begingroup$
And I find it interesting you mention code. I also think that Maths has a fair bit it can learn from computer programming. One of my professors said that programming ideally tries to be like Maths, but I've also come to think that there's a lot of merit in things going the other way, too.
$endgroup$
– The_Sympathizer
Jul 1 at 7:47
add a comment |
$begingroup$
All of those not on the ends are equal, as they all denote the same thing. The last one, I've never seen. The first one is "close" - I'm going to get to that.
However, there is actually a very good question you ask here - namely, if we have
$$fracdfdx$$
how can we write, with this notation, what the value at $x = 3$ is? And the "best" answer to this is that we have to make a clear distinction between three different things:
- the meaning of the symbol "$f(x)$",
- the meaning of the symbol "$f$", and
- what kind of "thing", "differentiation", is.
And this stuff typically isn't touched on as well as it could be, and when it needs to be. The symbol $f(x)$ is not a function, even though it often gets abused by calling it such. Rather, what it is is the value of the function at $x$, and if $x$ is a variable that hasn't been assigned a value yet, this value is also unspecified (except that it must be within the range of the function). That is, $f(x)$ is a symbol for a (possibly unspecified) number.
What $f$ is, on the other hand, is the function itself: the mathematical object that you can plug numbers into to get other numbers out. Thus, you should, ideally, never say "the function $f(x)$", when what you really mean is "the function $f$", according to the above understanding.
Now, for the last part. What "differentation" is, is what is called an operator, or a higher-order function: it is a function that takes in other functions as inputs and pops out functions. Such things are typically denoted by prefixing their symbol against the function symbol: hence here, you should have $fracddx f$, which is shortened to $fracdfdx$, and this forms a combined function symbol.
Hence, if you want the value at $x = 3$, say, I'd argue you should "best" write
$$left[fracdfdxright](3)$$
which means to evaluate the function denoted by the symbol $fracdfdx$ at 3. And this, thus, generalizes likewise to higher derivatives, e.g.
$$left[fracd^2 fdx^2right](3)$$
and so forth.
Of course, if you think about this even more, technically we should not ever write something like
$$fracdfdx + x$$
even though this is very common, e.g. in writing a differential equation, and you can "get away with it" in that people will generally know what you mean. This is because here, $x$ is a number, and $fracdfdx$ is a function. To be "absolutely" correct, you will have two choices: either demote $fracdfdx$ to a number, which will make it
$$left[fracdfdxright](x) + x$$
or, to promote $x$ to a function, in which case, if you don't want to give another function symbol, you have to use the anonymous function notation
$$fracdfdx + left(x mapsto xright)$$
Either of these would be a fully-correct mathematical expression (though keep in mind, they actually have a different, but related, meaning each! What is it?) But since they are a bit ungainly, we accept the "wrong" notation as a shorthand: it's called "abuse of notation". And there isn't anything wrong with that, as long as you clearly understand what is going on, which you can check by trying to mentally translate it into the strictly proper, if not typically used, notation. Just like learning a foreign human natural language, where that in typical speech we'll never always follow the rules of grammar perfectly, but we always try to learn the ideal forms first, when learning the mathematical language we should keep in mind the same principle.
answered Jun 30 at 23:53
The_SympathizerThe_Sympathizer
10k2 gold badges29 silver badges51 bronze badges
$endgroup$
All of those not on the ends are equal, as they all denote the same thing. The last one, I've never seen. The first one is "close" - I'm going to get to that.
However, there is actually a very good question you ask here - namely, if we have
$$fracdfdx$$
how can we write, with this notation, what the value at $x = 3$ is? And the "best" answer to this is that we have to make a clear distinction between three different things:
- the meaning of the symbol "$f(x)$",
- the meaning of the symbol "$f$", and
- what kind of "thing", "differentiation", is.
And this stuff typically isn't touched on as well as it could be, and when it needs to be. The symbol $f(x)$ is not a function, even though it often gets abused by calling it such. Rather, what it is is the value of the function at $x$, and if $x$ is a variable that hasn't been assigned a value yet, this value is also unspecified (except that it must be within the range of the function). That is, $f(x)$ is a symbol for a (possibly unspecified) number.
