Is the derivative an approximation? (Cost function, marginal cost, deviation in costs)












0












$begingroup$


Good evening everyone,



I wonder if the derivative is an approximation.



Because if I am calculating marginal costs there is always some small deviations regarding the marginal cost (additional cost in the case of producing one quantity more).



e.g.:
Let the cost function be
$C(x)=x^2+5x$ with $x$ denoting the quantity.



The derivative (here marginal cost function) in respect to $x$ is:



$C'(x) = 2x+5$.



Now for $C=1; C=2$ we get:



$C(1)=6$



$C(2)=14$



The Derivatives at $C=1;C=2$ are



$C'(1)=7$



$C'(2)=9$



So the change from (incurring costs in the case that we produce one quantity more) $C(1)$ to $C(2)$ is $8$.



But why is the derivative at position $C'(1)$ then 7 and not $8$ ?



It seems like the derivative is an approximation and not really "exact" because there is always a difference of 1 regarding $C(x+1)-C(x)$.



Hope someone can clearify this, because in the case of other functions ($3x^2+4$) the difference varies (in the case of $3x^2+4$ the difference is always 3).



It is that $C(x+1)-C(x)$ is the exact difference in costs whereby the derivative has always an deviation/error.



sincerely,
Monoid






functions derivatives intuition







share|cite|improve this question









$endgroup$












  • $begingroup$
    $C(x+1)-C(x)$ is the derivative at some point between $x$ and $x+1$ (and this the best that you can say in general, that is, for an arbitrary function).
    $endgroup$
    – John B
    Nov 30 '18 at 22:55










  • $begingroup$
    @Monoid it's really more the other way around, taking the difference between two costs with close x values is an approximation of the derivative. The derivative is the difference in the function as the difference in x goes to zero. Looking at it from your perspective, adding the derivative to the function to get to the next value is just doing a truncated version of the Taylor series.
    $endgroup$
    – Tyberius
    Nov 30 '18 at 22:58










  • $begingroup$
    A derivative of a function is NOT an approximation. It has different menaings depending on the nature of the function. First derivative of the cost function is the Marginal Cost. There is a good explanation of derivatives as it relates to Cost Function in here dummies.com/education/math/calculus/… and here:columbia.edu/itc/sipa/math/calc_econ_interp_u.html
    $endgroup$
    – NoChance
    Dec 1 '18 at 1:23


















0












$begingroup$


Good evening everyone,



I wonder if the derivative is an approximation.



Because if I am calculating marginal costs there is always some small deviations regarding the marginal cost (additional cost in the case of producing one quantity more).



e.g.:
Let the cost function be
$C(x)=x^2+5x$ with $x$ denoting the quantity.



The derivative (here marginal cost function) in respect to $x$ is:



$C'(x) = 2x+5$.



Now for $C=1; C=2$ we get:



$C(1)=6$



$C(2)=14$



The Derivatives at $C=1;C=2$ are



$C'(1)=7$



$C'(2)=9$



So the change from (incurring costs in the case that we produce one quantity more) $C(1)$ to $C(2)$ is $8$.



But why is the derivative at position $C'(1)$ then 7 and not $8$ ?



It seems like the derivative is an approximation and not really "exact" because there is always a difference of 1 regarding $C(x+1)-C(x)$.



Hope someone can clearify this, because in the case of other functions ($3x^2+4$) the difference varies (in the case of $3x^2+4$ the difference is always 3).



It is that $C(x+1)-C(x)$ is the exact difference in costs whereby the derivative has always an deviation/error.



sincerely,
Monoid






functions derivatives intuition







share|cite|improve this question









$endgroup$












  • $begingroup$
    $C(x+1)-C(x)$ is the derivative at some point between $x$ and $x+1$ (and this the best that you can say in general, that is, for an arbitrary function).
    $endgroup$
    – John B
    Nov 30 '18 at 22:55










  • $begingroup$
    @Monoid it's really more the other way around, taking the difference between two costs with close x values is an approximation of the derivative. The derivative is the difference in the function as the difference in x goes to zero. Looking at it from your perspective, adding the derivative to the function to get to the next value is just doing a truncated version of the Taylor series.
    $endgroup$
    – Tyberius
    Nov 30 '18 at 22:58










  • $begingroup$
    A derivative of a function is NOT an approximation. It has different menaings depending on the nature of the function. First derivative of the cost function is the Marginal Cost. There is a good explanation of derivatives as it relates to Cost Function in here dummies.com/education/math/calculus/… and here:columbia.edu/itc/sipa/math/calc_econ_interp_u.html
    $endgroup$
    – NoChance
    Dec 1 '18 at 1:23
















0












0








0





$begingroup$


Good evening everyone,



I wonder if the derivative is an approximation.



