Notation for the derivative of a function: $f'$ or $f'(x);$?












3












$begingroup$


The derivative of a function is often defined as $f'$ and $f'(x)$. So which one is it? $f'(x)$ is the output of the function $f'$, so why do I see people using $f'$ and $f'(x)$ interchangeably to refer to the derivative of a function?






calculus derivatives notation







share|cite|improve this question











$endgroup$












  • $begingroup$
    By the way, check the first bullet point here $ddot smile$
    $endgroup$
    – Git Gud
    Nov 26 '18 at 9:53










  • $begingroup$
    $f'$ is the function (therefore a map), $f'(x)$ is the value of the function in the point $x$ (therefore a number). Yes, people mix them up all the times.
    $endgroup$
    – gented
    Nov 26 '18 at 11:01








  • 1




    $begingroup$
    You are correct that $f$ is the function and $f(x)$ is the value of the function when evaluated at a point $x$ in its domain (ditto $f'$ and $f'(x)$). Many elementary texts blur this distinction in an attempt to "dumb down" the material. This causes no end of confusion later on, and you have done well to note the problem.
    $endgroup$
    – Xander Henderson
    Nov 26 '18 at 13:20










  • $begingroup$
    @J.Smith In case my answer would be deleted I let here the main reference I've found on that topic Calculus for Dummies
    $endgroup$
    – gimusi
    Nov 26 '18 at 14:36
















3












$begingroup$


The derivative of a function is often defined as $f'$ and $f'(x)$. So which one is it? $f'(x)$ is the output of the function $f'$, so why do I see people using $f'$ and $f'(x)$ interchangeably to refer to the derivative of a function?






calculus derivatives notation







share|cite|improve this question











$endgroup$












  • $begingroup$
    By the way, check the first bullet point here $ddot smile$
    $endgroup$
    – Git Gud
    Nov 26 '18 at 9:53










  • $begingroup$
    $f'$ is the function (therefore a map), $f'(x)$ is the value of the function in the point $x$ (therefore a number). Yes, people mix them up all the times.
    $endgroup$
    – gented
    Nov 26 '18 at 11:01








  • 1




    $begingroup$
    You are correct that $f$ is the function and $f(x)$ is the value of the function when evaluated at a point $x$ in its domain (ditto $f'$ and $f'(x)$). Many elementary texts blur this distinction in an attempt to "dumb down" the material. This causes no end of confusion later on, and you have done well to note the problem.
    $endgroup$
    – Xander Henderson
    Nov 26 '18 at 13:20










  • $begingroup$
    @J.Smith In case my answer would be deleted I let here the main reference I've found on that topic Calculus for Dummies
    $endgroup$
    – gimusi
    Nov 26 '18 at 14:36














3












3








3


0



$begingroup$


The derivative of a function is often defined as $f'$ and $f'(x)$. So which one is it? $f'(x)$ is the output of the function $f'$, so why do I see people using $f'$ and $f'(x)$ interchangeably to refer to the derivative of a function?






calculus derivatives notation







share|cite|improve this question











$endgroup$




The derivative of a function is often defined as $f'$ and $f'(x)$. So which one is it? $f'(x)$ is the output of the function $f'$, so why do I see people using $f'$ and $f'(x)$ interchangeably to refer to the derivative of a function?






calculus derivatives notation




calculus derivatives notation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 '18 at 12:12









amWhy

1




1










asked Nov 26 '18 at 8:25









J. SmithJ. Smith

241




241












  • $begingroup$
    By the way, check the first bullet point here $ddot smile$
    $endgroup$
    – Git Gud
    Nov 26 '18 at 9:53










  • $begingroup$
    $f'$ is the function (therefore a map), $f'(x)$ is the value of the function in the point $x$ (therefore a number). Yes, people mix them up all the times.
    $endgroup$
    – gented
    Nov 26 '18 at 11:01








  • 1




    $begingroup$
    You are correct that $f$ is the function and $f(x)$ is the value of the function when evaluated at a point $x$ in its domain (ditto $f'$ and $f'(x)$). Many elementary texts blur this distinction in an attempt to "dumb down" the material. This causes no end of confusion later on, and you have done well to note the problem.
    $endgroup$
    – Xander Henderson
    Nov 26 '18 at 13:20










  • $begingroup$
    @J.Smith In case my answer would be deleted I let here the main reference I've found on that topic Calculus for Dummies
    $endgroup$
    – gimusi
    Nov 26 '18 at 14:36


















  • $begingroup$
    By the way, check the first bullet point here $ddot smile$
    $endgroup$
    – Git Gud
    Nov 26 '18 at 9:53










  • $begingroup$
    $f'$ is the function (therefore a map), $f'(x)$ is the value of the function in the point $x$ (therefore a number). Yes, people mix them up all the times.
    $endgroup$
    – gented
    Nov 26 '18 at 11:01








  • 1




    $begingroup$
    You are correct that $f$ is the function and $f(x)$ is the value of the function when evaluated at a point $x$ in its domain (ditto $f'$ and $f'(x)$). Many elementary texts blur this distinction in an attempt to "dumb down" the material. This causes no end of confusion later on, and you have done well to note the problem.
    $endgroup$
    – Xander Henderson
    Nov 26 '18 at 13:20










  • $begingroup$
    @J.Smith In case my answer would be deleted I let here the main reference I've found on that topic Calculus for Dummies
    $endgroup$
    – gimusi
    Nov 26 '18 at 14:36
















$begingroup$
By the way, check the first bullet point here $ddot smile$
$endgroup$
– Git Gud
Nov 26 '18 at 9:53




$begingroup$
By the way, check the first bullet point here $ddot smile$
$endgroup$
– Git Gud
Nov 26 '18 at 9:53












$begingroup$
$f'$ is the function (therefore a map), $f'(x)$ is the value of the function in the point $x$ (therefore a number). Yes, people mix them up all the times.
$endgroup$
– gented
Nov 26 '18 at 11:01