What $f$ is, on the other hand, is the function itself: the mathematical object that you can plug numbers into to get other numbers out. Thus, you should, ideally, never say "the function $f(x)$", when what you really mean is "the function $f$", according to the above understanding.
Now, for the last part. What "differentation" is, is what is called an operator, or a higher-order function: it is a function that takes in other functions as inputs and pops out functions. Such things are typically denoted by prefixing their symbol against the function symbol: hence here, you should have $fracddx f$, which is shortened to $fracdfdx$, and this forms a combined function symbol.
Hence, if you want the value at $x = 3$, say, I'd argue you should "best" write
$$left[fracdfdxright](3)$$
which means to evaluate the function denoted by the symbol $fracdfdx$ at 3. And this, thus, generalizes likewise to higher derivatives, e.g.
$$left[fracd^2 fdx^2right](3)$$
and so forth.
Of course, if you think about this even more, technically we should not ever write something like
$$fracdfdx + x$$
even though this is very common, e.g. in writing a differential equation, and you can "get away with it" in that people will generally know what you mean. This is because here, $x$ is a number, and $fracdfdx$ is a function. To be "absolutely" correct, you will have two choices: either demote $fracdfdx$ to a number, which will make it
$$left[fracdfdxright](x) + x$$
or, to promote $x$ to a function, in which case, if you don't want to give another function symbol, you have to use the anonymous function notation
$$fracdfdx + left(x mapsto xright)$$
Either of these would be a fully-correct mathematical expression (though keep in mind, they actually have a different, but related, meaning each! What is it?) But since they are a bit ungainly, we accept the "wrong" notation as a shorthand: it's called "abuse of notation". And there isn't anything wrong with that, as long as you clearly understand what is going on, which you can check by trying to mentally translate it into the strictly proper, if not typically used, notation. Just like learning a foreign human natural language, where that in typical speech we'll never always follow the rules of grammar perfectly, but we always try to learn the ideal forms first, when learning the mathematical language we should keep in mind the same principle.
answered Jun 30 at 23:53
The_SympathizerThe_Sympathizer
10k2 gold badges29 silver badges51 bronze badges
edited Jun 30 at 23:59
answered Jun 30 at 23:53
The_SympathizerThe_Sympathizer
10k2 gold badges29 silver badges51 bronze badges
answered Jun 30 at 23:53
The_SympathizerThe_Sympathizer
10k2 gold badges29 silver badges51 bronze badges
answered Jun 30 at 23:53
The_SympathizerThe_Sympathizer
10k2 gold badges29 silver badges51 bronze badges
10k2 gold badges29 silver badges51 bronze badges
$begingroup$
You could also write $$fracmathrmdfmathrmdx + I qquadtextorqquad fracmathrmdfmathrmdx + operatornameid, $$ if you don't want to deal with the notation $xmapsto x$ in the middle of an expression.
$endgroup$
– Xander Henderson
Jul 1 at 0:03
$begingroup$
For me the $fracd^2dx^2f(3)$ is just as clear as the $left[fracd^2dx^2fright](3)$ because the order of the operators can matter in math. What I find the real problem here, that people can confuse $fracd^2 f(3)dx^2$ even to $fracd^2 cdot f(3)d cdot x^2$if we use sugar syntax like $fracd^2 fdx^2$ instead of $fracd^2dx^2f$. Even the $fracd^2dx^2f$ is not the best, but at least it gives a hint of what we are talking about.
$endgroup$
– inf3rno
Jul 1 at 6:11
$begingroup$
Maybe my question is not perfect, sometimes I just feel that something is wrong, but I cannot explain why until I think about it more. It's hard to write a good question. I like how you explained the difference between $f$ and $f(x)$ in your answer. I think being strict with notations is important, because it makes easier to understand what we are doing. It is just like coding in team, you write quality code, because others need to understand it and because if you read your own code again after a few months you might end up scratching your head too. It actually happened to me a few times. :D
$endgroup$
– inf3rno
Jul 1 at 6:13
1
$begingroup$
@inf3rno : Yes, you can also use $fracddx f(x)$ to reference the value of the derivative at the point $x$, as the operator binds more tightly to the function symbol, than to the evaluation. Also, I would not suggest to write $fracd^2 f(3)dx^2$ - either $fracd^2 fdx(3)$, or $fracd^2dx f(3)$, or as given with the brackets.
