Answer
We use the linear approximation, Eq. (5) to show that
$\frac{{\Delta I}}{I} \approx a\frac{{\Delta x}}{x} + b\frac{{\Delta y}}{y}$
Work Step by Step
We are given $I = {x^a}{y^b}$.
So, the partial derivatives with respect to $x$ and $y$ are
$\frac{{\partial I}}{{\partial x}} = a{x^{a - 1}}{y^b}$ ${\ \ }$ and ${\ \ }$ $\frac{{\partial I}}{{\partial y}} = b{x^a}{y^{b - 1}}$,
respectively.
Using the linear approximation, Eq. (5) the change in $I$ is estimated to be
$\Delta I \approx \frac{{\partial I}}{{\partial x}}\Delta x + \frac{{\partial I}}{{\partial y}}\Delta y$
$\Delta I \approx a{x^{a - 1}}{y^b}\Delta x + b{x^a}{y^{b - 1}}\Delta y$
Since $I = {x^a}{y^b}$, so $a{x^{a - 1}}{y^b} = \frac{{aI}}{x}$ and $b{x^a}{y^{b - 1}} = \frac{{bI}}{y}$.
Thus, $\Delta I \approx \frac{{aI}}{x}\Delta x + \frac{{bI}}{y}\Delta y$.
Hence,
$\frac{{\Delta I}}{I} \approx a\frac{{\Delta x}}{x} + b\frac{{\Delta y}}{y}$