$begingroup$
$f'$ is the function (therefore a map), $f'(x)$ is the value of the function in the point $x$ (therefore a number). Yes, people mix them up all the times.
$endgroup$
– gented
Nov 26 '18 at 11:01






1




1




$begingroup$
You are correct that $f$ is the function and $f(x)$ is the value of the function when evaluated at a point $x$ in its domain (ditto $f'$ and $f'(x)$). Many elementary texts blur this distinction in an attempt to "dumb down" the material. This causes no end of confusion later on, and you have done well to note the problem.
$endgroup$
– Xander Henderson
Nov 26 '18 at 13:20




$begingroup$
You are correct that $f$ is the function and $f(x)$ is the value of the function when evaluated at a point $x$ in its domain (ditto $f'$ and $f'(x)$). Many elementary texts blur this distinction in an attempt to "dumb down" the material. This causes no end of confusion later on, and you have done well to note the problem.
$endgroup$
– Xander Henderson
Nov 26 '18 at 13:20












$begingroup$
@J.Smith In case my answer would be deleted I let here the main reference I've found on that topic Calculus for Dummies
$endgroup$
– gimusi
Nov 26 '18 at 14:36




$begingroup$
@J.Smith In case my answer would be deleted I let here the main reference I've found on that topic Calculus for Dummies
$endgroup$
– gimusi
Nov 26 '18 at 14:36










4 Answers
4






active

oldest

votes


















10












$begingroup$

By definition a function is a triple $(f,D,C)$, which is very often denoted by $f colon D to C$, where $C,D$ are two sets and $f$ associates to each element of $D$ one and only one element of $C$.



So when it is clear what $C$ and $D$ are, or in cases where it is not possible or not necessary to write them down, you just write $f$. The expression $f(x)$ denotes the element in $C$ which $x in D$ is mapped to. So $f$ is a function, $f(x)$ is an element of $C$, two completely different things.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Nice answer! (+1)
    $endgroup$
    – Robert Z
    Nov 26 '18 at 8:52










  • $begingroup$
    Thank you @RobertZ.
    $endgroup$
    – Gibbs
    Nov 26 '18 at 8:59






  • 2




    $begingroup$
    I agree with the forst part (+1) but I would be more relaxed for the second part, I think that using f(x) to indicate the function can be tolerated.
    $endgroup$
    – gimusi
    Nov 26 '18 at 9:22






  • 1




    $begingroup$
    I think it is a good habit for a beginner to stick to the definitions. After some practice and experience one becomes conscious on where conventions can be relaxed and some abuse of language might be ok.
    $endgroup$
    – Gibbs
    Nov 26 '18 at 10:06










  • $begingroup$
    The distinction between $f$ and $f(x) $ is important but "let $f(x) =xsin(1/x)$" has almost become a routine shorthand for "let $f:mathbb{R} setminus{0}tomathbb {R} $ be a function defined by $f(x) =xsin(1/x)$".
    $endgroup$
    – Paramanand Singh
    Nov 27 '18 at 5:05





















3












$begingroup$

$f$ denotes the function and $f(x)$ the output of the function when evaluated at $x$.



This convention does not differ for the derivative.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    f’ denotes the derivative and f’(x) denotes the output of the derivative, which is the instantaneous rate of change of a function at any point. Correct?
    $endgroup$
    – J. Smith
    Nov 26 '18 at 10:08












  • $begingroup$
    @J.Smith: you get it.
    $endgroup$
    – Yves Daoust
    Nov 26 '18 at 10:13










  • $begingroup$
    @YvesDaoust I think we could be more "relaxed" with that definition and notation, no one will be wound considering the function $f(x)$ :)
    $endgroup$
    – gimusi
    Nov 26 '18 at 10:17






  • 2




    $begingroup$
    @gimusi: this would reduce the expressive power (not possible to distinguish the function and the value) and create ambiguities. So, no.
    $endgroup$
    – Yves Daoust
    Nov 26 '18 at 10:20










  • $begingroup$
    @YvesDaoust Also f creates ambiguity if we do not specify the domain and the codomain, therefore anytime we refer to a function we should use $f:Ato B$. I don't thing it would be a useful notation. $f(x)$ to indicate the funtion can be used many times without any ambiguity. Of course we ca suggest to do not use that but it is a matter of preferences and not a law.
    $endgroup$
    – gimusi
    Nov 26 '18 at 10:23



















2












$begingroup$

I would read $f'(x)$ as "the function $f'$ applied to the element $x$ of the domain". This gives us a a new element in the range. Meanwhile I read $f'$ as a relation, it tells us which elements are mapped to which other elements. The prime just tells us that it is relation to some other function $f$ in a very specific way (derivation).



Example:
Our 'input' set is ${1,2,3 }$ our output set is ${A,B,C,D}$
$$f={(1,C),(2,A),(3,D) }$$
So we now know that $f(1)=C$ and $f(2)=A$. Notice that the element $B$ is not reached and this function is not surjective.



What you should take from this finite example is that a function is a rule that tells us which elements are in a way "paired", while $f(x)$ tells us about a specific pair. However sometimes people just represent the function like this by saying:



For arbitrary $x$ (so in our example $1$,$2$ or $3$), $f(x)$ is given by $dots$
This is indeed another representation of the same idea, but mathematicians ofter prefer the "relation" idea.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    The derivative of the function $f$ is $f'$. People usually make the mistake of saying that it is $f'(x)$, just like they talk about, say, the function $sin(x)$, when, in fact, they should be talking about the $sin$ function.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So when people say 2x is the derivative of x^2, is that incorrect? Also, am I correct in saying that the derivative is the function of the form y=f’(x) whose output represents the instantaneous rate of change at any point of a function or the slope of the tangent line to a point on a curve?
      $endgroup$
      – J. Smith
      Nov 26 '18 at 8:38






