Example 3.23. Calculate $\Delta f\left(P_{0}\right), d f\left(P_{0}\right)$, if $f(x, y)=$ $=x^{2} y, x_{0}=5, y_{0}=4, \Delta x=0.1, \Delta y=-0.2$.
Solution. By the definition of the total increment, we have

$$
\begin{gathered}
\Delta f\left(x_{0}, y_{0}\right)=f\left(x_{0}+\Delta x, y_{0}+\Delta y\right)-f\left(x_{0}, y_{0}\right)= \\
=\left(x_{0}+\Delta x\right)^{2}\left(y_{0}+\Delta y\right)-x_{0}^{2} y_{0}= \\
=\left(x_{0}^{2}+2 x_{0} \Delta x+\Delta x^{2}\right)\left(y_{0}+\Delta y\right)-x_{0}^{2} y_{0}= \\
=x_{0}^{2} y_{0}+x_{0}^{2} \Delta y+2 x_{0} y_{0} \Delta x+2 x_{0} \Delta x \Delta y+y_{0} \Delta x^{2}+\Delta x^{2} \Delta y-x_{0}^{2} y_{0}= \\
=2 x_{0} y_{0} \Delta x+x_{0}^{2} \Delta y+y_{0} \Delta x^{2}+2 x_{0} \Delta x \Delta y+\Delta y \Delta x^{2}= \\
=2 \cdot 20 \cdot 0.1-25 \cdot 0.2+4 \cdot 0.01-10 \cdot 0.02-0.2 \cdot 0.01=-0.802
\end{gathered}
$$

Note that the first two terms in the obtained expression are $d f\left(x_{0}, y_{0}\right)$, the next two are $-\frac{1}{2!} d^{2} f\left(x_{0}, y_{0}\right)$, and the last one is $-\frac{1}{3!} d^{3} f\left(x_{0}, y_{0}\right)$, since according to (3.5) and (3.6) we have

$$
\begin{gathered}
d^{2} f=f_{x x}^{\prime \prime} \Delta x^{2}+2 f_{x y}^{\prime \prime} \Delta x \Delta y=2 y_{0} \Delta x^{2}+2 x_{0} \Delta x \Delta y ;\left(f_{y y}^{\prime \prime}=0\right) \\
d^{3} f=3 f_{x^{2} y}^{\prime \prime \prime} \Delta x^{2} \Delta y=6 \Delta x^{2} \Delta y=3!\Delta x^{2} \Delta y
\end{gathered}
$$

(all other partial derivatives in (3.7) in this case are zero).

Thus, $\Delta f=d f+\frac{1}{2!} d^{2} f+\frac{1}{3!} d^{3} f$ (all other differentials of higher order in this case are zero).

Let's find the total differential

$$
\begin{gathered}
d f\left(x_{0}, y_{0}\right)=f_{x}^{\prime}\left(x_{0}, y_{0}\right) \Delta x+f_{y}^{\prime}\left(x_{0}, y_{0}\right) \Delta y=2 x_{0} y_{0} \Delta x+x_{0}^{2} \Delta y= \\
=2 \cdot 20 \cdot 0.1-25 \cdot 0.2=-1
\end{gathered}
$$