Answer
$\Delta N \approx \frac{{\partial N}}{{\partial R}}{|_{\left( {L,R,d} \right) = \left( {0.5,0.15,8} \right)}}\Delta R \simeq 1109.06$
Work Step by Step
From Example 5 we obtain the average number $N$ of cycles before the chip fails given by
$N = {\left( {\frac{{2200R}}{{Ld}}} \right)^{1.9}}$
If $L=0.5$, $d=8$ and $R$ is increased from $0.15$ to $0.17$, we estimate the change $\Delta N$ by the linear approximation
$\Delta N \approx \frac{{\partial N}}{{\partial R}}\Delta R$
Since $L$ and $d$ are constant, the partial derivative is
$\frac{{\partial N}}{{\partial R}} = \frac{\partial }{{\partial R}}{\left( {\frac{{2200R}}{{Ld}}} \right)^{1.9}} = 1.9{\left( {\frac{{2200}}{{Ld}}} \right)^{1.9}}{R^{0.9}}$
So,
$\frac{{\partial N}}{{\partial R}}{|_{\left( {L,R,d} \right) = \left( {0.5,0.15,8} \right)}} = 1.9{\left( {\frac{{2200}}{{0.5\cdot8}}} \right)^{1.9}}{0.15^{0.9}} \simeq 55453$
Now, we evaluate $\Delta N$
$\Delta N \approx \frac{{\partial N}}{{\partial R}}{|_{\left( {L,R,d} \right) = \left( {0.5,0.15,8} \right)}}\Delta R = 55453\cdot\left( {0.17 - 0.15} \right) \simeq 1109.06$