    • 1




      $begingroup$
      It is an abuse of language. It would be correct to say that $(operatorname{Id}^2)'=2operatorname{Id}$ and, of course, that's what people mean when they say that the derivative of $x^2$ is $2x$. Concerning your second question, the answer is affirmative.
      $endgroup$
      – José Carlos Santos
      Nov 26 '18 at 8:42










    • $begingroup$
      This is too far away from the original question and I suggest that you post it as another question. But, for me (and, I think, for most users of this forum)$$f'(x)=lim_{yto x}frac{f(y)-f(x)}{y-x}.$$
      $endgroup$
      – José Carlos Santos
      Nov 26 '18 at 8:59






    • 1




      $begingroup$
      I have to disagree with José when he says that "of course, that's what people mean when they say that the derivative of $x^2$ is $2x$". I agree that's what people mean when they know what they're talking about and understand the distinction outlined in several answers in this question. The thing is, most people that say this do not understand this distinction, at least I remember a time (pre-university) where I didn't since I didn't even know of a proper definition of function.
      $endgroup$
      – Git Gud
      Nov 26 '18 at 9:47












    • $begingroup$
      @GitGud Well, okay, perhaps that I was more than a bit optimistic here…
      $endgroup$
      – José Carlos Santos
      Nov 26 '18 at 9:48











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014033%2fnotation-for-the-derivative-of-a-function-f-or-fx%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    10












    $begingroup$

    By definition a function is a triple $(f,D,C)$, which is very often denoted by $f colon D to C$, where $C,D$ are two sets and $f$ associates to each element of $D$ one and only one element of $C$.



    So when it is clear what $C$ and $D$ are, or in cases where it is not possible or not necessary to write them down, you just write $f$. The expression $f(x)$ denotes the element in $C$ which $x in D$ is mapped to. So $f$ is a function, $f(x)$ is an element of $C$, two completely different things.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Nice answer! (+1)
      $endgroup$
      – Robert Z
      Nov 26 '18 at 8:52










    • $begingroup$
      Thank you @RobertZ.
      $endgroup$
      – Gibbs
      Nov 26 '18 at 8:59






    • 2




      $begingroup$
      I agree with the forst part (+1) but I would be more relaxed for the second part, I think that using f(x) to indicate the function can be tolerated.
      $endgroup$
      – gimusi
      Nov 26 '18 at 9:22






    • 1




      $begingroup$
      I think it is a good habit for a beginner to stick to the definitions. After some practice and experience one becomes conscious on where conventions can be relaxed and some abuse of language might be ok.
      $endgroup$
      – Gibbs
      Nov 26 '18 at 10:06










    • $begingroup$
      The distinction between $f$ and $f(x) $ is important but "let $f(x) =xsin(1/x)$" has almost become a routine shorthand for "let $f:mathbb{R} setminus{0}tomathbb {R} $ be a function defined by $f(x) =xsin(1/x)$".
      $endgroup$
      – Paramanand Singh
      Nov 27 '18 at 5:05


















    10












    $begingroup$

    By definition a function is a triple $(f,D,C)$, which is very often denoted by $f colon D to C$, where $C,D$ are two sets and $f$ associates to each element of $D$ one and only one element of $C$.



    So when it is clear what $C$ and $D$ are, or in cases where it is not possible or not necessary to write them down, you just write $f$. The expression $f(x)$ denotes the element in $C$ which $x in D$ is mapped to. So $f$ is a function, $f(x)$ is an element of $C$, two completely different things.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Nice answer! (+1)
      $endgroup$
      – Robert Z
      Nov 26 '18 at 8:52










    • $begingroup$
      Thank you @RobertZ.
      $endgroup$
      – Gibbs
      Nov 26 '18 at 8:59






    • 2




      $begingroup$
      I agree with the forst part (+1) but I would be more relaxed for the second part, I think that using f(x) to indicate the function can be tolerated.
      $endgroup$
      – gimusi
      Nov 26 '18 at 9:22






    • 1




      $begingroup$
      I think it is a good habit for a beginner to stick to the definitions. After some practice and experience one becomes conscious on where conventions can be relaxed and some abuse of language might be ok.
      $endgroup$
      – Gibbs
      Nov 26 '18 at 10:06










    • $begingroup$
      The distinction between $f$ and $f(x) $ is important but "let $f(x) =xsin(1/x)$" has almost become a routine shorthand for "let $f:mathbb{R} setminus{0}tomathbb {R} $ be a function defined by $f(x) =xsin(1/x)$".
      $endgroup$
      – Paramanand Singh
      Nov 27 '18 at 5:05
















    10












    10








    10





    $begingroup$

    By definition a function is a triple $(f,D,C)$, which is very often denoted by $f colon D to C$, where $C,D$ are two sets and $f$ associates to each element of $D$ one and only one element of $C$.



    So when it is clear what $C$ and $D$ are, or in cases where it is not possible or not necessary to write them down, you just write $f$. The expression $f(x)$ denotes the element in $C$ which $x in D$ is mapped to. So $f$ is a function, $f(x)$ is an element of $C$, two completely different things.






    share|cite|improve this answer











    $endgroup$



    By definition a function is a triple $(f,D,C)$, which is very often denoted by $f colon D to C$, where $C,D$ are two sets and $f$ associates to each element of $D$ one and only one element of $C$.



