1002. Find the total increment and the total differential of the function $z=x^{2}+x y$ at $x=2 ; y=1 ; \Delta x=0.01 ; \Delta y=0.02$.
Solution. We will use formula (1) and the definition of the total differential of the function $z=f(x, y)$:

$$
\begin{gathered}
\Delta z=(x+\Delta x)^{2}+(x+\Delta x)(y+\Delta y)-\left(x^{2}+x y\right)= \\
=(2 x+y) \Delta x+x \Delta y+\Delta x^{2}+\Delta x \Delta y
\end{gathered}
$$

from which

$$
d z=(2 x+y) \Delta x+x \Delta y
$$

Substituting the given values of $x, y, \Delta x, \Delta y$ into these expressions, we get:

$\Delta z=(2 \cdot 2+1) 0.01+2 \cdot 0.02+(0.01)^{2}+0.01 \cdot 0.02=0.0903$ $d z=(2 \cdot 2+1) 0.01+2 \cdot 0.02=0.09$.