In Table 4.2, Hughes & Hase claim that for the function $Z(A,B) = k\frac{A^n}{B^m}$, the fractional uncertainty in $Z$ is given by $$\frac{\alpha_z}{Z} = \sqrt{\left(n\frac{\alpha_A}{A}\right)^2 + \left(m\frac{\alpha_B}{B}\right)^2}. $$ Use the calculus approach to prove this result.
We know that
$$ \alpha_Z = \sqrt{\left(\alpha_Z^A\right)^2 + \left(\alpha_Z^B\right)^2}, $$where $\alpha_Z^A$ is the uncertainty in $Z$ due to the uncertainty in $A$, and $\alpha_Z^B$ is the uncertainty in $Z$ due to the uncertainty in $B$. The calculus approach says that
$$ \alpha_Z^A = \left(\frac{\partial Z}{\partial A}\right)\alpha_A\quad\mbox{and}\quad \alpha_Z^B= \left(\frac{\partial Z}{\partial B}\right)\alpha_B. $$If $Z = k\frac{A^n}{B^m}$, then
$$ \alpha^A_Z = nk\frac{A^{n-1}}{B^m} \alpha_A = n\frac{Z}{A}\alpha_A,\quad\mbox{and}\quad \alpha^B_Z = -m\frac{A^n}{B^{m+1}}\alpha_B = -m\frac{Z}{B}\alpha_B. $$Combining these results
\begin{eqnarray*} \alpha_Z &=& \sqrt{\left(\alpha_Z^A\right)^2 + \left(\alpha_Z^B\right)^2} \\ &=& \sqrt{\left(n\frac{Z}{A}\alpha_A\right)^2 + \left(-m\frac{Z}{B}\alpha_B\right)^2} \end{eqnarray*}or
$$ \frac{\alpha_Z}{Z}= \sqrt{\left(n\frac{\alpha_A}{A}\right)^2 + \left(m\frac{\alpha_B}{B}\right)^2}, $$as desired.