    So when it is clear what $C$ and $D$ are, or in cases where it is not possible or not necessary to write them down, you just write $f$. The expression $f(x)$ denotes the element in $C$ which $x in D$ is mapped to. So $f$ is a function, $f(x)$ is an element of $C$, two completely different things.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 26 '18 at 9:00

























    answered Nov 26 '18 at 8:39









    GibbsGibbs

    5,2623827




    5,2623827












    • $begingroup$
      Nice answer! (+1)
      $endgroup$
      – Robert Z
      Nov 26 '18 at 8:52










    • $begingroup$
      Thank you @RobertZ.
      $endgroup$
      – Gibbs
      Nov 26 '18 at 8:59






    • 2




      $begingroup$
      I agree with the forst part (+1) but I would be more relaxed for the second part, I think that using f(x) to indicate the function can be tolerated.
      $endgroup$
      – gimusi
      Nov 26 '18 at 9:22






    • 1




      $begingroup$
      I think it is a good habit for a beginner to stick to the definitions. After some practice and experience one becomes conscious on where conventions can be relaxed and some abuse of language might be ok.
      $endgroup$
      – Gibbs
      Nov 26 '18 at 10:06










    • $begingroup$
      The distinction between $f$ and $f(x) $ is important but "let $f(x) =xsin(1/x)$" has almost become a routine shorthand for "let $f:mathbb{R} setminus{0}tomathbb {R} $ be a function defined by $f(x) =xsin(1/x)$".
      $endgroup$
      – Paramanand Singh
      Nov 27 '18 at 5:05




















    • $begingroup$
      Nice answer! (+1)
      $endgroup$
      – Robert Z
      Nov 26 '18 at 8:52










    • $begingroup$
      Thank you @RobertZ.
      $endgroup$
      – Gibbs
      Nov 26 '18 at 8:59






    • 2




      $begingroup$
      I agree with the forst part (+1) but I would be more relaxed for the second part, I think that using f(x) to indicate the function can be tolerated.
      $endgroup$
      – gimusi
      Nov 26 '18 at 9:22






    • 1




      $begingroup$
      I think it is a good habit for a beginner to stick to the definitions. After some practice and experience one becomes conscious on where conventions can be relaxed and some abuse of language might be ok.
      $endgroup$
      – Gibbs
      Nov 26 '18 at 10:06










    • $begingroup$
      The distinction between $f$ and $f(x) $ is important but "let $f(x) =xsin(1/x)$" has almost become a routine shorthand for "let $f:mathbb{R} setminus{0}tomathbb {R} $ be a function defined by $f(x) =xsin(1/x)$".
      $endgroup$
      – Paramanand Singh
      Nov 27 '18 at 5:05


















    $begingroup$
    Nice answer! (+1)
    $endgroup$
    – Robert Z
    Nov 26 '18 at 8:52




    $begingroup$
    Nice answer! (+1)
    $endgroup$
    – Robert Z
    Nov 26 '18 at 8:52












    $begingroup$
    Thank you @RobertZ.
    $endgroup$
    – Gibbs
    Nov 26 '18 at 8:59




    $begingroup$
    Thank you @RobertZ.
    $endgroup$
    – Gibbs
    Nov 26 '18 at 8:59




    2




    2




    $begingroup$
    I agree with the forst part (+1) but I would be more relaxed for the second part, I think that using f(x) to indicate the function can be tolerated.
    $endgroup$
    – gimusi
    Nov 26 '18 at 9:22




    $begingroup$
    I agree with the forst part (+1) but I would be more relaxed for the second part, I think that using f(x) to indicate the function can be tolerated.
    $endgroup$
    – gimusi
    Nov 26 '18 at 9:22




    1




    1




    $begingroup$
    I think it is a good habit for a beginner to stick to the definitions. After some practice and experience one becomes conscious on where conventions can be relaxed and some abuse of language might be ok.
    $endgroup$
    – Gibbs
    Nov 26 '18 at 10:06




    $begingroup$
    I think it is a good habit for a beginner to stick to the definitions. After some practice and experience one becomes conscious on where conventions can be relaxed and some abuse of language might be ok.
    $endgroup$
    – Gibbs
    Nov 26 '18 at 10:06












    $begingroup$
    The distinction between $f$ and $f(x) $ is important but "let $f(x) =xsin(1/x)$" has almost become a routine shorthand for "let $f:mathbb{R} setminus{0}tomathbb {R} $ be a function defined by $f(x) =xsin(1/x)$".
    $endgroup$
    – Paramanand Singh
    Nov 27 '18 at 5:05






    $begingroup$
    The distinction between $f$ and $f(x) $ is important but "let $f(x) =xsin(1/x)$" has almost become a routine shorthand for "let $f:mathbb{R} setminus{0}tomathbb {R} $ be a function defined by $f(x) =xsin(1/x)$".
    $endgroup$
    – Paramanand Singh
    Nov 27 '18 at 5:05













    3












    $begingroup$

    $f$ denotes the function and $f(x)$ the output of the function when evaluated at $x$.



    This convention does not differ for the derivative.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      f’ denotes the derivative and f’(x) denotes the output of the derivative, which is the instantaneous rate of change of a function at any point. Correct?
      $endgroup$
      – J. Smith
      Nov 26 '18 at 10:08












    • $begingroup$
      @J.Smith: you get it.
      $endgroup$
      – Yves Daoust
      Nov 26 '18 at 10:13










    • $begingroup$
      @YvesDaoust I think we could be more "relaxed" with that definition and notation, no one will be wound considering the function $f(x)$ :)
      $endgroup$
      – gimusi
      Nov 26 '18 at 10:17






    • 2




      $begingroup$
      @gimusi: this would reduce the expressive power (not possible to distinguish the function and the value) and create ambiguities. So, no.
      $endgroup$
      – Yves Daoust
      Nov 26 '18 at 10:20










    • $begingroup$
      @YvesDaoust Also f creates ambiguity if we do not specify the domain and the codomain, therefore anytime we refer to a function we should use $f:Ato B$. I don't thing it would be a useful notation. $f(x)$ to indicate the funtion can be used many times without any ambiguity. Of course we ca suggest to do not use that but it is a matter of preferences and not a law.
      $endgroup$
      – gimusi
      Nov 26 '18 at 10:23
















    3












    $begingroup$

    $f$ denotes the function and $f(x)$ the output of the function when evaluated at $x$.