$endgroup$
– The_Sympathizer
Jul 1 at 7:46
1
$begingroup$
And I find it interesting you mention code. I also think that Maths has a fair bit it can learn from computer programming. One of my professors said that programming ideally tries to be like Maths, but I've also come to think that there's a lot of merit in things going the other way, too.
$endgroup$
– The_Sympathizer
Jul 1 at 7:47
add a comment |
$begingroup$
You could also write $$fracmathrmdfmathrmdx + I qquadtextorqquad fracmathrmdfmathrmdx + operatornameid, $$ if you don't want to deal with the notation $xmapsto x$ in the middle of an expression.
$endgroup$
– Xander Henderson
Jul 1 at 0:03
$begingroup$
For me the $fracd^2dx^2f(3)$ is just as clear as the $left[fracd^2dx^2fright](3)$ because the order of the operators can matter in math. What I find the real problem here, that people can confuse $fracd^2 f(3)dx^2$ even to $fracd^2 cdot f(3)d cdot x^2$if we use sugar syntax like $fracd^2 fdx^2$ instead of $fracd^2dx^2f$. Even the $fracd^2dx^2f$ is not the best, but at least it gives a hint of what we are talking about.
$endgroup$
– inf3rno
Jul 1 at 6:11
$begingroup$
Maybe my question is not perfect, sometimes I just feel that something is wrong, but I cannot explain why until I think about it more. It's hard to write a good question. I like how you explained the difference between $f$ and $f(x)$ in your answer. I think being strict with notations is important, because it makes easier to understand what we are doing. It is just like coding in team, you write quality code, because others need to understand it and because if you read your own code again after a few months you might end up scratching your head too. It actually happened to me a few times. :D
$endgroup$
– inf3rno
Jul 1 at 6:13
1
$begingroup$
@inf3rno : Yes, you can also use $fracddx f(x)$ to reference the value of the derivative at the point $x$, as the operator binds more tightly to the function symbol, than to the evaluation. Also, I would not suggest to write $fracd^2 f(3)dx^2$ - either $fracd^2 fdx(3)$, or $fracd^2dx f(3)$, or as given with the brackets.
$endgroup$
– The_Sympathizer
Jul 1 at 7:46
1
$begingroup$
And I find it interesting you mention code. I also think that Maths has a fair bit it can learn from computer programming. One of my professors said that programming ideally tries to be like Maths, but I've also come to think that there's a lot of merit in things going the other way, too.
$endgroup$
– The_Sympathizer
Jul 1 at 7:47
$begingroup$
You could also write $$fracmathrmdfmathrmdx + I qquadtextorqquad fracmathrmdfmathrmdx + operatornameid, $$ if you don't want to deal with the notation $xmapsto x$ in the middle of an expression.
$endgroup$
– Xander Henderson
Jul 1 at 0:03
$begingroup$
You could also write $$fracmathrmdfmathrmdx + I qquadtextorqquad fracmathrmdfmathrmdx + operatornameid, $$ if you don't want to deal with the notation $xmapsto x$ in the middle of an expression.
$endgroup$
– Xander Henderson
Jul 1 at 0:03
$begingroup$
For me the $fracd^2dx^2f(3)$ is just as clear as the $left[fracd^2dx^2fright](3)$ because the order of the operators can matter in math. What I find the real problem here, that people can confuse $fracd^2 f(3)dx^2$ even to $fracd^2 cdot f(3)d cdot x^2$if we use sugar syntax like $fracd^2 fdx^2$ instead of $fracd^2dx^2f$. Even the $fracd^2dx^2f$ is not the best, but at least it gives a hint of what we are talking about.
$endgroup$
– inf3rno
Jul 1 at 6:11
$begingroup$
For me the $fracd^2dx^2f(3)$ is just as clear as the $left[fracd^2dx^2fright](3)$ because the order of the operators can matter in math. What I find the real problem here, that people can confuse $fracd^2 f(3)dx^2$ even to $fracd^2 cdot f(3)d cdot x^2$if we use sugar syntax like $fracd^2 fdx^2$ instead of $fracd^2dx^2f$. Even the $fracd^2dx^2f$ is not the best, but at least it gives a hint of what we are talking about.