Because if I am calculating marginal costs there is always some small deviations regarding the marginal cost (additional cost in the case of producing one quantity more).



e.g.:
Let the cost function be
$C(x)=x^2+5x$ with $x$ denoting the quantity.



The derivative (here marginal cost function) in respect to $x$ is:



$C'(x) = 2x+5$.



Now for $C=1; C=2$ we get:



$C(1)=6$



$C(2)=14$



The Derivatives at $C=1;C=2$ are



$C'(1)=7$



$C'(2)=9$



So the change from (incurring costs in the case that we produce one quantity more) $C(1)$ to $C(2)$ is $8$.



But why is the derivative at position $C'(1)$ then 7 and not $8$ ?



It seems like the derivative is an approximation and not really "exact" because there is always a difference of 1 regarding $C(x+1)-C(x)$.



Hope someone can clearify this, because in the case of other functions ($3x^2+4$) the difference varies (in the case of $3x^2+4$ the difference is always 3).



It is that $C(x+1)-C(x)$ is the exact difference in costs whereby the derivative has always an deviation/error.



sincerely,
Monoid






functions derivatives intuition







share|cite|improve this question









$endgroup$




Good evening everyone,



I wonder if the derivative is an approximation.



Because if I am calculating marginal costs there is always some small deviations regarding the marginal cost (additional cost in the case of producing one quantity more).



e.g.:
Let the cost function be
$C(x)=x^2+5x$ with $x$ denoting the quantity.



The derivative (here marginal cost function) in respect to $x$ is:



$C'(x) = 2x+5$.



Now for $C=1; C=2$ we get:



$C(1)=6$



$C(2)=14$



The Derivatives at $C=1;C=2$ are



$C'(1)=7$



$C'(2)=9$



So the change from (incurring costs in the case that we produce one quantity more) $C(1)$ to $C(2)$ is $8$.



But why is the derivative at position $C'(1)$ then 7 and not $8$ ?



It seems like the derivative is an approximation and not really "exact" because there is always a difference of 1 regarding $C(x+1)-C(x)$.



Hope someone can clearify this, because in the case of other functions ($3x^2+4$) the difference varies (in the case of $3x^2+4$ the difference is always 3).



It is that $C(x+1)-C(x)$ is the exact difference in costs whereby the derivative has always an deviation/error.



sincerely,
Monoid






functions derivatives intuition




functions derivatives intuition






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 30 '18 at 22:46









MonoidMonoid

11




11












  • $begingroup$
    $C(x+1)-C(x)$ is the derivative at some point between $x$ and $x+1$ (and this the best that you can say in general, that is, for an arbitrary function).
    $endgroup$
    – John B
    Nov 30 '18 at 22:55










  • $begingroup$
    @Monoid it's really more the other way around, taking the difference between two costs with close x values is an approximation of the derivative. The derivative is the difference in the function as the difference in x goes to zero. Looking at it from your perspective, adding the derivative to the function to get to the next value is just doing a truncated version of the Taylor series.
    $endgroup$
    – Tyberius
    Nov 30 '18 at 22:58










  • $begingroup$
    A derivative of a function is NOT an approximation. It has different menaings depending on the nature of the function. First derivative of the cost function is the Marginal Cost. There is a good explanation of derivatives as it relates to Cost Function in here dummies.com/education/math/calculus/… and here:columbia.edu/itc/sipa/math/calc_econ_interp_u.html
    $endgroup$
    – NoChance
    Dec 1 '18 at 1:23




















  • $begingroup$
    $C(x+1)-C(x)$ is the derivative at some point between $x$ and $x+1$ (and this the best that you can say in general, that is, for an arbitrary function).
    $endgroup$
    – John B
    Nov 30 '18 at 22:55










  • $begingroup$
    @Monoid it's really more the other way around, taking the difference between two costs with close x values is an approximation of the derivative. The derivative is the difference in the function as the difference in x goes to zero. Looking at it from your perspective, adding the derivative to the function to get to the next value is just doing a truncated version of the Taylor series.
    $endgroup$
    – Tyberius
    Nov 30 '18 at 22:58