    This convention does not differ for the derivative.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      f’ denotes the derivative and f’(x) denotes the output of the derivative, which is the instantaneous rate of change of a function at any point. Correct?
      $endgroup$
      – J. Smith
      Nov 26 '18 at 10:08












    • $begingroup$
      @J.Smith: you get it.
      $endgroup$
      – Yves Daoust
      Nov 26 '18 at 10:13










    • $begingroup$
      @YvesDaoust I think we could be more "relaxed" with that definition and notation, no one will be wound considering the function $f(x)$ :)
      $endgroup$
      – gimusi
      Nov 26 '18 at 10:17






    • 2




      $begingroup$
      @gimusi: this would reduce the expressive power (not possible to distinguish the function and the value) and create ambiguities. So, no.
      $endgroup$
      – Yves Daoust
      Nov 26 '18 at 10:20










    • $begingroup$
      @YvesDaoust Also f creates ambiguity if we do not specify the domain and the codomain, therefore anytime we refer to a function we should use $f:Ato B$. I don't thing it would be a useful notation. $f(x)$ to indicate the funtion can be used many times without any ambiguity. Of course we ca suggest to do not use that but it is a matter of preferences and not a law.
      $endgroup$
      – gimusi
      Nov 26 '18 at 10:23














    3












    3








    3





    $begingroup$

    $f$ denotes the function and $f(x)$ the output of the function when evaluated at $x$.



    This convention does not differ for the derivative.






    share|cite|improve this answer









    $endgroup$



    $f$ denotes the function and $f(x)$ the output of the function when evaluated at $x$.



    This convention does not differ for the derivative.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 26 '18 at 10:07









    Yves DaoustYves Daoust

    127k673226




    127k673226












    • $begingroup$
      f’ denotes the derivative and f’(x) denotes the output of the derivative, which is the instantaneous rate of change of a function at any point. Correct?
      $endgroup$
      – J. Smith
      Nov 26 '18 at 10:08












    • $begingroup$
      @J.Smith: you get it.
      $endgroup$
      – Yves Daoust
      Nov 26 '18 at 10:13










    • $begingroup$
      @YvesDaoust I think we could be more "relaxed" with that definition and notation, no one will be wound considering the function $f(x)$ :)
      $endgroup$
      – gimusi
      Nov 26 '18 at 10:17






    • 2




      $begingroup$
      @gimusi: this would reduce the expressive power (not possible to distinguish the function and the value) and create ambiguities. So, no.
      $endgroup$
      – Yves Daoust
      Nov 26 '18 at 10:20










    • $begingroup$
      @YvesDaoust Also f creates ambiguity if we do not specify the domain and the codomain, therefore anytime we refer to a function we should use $f:Ato B$. I don't thing it would be a useful notation. $f(x)$ to indicate the funtion can be used many times without any ambiguity. Of course we ca suggest to do not use that but it is a matter of preferences and not a law.
      $endgroup$
      – gimusi
      Nov 26 '18 at 10:23


















    • $begingroup$
      f’ denotes the derivative and f’(x) denotes the output of the derivative, which is the instantaneous rate of change of a function at any point. Correct?
      $endgroup$
      – J. Smith
      Nov 26 '18 at 10:08












    • $begingroup$
      @J.Smith: you get it.
      $endgroup$
      – Yves Daoust
      Nov 26 '18 at 10:13










    • $begingroup$
      @YvesDaoust I think we could be more "relaxed" with that definition and notation, no one will be wound considering the function $f(x)$ :)
      $endgroup$
      – gimusi
      Nov 26 '18 at 10:17






    • 2




      $begingroup$
      @gimusi: this would reduce the expressive power (not possible to distinguish the function and the value) and create ambiguities. So, no.
      $endgroup$
      – Yves Daoust
      Nov 26 '18 at 10:20










    • $begingroup$
      @YvesDaoust Also f creates ambiguity if we do not specify the domain and the codomain, therefore anytime we refer to a function we should use $f:Ato B$. I don't thing it would be a useful notation. $f(x)$ to indicate the funtion can be used many times without any ambiguity. Of course we ca suggest to do not use that but it is a matter of preferences and not a law.
      $endgroup$
      – gimusi
      Nov 26 '18 at 10:23
















    $begingroup$
    f’ denotes the derivative and f’(x) denotes the output of the derivative, which is the instantaneous rate of change of a function at any point. Correct?
    $endgroup$
    – J. Smith
    Nov 26 '18 at 10:08






    $begingroup$
    f’ denotes the derivative and f’(x) denotes the output of the derivative, which is the instantaneous rate of change of a function at any point. Correct?
    $endgroup$
    – J. Smith
    Nov 26 '18 at 10:08














    $begingroup$
    @J.Smith: you get it.
    $endgroup$
    – Yves Daoust
    Nov 26 '18 at 10:13




    $begingroup$
    @J.Smith: you get it.
    $endgroup$
    – Yves Daoust
    Nov 26 '18 at 10:13












    $begingroup$
    @YvesDaoust I think we could be more "relaxed" with that definition and notation, no one will be wound considering the function $f(x)$ :)
    $endgroup$
    – gimusi
    Nov 26 '18 at 10:17




    $begingroup$
    @YvesDaoust I think we could be more "relaxed" with that definition and notation, no one will be wound considering the function $f(x)$ :)
    $endgroup$
    – gimusi
    Nov 26 '18 at 10:17




    2




    2




    $begingroup$
    @gimusi: this would reduce the expressive power (not possible to distinguish the function and the value) and create ambiguities. So, no.
    $endgroup$
    – Yves Daoust
    Nov 26 '18 at 10:20




    $begingroup$
    @gimusi: this would reduce the expressive power (not possible to distinguish the function and the value) and create ambiguities. So, no.
    $endgroup$
    – Yves Daoust
    Nov 26 '18 at 10:20












    $begingroup$
    @YvesDaoust Also f creates ambiguity if we do not specify the domain and the codomain, therefore anytime we refer to a function we should use $f:Ato B$. I don't thing it would be a useful notation. $f(x)$ to indicate the funtion can be used many times without any ambiguity. Of course we ca suggest to do not use that but it is a matter of preferences and not a law.
    $endgroup$
    – gimusi
    Nov 26 '18 at 10:23




    $begingroup$
    @YvesDaoust Also f creates ambiguity if we do not specify the domain and the codomain, therefore anytime we refer to a function we should use $f:Ato B$. I don't thing it would be a useful notation. $f(x)$ to indicate the funtion can be used many times without any ambiguity. Of course we ca suggest to do not use that but it is a matter of preferences and not a law.
    $endgroup$
    – gimusi
    Nov 26 '18 at 10:23











    2












    $begingroup$

    I would read $f'(x)$ as "the function $f'$ applied to the element $x$ of the domain". This gives us a a new element in the range. Meanwhile I read $f'$ as a relation, it tells us which elements are mapped to which other elements. The prime just tells us that it is relation to some other function $f$ in a very specific way (derivation).



