Given, $c=15300 \pm 100$. Then what is the maximum absolute error in $c^3$?
My attempt:
Let $u = c^3$, then maximum absolute error in $u$ is $\Delta u = \frac{du}{dc}\times \Delta c = 3c^2\times \Delta c = 3\times 15300^2 \times 100 = 70.23\times 10^9$.
I want to know if my attempt is correct because my textbook gives completely different answer, $5.766\times 10^{12}$.
Please anyone help me clear this doubt. Thanks in advance.
The formula you mention is a linear approximation of the error, so it does not provide an exact bound. Also, when applying the formula, you are assuming that $c=15300$, which is not the case. What you could say using that formula is that
$$ |\Delta u| \approx |f'(c) \Delta c| \leq 3c^2 |\Delta c| \leq 3(15300+100)^2 \cdot 200 \approx 1.42 \times 10^{11} $$
However, as it was pointed out, since $c \in [15200, 15400]$, surely $c^3 \in [ 15200^3, 15400^3]$ ando so, if you approximate $c^3$ by $15300^3$, the maximum absolute error is approx. $7.07 \times 10^{10}$.