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
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.