    Example:
    Our 'input' set is ${1,2,3 }$ our output set is ${A,B,C,D}$
    $$f={(1,C),(2,A),(3,D) }$$
    So we now know that $f(1)=C$ and $f(2)=A$. Notice that the element $B$ is not reached and this function is not surjective.



    What you should take from this finite example is that a function is a rule that tells us which elements are in a way "paired", while $f(x)$ tells us about a specific pair. However sometimes people just represent the function like this by saying:



    For arbitrary $x$ (so in our example $1$,$2$ or $3$), $f(x)$ is given by $dots$
    This is indeed another representation of the same idea, but mathematicians ofter prefer the "relation" idea.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      I would read $f'(x)$ as "the function $f'$ applied to the element $x$ of the domain". This gives us a a new element in the range. Meanwhile I read $f'$ as a relation, it tells us which elements are mapped to which other elements. The prime just tells us that it is relation to some other function $f$ in a very specific way (derivation).



      Example:
      Our 'input' set is ${1,2,3 }$ our output set is ${A,B,C,D}$
      $$f={(1,C),(2,A),(3,D) }$$
      So we now know that $f(1)=C$ and $f(2)=A$. Notice that the element $B$ is not reached and this function is not surjective.



      What you should take from this finite example is that a function is a rule that tells us which elements are in a way "paired", while $f(x)$ tells us about a specific pair. However sometimes people just represent the function like this by saying:



      For arbitrary $x$ (so in our example $1$,$2$ or $3$), $f(x)$ is given by $dots$
      This is indeed another representation of the same idea, but mathematicians ofter prefer the "relation" idea.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        I would read $f'(x)$ as "the function $f'$ applied to the element $x$ of the domain". This gives us a a new element in the range. Meanwhile I read $f'$ as a relation, it tells us which elements are mapped to which other elements. The prime just tells us that it is relation to some other function $f$ in a very specific way (derivation).



        Example:
        Our 'input' set is ${1,2,3 }$ our output set is ${A,B,C,D}$
        $$f={(1,C),(2,A),(3,D) }$$
        So we now know that $f(1)=C$ and $f(2)=A$. Notice that the element $B$ is not reached and this function is not surjective.



        What you should take from this finite example is that a function is a rule that tells us which elements are in a way "paired", while $f(x)$ tells us about a specific pair. However sometimes people just represent the function like this by saying:



        For arbitrary $x$ (so in our example $1$,$2$ or $3$), $f(x)$ is given by $dots$
        This is indeed another representation of the same idea, but mathematicians ofter prefer the "relation" idea.






        share|cite|improve this answer











        $endgroup$



        I would read $f'(x)$ as "the function $f'$ applied to the element $x$ of the domain". This gives us a a new element in the range. Meanwhile I read $f'$ as a relation, it tells us which elements are mapped to which other elements. The prime just tells us that it is relation to some other function $f$ in a very specific way (derivation).



        Example:
        Our 'input' set is ${1,2,3 }$ our output set is ${A,B,C,D}$
        $$f={(1,C),(2,A),(3,D) }$$
        So we now know that $f(1)=C$ and $f(2)=A$. Notice that the element $B$ is not reached and this function is not surjective.



        What you should take from this finite example is that a function is a rule that tells us which elements are in a way "paired", while $f(x)$ tells us about a specific pair. However sometimes people just represent the function like this by saying:



        For arbitrary $x$ (so in our example $1$,$2$ or $3$), $f(x)$ is given by $dots$
        This is indeed another representation of the same idea, but mathematicians ofter prefer the "relation" idea.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 26 '18 at 20:51









        amWhy

        1




        1










        answered Nov 26 '18 at 8:49









        Wesley StrikWesley Strik

        2,007423




        2,007423























            1












            $begingroup$

            The derivative of the function $f$ is $f'$. People usually make the mistake of saying that it is $f'(x)$, just like they talk about, say, the function $sin(x)$, when, in fact, they should be talking about the $sin$ function.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              So when people say 2x is the derivative of x^2, is that incorrect? Also, am I correct in saying that the derivative is the function of the form y=f’(x) whose output represents the instantaneous rate of change at any point of a function or the slope of the tangent line to a point on a curve?
              $endgroup$
              – J. Smith
              Nov 26 '18 at 8:38






            • 1




              $begingroup$
              It is an abuse of language. It would be correct to say that $(operatorname{Id}^2)'=2operatorname{Id}$ and, of course, that's what people mean when they say that the derivative of $x^2$ is $2x$. Concerning your second question, the answer is affirmative.
              $endgroup$
              – José Carlos Santos
              Nov 26 '18 at 8:42










            • $begingroup$
              This is too far away from the original question and I suggest that you post it as another question. But, for me (and, I think, for most users of this forum)$$f'(x)=lim_{yto x}frac{f(y)-f(x)}{y-x}.$$
              $endgroup$
              – José Carlos Santos
              Nov 26 '18 at 8:59






