Is the derivative an approximation? (Cost function, marginal cost, deviation in costs)
$begingroup$
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
asked Nov 30 '18 at 22:46
MonoidMonoid
11
$endgroup$
add a comment |
$begingroup$
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
asked Nov 30 '18 at 22:46
MonoidMonoid
11
$endgroup$
$begingroup$
$C(x+1)-C(x)$ 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
add a comment |
$begingroup$
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
asked Nov 30 '18 at 22:46
MonoidMonoid
11
$endgroup$
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
asked Nov 30 '18 at 22:46
MonoidMonoid
11
asked Nov 30 '18 at 22:46
MonoidMonoid
11
asked Nov 30 '18 at 22:46
MonoidMonoid
11
asked Nov 30 '18 at 22:46
MonoidMonoid
11
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
add a comment |
$begingroup$
$C(x+1)-C(x)$ 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
add a comment |
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.
answered Nov 30 '18 at 23:37
JB071098JB071098
363212
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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.
answered Nov 30 '18 at 23:37
JB071098JB071098
363212
$endgroup$
add a comment |
$begingroup$
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.
answered Nov 30 '18 at 23:37
JB071098JB071098
363212
$endgroup$
add a comment |
$begingroup$
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.
answered Nov 30 '18 at 23:37
JB071098JB071098
363212
$endgroup$
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.
answered Nov 30 '18 at 23:37
JB071098JB071098
363212
answered Nov 30 '18 at 23:37
JB071098JB071098
363212
answered Nov 30 '18 at 23:37
JB071098JB071098
363212
answered Nov 30 '18 at 23:37
JB071098JB071098
363212
363212
add a comment |
add a comment |
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$begingroup$
$C(x+1)-C(x)$ 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