Example 5. Determine the increments of the function $z=x y$, when $x$ and $y$ change from the point $M_{0}(1,2)$ to the points: $M_{1}(1,1 ; 2), \quad M_{2}(1 ; 1,9)$, $M_{3}(1,1 ; 2,2)$.
## Solution.

1. When $x$ and $y$ change from point $M_{0}(1,2)$ to point $M_{1}(1.1, 2)$, only the argument $x$ undergoes an increment, with $\Delta x=1.1-1=0.1$. The partial increment of the function with respect to $x$ is determined by formula (6.1):

$$
\begin{gathered}
\Delta_{x} z=(x+\Delta x) y-x y=x y+\Delta x y-x y=y \Delta x \\
\Delta_{x} z=y \Delta x, \Delta_{x} z=2 \cdot 0.1=0.2
\end{gathered}
$$

2. When $x$ and $y$ change from point $M_{0}(1,2)$ to point $M_{2}(1, 1.9)$, only the argument $y$ undergoes an increment, with $\Delta y=1.9-2=-0.1$.

The partial increment of the function with respect to $y$ is calculated using formula (6.2):

$$
\begin{gathered}
\Delta_{y} z=x(y+\Delta y)-x y=x y+x \Delta y-x y=x \Delta y \\
\Delta_{y} z=x \Delta y, \Delta_{y} z=1(-0.1)=-0.1
\end{gathered}
$$

3. In the last case (when changing from $M_{0}$ to $M_{3}$), both arguments undergo increments, with $\Delta x=1.1-1=0.1 ; \quad \Delta y=$ $=2.2-2=0.2$.

The total increment of the function is determined by formula (6.3):

$$
\begin{gathered}
\Delta z=(x+\Delta x)(y+\Delta y)-x y=x y+x \Delta y+y \Delta x+\Delta x \Delta y-x y= \\
=x \Delta y+y \Delta x+\Delta x \Delta y, \Delta z=1 \cdot 0.2+2 \cdot 0.1+0.1 \cdot 0.2=0.42
\end{gathered}
$$

Remark. From the formulas $\Delta_{x} z=y \Delta x, \Delta_{y} z=x \Delta y, \Delta z=x \Delta y+y \Delta x+$ $+\Delta x \Delta y$, it is clear that $\Delta z \neq \Delta_{x} z+\Delta_{y} z$.