## Thomas' Calculus 13th Edition

$$\approx 15.83 \%$$
Re-write as: $dI=\dfrac{dV}{R}-\dfrac{V \ dR}{R^2}$ $dI(24,100)=\dfrac{dV}{100}-\dfrac{24 \ dR}{(100)^2}$ Set $v=-1, dR=-20$ $\implies dI=-0.01+480(0.0001)=0.038$ Estimated change in $I$ can be computed as: $$I=\dfrac{dI}{I} \times 100 \\ =\dfrac{0.038}{0.24} (100) \approx 15.83 \%$$