            • 1




              $begingroup$
              I have to disagree with José when he says that "of course, that's what people mean when they say that the derivative of $x^2$ is $2x$". I agree that's what people mean when they know what they're talking about and understand the distinction outlined in several answers in this question. The thing is, most people that say this do not understand this distinction, at least I remember a time (pre-university) where I didn't since I didn't even know of a proper definition of function.
              $endgroup$
              – Git Gud
              Nov 26 '18 at 9:47












            • $begingroup$
              @GitGud Well, okay, perhaps that I was more than a bit optimistic here…
              $endgroup$
              – José Carlos Santos
              Nov 26 '18 at 9:48
















            1












            $begingroup$

            The derivative of the function $f$ is $f'$. People usually make the mistake of saying that it is $f'(x)$, just like they talk about, say, the function $sin(x)$, when, in fact, they should be talking about the $sin$ function.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              So when people say 2x is the derivative of x^2, is that incorrect? Also, am I correct in saying that the derivative is the function of the form y=f’(x) whose output represents the instantaneous rate of change at any point of a function or the slope of the tangent line to a point on a curve?
              $endgroup$
              – J. Smith
              Nov 26 '18 at 8:38






            • 1




              $begingroup$
              It is an abuse of language. It would be correct to say that $(operatorname{Id}^2)'=2operatorname{Id}$ and, of course, that's what people mean when they say that the derivative of $x^2$ is $2x$. Concerning your second question, the answer is affirmative.
              $endgroup$
              – José Carlos Santos
              Nov 26 '18 at 8:42










            • $begingroup$
              This is too far away from the original question and I suggest that you post it as another question. But, for me (and, I think, for most users of this forum)$$f'(x)=lim_{yto x}frac{f(y)-f(x)}{y-x}.$$
              $endgroup$
              – José Carlos Santos
              Nov 26 '18 at 8:59






            • 1




              $begingroup$
              I have to disagree with José when he says that "of course, that's what people mean when they say that the derivative of $x^2$ is $2x$". I agree that's what people mean when they know what they're talking about and understand the distinction outlined in several answers in this question. The thing is, most people that say this do not understand this distinction, at least I remember a time (pre-university) where I didn't since I didn't even know of a proper definition of function.
              $endgroup$
              – Git Gud
              Nov 26 '18 at 9:47












            • $begingroup$
              @GitGud Well, okay, perhaps that I was more than a bit optimistic here…
              $endgroup$
              – José Carlos Santos
              Nov 26 '18 at 9:48














            1












            1








            1





            $begingroup$

            The derivative of the function $f$ is $f'$. People usually make the mistake of saying that it is $f'(x)$, just like they talk about, say, the function $sin(x)$, when, in fact, they should be talking about the $sin$ function.






            share|cite|improve this answer









            $endgroup$



            The derivative of the function $f$ is $f'$. People usually make the mistake of saying that it is $f'(x)$, just like they talk about, say, the function $sin(x)$, when, in fact, they should be talking about the $sin$ function.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 26 '18 at 8:27









            José Carlos SantosJosé Carlos Santos

            160k22127232




            160k22127232












            • $begingroup$
              So when people say 2x is the derivative of x^2, is that incorrect? Also, am I correct in saying that the derivative is the function of the form y=f’(x) whose output represents the instantaneous rate of change at any point of a function or the slope of the tangent line to a point on a curve?
              $endgroup$
              – J. Smith
              Nov 26 '18 at 8:38






            • 1




              $begingroup$
              It is an abuse of language. It would be correct to say that $(operatorname{Id}^2)'=2operatorname{Id}$ and, of course, that's what people mean when they say that the derivative of $x^2$ is $2x$. Concerning your second question, the answer is affirmative.
              $endgroup$
              – José Carlos Santos
              Nov 26 '18 at 8:42










            • $begingroup$
              This is too far away from the original question and I suggest that you post it as another question. But, for me (and, I think, for most users of this forum)$$f'(x)=lim_{yto x}frac{f(y)-f(x)}{y-x}.$$
              $endgroup$
              – José Carlos Santos
              Nov 26 '18 at 8:59






            • 1




              $begingroup$
              I have to disagree with José when he says that "of course, that's what people mean when they say that the derivative of $x^2$ is $2x$". I agree that's what people mean when they know what they're talking about and understand the distinction outlined in several answers in this question. The thing is, most people that say this do not understand this distinction, at least I remember a time (pre-university) where I didn't since I didn't even know of a proper definition of function.
              $endgroup$
              – Git Gud
              Nov 26 '18 at 9:47












            • $begingroup$
              @GitGud Well, okay, perhaps that I was more than a bit optimistic here…
              $endgroup$
              – José Carlos Santos
              Nov 26 '18 at 9:48


















            • $begingroup$
              So when people say 2x is the derivative of x^2, is that incorrect? Also, am I correct in saying that the derivative is the function of the form y=f’(x) whose output represents the instantaneous rate of change at any point of a function or the slope of the tangent line to a point on a curve?
              $endgroup$
              – J. Smith
              Nov 26 '18 at 8:38






            • 1




              $begingroup$
              It is an abuse of language. It would be correct to say that $(operatorname{Id}^2)'=2operatorname{Id}$ and, of course, that's what people mean when they say that the derivative of $x^2$ is $2x$. Concerning your second question, the answer is affirmative.
              $endgroup$
              – José Carlos Santos
              Nov 26 '18 at 8:42










            • $begingroup$
              This is too far away from the original question and I suggest that you post it as another question. But, for me (and, I think, for most users of this forum)$$f'(x)=lim_{yto x}frac{f(y)-f(x)}{y-x}.$$
              $endgroup$
              – José Carlos Santos
              Nov 26 '18 at 8:59






            • 1




              $begingroup$
              I have to disagree with José when he says that "of course, that's what people mean when they say that the derivative of $x^2$ is $2x$". I agree that's what people mean when they know what they're talking about and understand the distinction outlined in several answers in this question. The thing is, most people that say this do not understand this distinction, at least I remember a time (pre-university) where I didn't since I didn't even know of a proper definition of function.
              $endgroup$
              – Git Gud
              Nov 26 '18 at 9:47