  • $begingroup$
    A derivative of a function is NOT an approximation. It has different menaings depending on the nature of the function. First derivative of the cost function is the Marginal Cost. There is a good explanation of derivatives as it relates to Cost Function in here dummies.com/education/math/calculus/… and here:columbia.edu/itc/sipa/math/calc_econ_interp_u.html
    $endgroup$
    – NoChance
    Dec 1 '18 at 1:23


















$begingroup$
$C(x+1)-C(x)$ is the derivative at some point between $x$ and $x+1$ (and this the best that you can say in general, that is, for an arbitrary function).
$endgroup$
– John B
Nov 30 '18 at 22:55




$begingroup$
$C(x+1)-C(x)$ is the derivative at some point between $x$ and $x+1$ (and this the best that you can say in general, that is, for an arbitrary function).
$endgroup$
– John B
Nov 30 '18 at 22:55












$begingroup$
@Monoid it's really more the other way around, taking the difference between two costs with close x values is an approximation of the derivative. The derivative is the difference in the function as the difference in x goes to zero. Looking at it from your perspective, adding the derivative to the function to get to the next value is just doing a truncated version of the Taylor series.
$endgroup$
– Tyberius
Nov 30 '18 at 22:58




$begingroup$
@Monoid it's really more the other way around, taking the difference between two costs with close x values is an approximation of the derivative. The derivative is the difference in the function as the difference in x goes to zero. Looking at it from your perspective, adding the derivative to the function to get to the next value is just doing a truncated version of the Taylor series.
$endgroup$
– Tyberius
Nov 30 '18 at 22:58












$begingroup$
A derivative of a function is NOT an approximation. It has different menaings depending on the nature of the function. First derivative of the cost function is the Marginal Cost. There is a good explanation of derivatives as it relates to Cost Function in here dummies.com/education/math/calculus/… and here:columbia.edu/itc/sipa/math/calc_econ_interp_u.html
$endgroup$
– NoChance
Dec 1 '18 at 1:23






$begingroup$
A derivative of a function is NOT an approximation. It has different menaings depending on the nature of the function. First derivative of the cost function is the Marginal Cost. There is a good explanation of derivatives as it relates to Cost Function in here dummies.com/education/math/calculus/… and here:columbia.edu/itc/sipa/math/calc_econ_interp_u.html
$endgroup$
– NoChance
Dec 1 '18 at 1:23












1 Answer
1






active

oldest

votes


















1












$begingroup$

The derivative is actually by definition a limit. The reason that your intuition tells you that $C'(1)$ should be $8$ is because that is the slope of the secant line connecting $C(7)$ and $C(8). The purple line below might help you picture what this looks like intuitively.



The green line represents the derivative "slope" at $x=1$, and the red line represents the slope of the line connecting the two points. Notice that it's not as steep as the red line.






share|cite|improve this answer









$endgroup$













    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%2f3020757%2fis-the-derivative-an-approximation-cost-function-marginal-cost-deviation-in%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    The derivative is actually by definition a limit. The reason that your intuition tells you that $C'(1)$ should be $8$ is because that is the slope of the secant line connecting $C(7)$ and $C(8). The purple line below might help you picture what this looks like intuitively.



    The green line represents the derivative "slope" at $x=1$, and the red line represents the slope of the line connecting the two points. Notice that it's not as steep as the red line.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The derivative is actually by definition a limit. The reason that your intuition tells you that $C'(1)$ should be $8$ is because that is the slope of the secant line connecting $C(7)$ and $C(8). The purple line below might help you picture what this looks like intuitively.



      The green line represents the derivative "slope" at $x=1$, and the red line represents the slope of the line connecting the two points. Notice that it's not as steep as the red line.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The derivative is actually by definition a limit. The reason that your intuition tells you that $C'(1)$ should be $8$ is because that is the slope of the secant line connecting $C(7)$ and $C(8). The purple line below might help you picture what this looks like intuitively.



        The green line represents the derivative "slope" at $x=1$, and the red line represents the slope of the line connecting the two points. Notice that it's not as steep as the red line.






        share|cite|improve this answer









        $endgroup$



        The derivative is actually by definition a limit. The reason that your intuition tells you that $C'(1)$ should be $8$ is because that is the slope of the secant line connecting $C(7)$ and $C(8). The purple line below might help you picture what this looks like intuitively.



        The green line represents the derivative "slope" at $x=1$, and the red line represents the slope of the line connecting the two points. Notice that it's not as steep as the red line.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 23:37









        JB071098JB071098

        370212




        370212






























            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%2f3020757%2fis-the-derivative-an-approximation-cost-function-marginal-cost-deviation-in%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

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
            Read more