$endgroup$
– inf3rno
Jul 1 at 6:11
$begingroup$
Maybe my question is not perfect, sometimes I just feel that something is wrong, but I cannot explain why until I think about it more. It's hard to write a good question. I like how you explained the difference between $f$ and $f(x)$ in your answer. I think being strict with notations is important, because it makes easier to understand what we are doing. It is just like coding in team, you write quality code, because others need to understand it and because if you read your own code again after a few months you might end up scratching your head too. It actually happened to me a few times. :D
$endgroup$
– inf3rno
Jul 1 at 6:13
$begingroup$
Maybe my question is not perfect, sometimes I just feel that something is wrong, but I cannot explain why until I think about it more. It's hard to write a good question. I like how you explained the difference between $f$ and $f(x)$ in your answer. I think being strict with notations is important, because it makes easier to understand what we are doing. It is just like coding in team, you write quality code, because others need to understand it and because if you read your own code again after a few months you might end up scratching your head too. It actually happened to me a few times. :D
$endgroup$
– inf3rno
Jul 1 at 6:13
1
1
$begingroup$
@inf3rno : Yes, you can also use $fracddx f(x)$ to reference the value of the derivative at the point $x$, as the operator binds more tightly to the function symbol, than to the evaluation. Also, I would not suggest to write $fracd^2 f(3)dx^2$ - either $fracd^2 fdx(3)$, or $fracd^2dx f(3)$, or as given with the brackets.
$endgroup$
– The_Sympathizer
Jul 1 at 7:46
$begingroup$
@inf3rno : Yes, you can also use $fracddx f(x)$ to reference the value of the derivative at the point $x$, as the operator binds more tightly to the function symbol, than to the evaluation. Also, I would not suggest to write $fracd^2 f(3)dx^2$ - either $fracd^2 fdx(3)$, or $fracd^2dx f(3)$, or as given with the brackets.
$endgroup$
– The_Sympathizer
Jul 1 at 7:46
1
1
$begingroup$
And I find it interesting you mention code. I also think that Maths has a fair bit it can learn from computer programming. One of my professors said that programming ideally tries to be like Maths, but I've also come to think that there's a lot of merit in things going the other way, too.
$endgroup$
– The_Sympathizer
Jul 1 at 7:47
$begingroup$
And I find it interesting you mention code. I also think that Maths has a fair bit it can learn from computer programming. One of my professors said that programming ideally tries to be like Maths, but I've also come to think that there's a lot of merit in things going the other way, too.
$endgroup$
– The_Sympathizer
Jul 1 at 7:47
add a comment |
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$begingroup$
The correct answer has already been given but note that $fracdf(3)dx=0$ since you're differentiating a constant. And $fracdf(3)d3$ is meaningless.
$endgroup$
– Deepak
Jun 30 at 6:41
4
$begingroup$
The last two examples are really one. It's not
d/dxand[f(x)]butd/dx [f(x)]- I don't know whether that was a transcription error (in which case please edit) or you are genuinely thinking square bracket notation indicates differentiation.$endgroup$
– abligh
Jun 30 at 14:42
1
$begingroup$
Referring to the linked page, the example
d/dx [f(x)]really is one example, not two, and it suffers from the fact that the writer of that page did not know how to typeset the mathematics in the usual way (or did not think it worth the trouble). I don't think I have ever seen $d/dx, [f(x)]$ in a regular textbook, but I have seen $fracddx[f(x)].$$endgroup$
– David K
Jun 30 at 15:53
1
$begingroup$
For a function of only one variable, right? Because once you get into a function of several variables, you can't use $f'$ notation, you either have to use subscript $f_x, f_y, f_xx, f_xy,...$ or partial operators $partial/partial x, partial/partial y, partial^2/partial x^2 ...$ or come up with a vector formulation, or something else.
$endgroup$
– smci
Jun 30 at 19:22
$begingroup$
@abligh Thanks, it is just a typo. I did not think much about what I am copy pasting. :-)
$endgroup$
– inf3rno
Jul 1 at 6:24