            • $begingroup$
              @GitGud Well, okay, perhaps that I was more than a bit optimistic here…
              $endgroup$
              – José Carlos Santos
              Nov 26 '18 at 9:48
















            $begingroup$
            So when people say 2x is the derivative of x^2, is that incorrect? Also, am I correct in saying that the derivative is the function of the form y=f’(x) whose output represents the instantaneous rate of change at any point of a function or the slope of the tangent line to a point on a curve?
            $endgroup$
            – J. Smith
            Nov 26 '18 at 8:38




            $begingroup$
            So when people say 2x is the derivative of x^2, is that incorrect? Also, am I correct in saying that the derivative is the function of the form y=f’(x) whose output represents the instantaneous rate of change at any point of a function or the slope of the tangent line to a point on a curve?
            $endgroup$
            – J. Smith
            Nov 26 '18 at 8:38




            1




            1




            $begingroup$
            It is an abuse of language. It would be correct to say that $(operatorname{Id}^2)'=2operatorname{Id}$ and, of course, that's what people mean when they say that the derivative of $x^2$ is $2x$. Concerning your second question, the answer is affirmative.
            $endgroup$
            – José Carlos Santos
            Nov 26 '18 at 8:42




            $begingroup$
            It is an abuse of language. It would be correct to say that $(operatorname{Id}^2)'=2operatorname{Id}$ and, of course, that's what people mean when they say that the derivative of $x^2$ is $2x$. Concerning your second question, the answer is affirmative.
            $endgroup$
            – José Carlos Santos
            Nov 26 '18 at 8:42












            $begingroup$
            This is too far away from the original question and I suggest that you post it as another question. But, for me (and, I think, for most users of this forum)$$f'(x)=lim_{yto x}frac{f(y)-f(x)}{y-x}.$$
            $endgroup$
            – José Carlos Santos
            Nov 26 '18 at 8:59




            $begingroup$
            This is too far away from the original question and I suggest that you post it as another question. But, for me (and, I think, for most users of this forum)$$f'(x)=lim_{yto x}frac{f(y)-f(x)}{y-x}.$$
            $endgroup$
            – José Carlos Santos
            Nov 26 '18 at 8:59




            1




            1




            $begingroup$
            I have to disagree with José when he says that "of course, that's what people mean when they say that the derivative of $x^2$ is $2x$". I agree that's what people mean when they know what they're talking about and understand the distinction outlined in several answers in this question. The thing is, most people that say this do not understand this distinction, at least I remember a time (pre-university) where I didn't since I didn't even know of a proper definition of function.
            $endgroup$
            – Git Gud
            Nov 26 '18 at 9:47






            $begingroup$
            I have to disagree with José when he says that "of course, that's what people mean when they say that the derivative of $x^2$ is $2x$". I agree that's what people mean when they know what they're talking about and understand the distinction outlined in several answers in this question. The thing is, most people that say this do not understand this distinction, at least I remember a time (pre-university) where I didn't since I didn't even know of a proper definition of function.
            $endgroup$
            – Git Gud
            Nov 26 '18 at 9:47














            $begingroup$
            @GitGud Well, okay, perhaps that I was more than a bit optimistic here…
            $endgroup$
            – José Carlos Santos
            Nov 26 '18 at 9:48




            $begingroup$
            @GitGud Well, okay, perhaps that I was more than a bit optimistic here…
            $endgroup$
            – José Carlos Santos
            Nov 26 '18 at 9:48


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014033%2fnotation-for-the-derivative-of-a-function-f-or-fx%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Contact image not getting when fetch all contact list from iPhone by CNContact

            Image
            up vote 0 down vote favorite I know this question already asked but not getting solution. From this code I will get all the information from the contact but image not found when open vcf files on mac os, also not getting when share this file. I use this stackoverflow link here but It's not help full. var contacts = [CNContact]() let keys = [CNContactVCardSerialization.descriptorForRequiredKeys() ] as [Any] let request = CNContactFetchRequest(keysToFetch: keys as! [CNKeyDescriptor]) do { try self.contactStore.enumerateContacts(with: request) { (contact, stop) in // Array containing all unified contacts from everywhere contacts.append(contact) } } catch { print("unable to fetch contacts") } do { let data = try
            Read more

            count number of partitions of a set with n elements into k subsets

            Image
            up vote 6 down vote favorite 1 This program is for count number of partitions of a set with n elements into k subsets I am confusing here return k*countP(n-1, k) + countP(n-1, k-1); can some one explain what is happening here? why we are multiplying with k? NOTE->I know this is not the best way to calculate number of partitions that would be DP // A C++ program to count number of partitions // of a set with n elements into k subsets #include<iostream> using namespace std; // Returns count of different partitions of n // elements in k subsets int countP(int n, int k) { // Base cases if (n == 0 || k == 0 || k > n) return 0; if (k == 1 || k == n) return 1; // S(n+1, k) = k*S(n, k) + S(n, k-1) return k*countP(n-1, k) + countP(n-1, k-1); } // Driver program int ma
            Read more

            A CLEAN and SIMPLE way to add appendices to Table of Contents and bookmarks

            Image
            0 I found a lot of questions abount appendices and ToC. Many users want appendices to be grouped in an Appendix part, however some problems arise with ToC, hyperref, PDF viewer bookmarks, and so on. There are different solutions which require extra packages, command patching and other extra code, however none of them satisfies me. I almost found an easy way to accomplish a good result, where appendices are added to bookmarks in the right way and hyperref links point to the right page. However, the number of the "Appendix" part page is wrong (it's the number of appendix A). Is there any EASY way to fix that? This is a MWE: documentclass{book} usepackage[nottoc,notlot,notlof]{tocbibind} usepackage{hyperref} begin{document} frontmatter tableofcontents mainmatter part{First} chapter{
            Read more