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
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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









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    active

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    oldest

    votes









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    $begingroup$

    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

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