Example 3.24. Replacing the increment of the function with its differential, approximately calculate $a=\frac{2.04^{2}}{\sqrt[3]{0.97 \sqrt[5]{1.02^{2}}}}$.
Solution. For the approximate calculation of this number, consider the function of three independent variables $f(x, y, z)=\frac{x^{2}}{\sqrt[3]{y \sqrt[5]{z^{2}}}}$. Its value at $x=2.04, y=0.97$, $z=1.02$ is the desired number $a$. To find $a$, note that at the point $P_{0}(2,1,1)$, that is, when $x_{0}=2, y_{0}=1$, $z_{0}=1$, the value $f\left(x_{0}, y_{0}, z_{0}\right)=\frac{2^{2}}{\sqrt[3]{1 \sqrt[5]{1^{2}}}}=4$. Next, in accordance with formula (3.10) for calculating $f(x, y, z)$, we find the partial derivatives. For this, it is convenient to represent the function $f(x, y, z)$ as $f(x, y, z)=x^{2} y^{-1 / 3} z^{-2 / 15}$. Then

$$
\begin{gathered}
f_{x}^{\prime}=2 x y^{-1 / 3} z^{-2 / 15} ; f_{y}^{\prime}=-\frac{1}{3} x^{2} y^{-4 / 3} z^{-2 / 15} \\
f_{z}^{\prime}=-\frac{2}{5} x^{2} y^{-1 / 3} z^{-17 / 15}
\end{gathered}
$$

We find the values of the partial derivatives at the point

$$
P_{0}: f_{x}^{\prime}=0 ; f_{y}^{\prime}=-\frac{4}{3} ; f_{z}^{\prime}=-\frac{8}{15}
$$

It is also clear that

$$
\begin{gathered}
x-x_{0}=2.04-2=0.04 ; y-y_{0}=0.97-1=-0.03 \\
z-z_{0}=1.02-1=0.02
\end{gathered}
$$

Thus

$$
\begin{aligned}
\frac{2.04^{2}}{\sqrt[3]{0.97 \sqrt[5]{1.02^{2}}}} & =\left.f(x, y, z)\right|_{\substack{x=2.04 \\
y=0.97 \\
z=1.02}} \approx \\
& \approx 4+4 \cdot 0.04+\frac{4}{3} \cdot (-0.03)-\frac{8}{15} \cdot 0.02 \approx 4.16
\end{aligned}
$$

In applications, it is often necessary to find a function of several variables from its total differential, that is, starting from an expression of the form $d f(x, y)=P(x, y) d x+$ $+Q(x, y) d y$, where $P(x, y)$ and $Q(x, y)$ are defined and continuous along with their partial derivatives $\frac{\partial P}{\partial y}$ and $\frac{\partial Q}{\partial x}$ in a closed, bounded, simply connected domain (i.e., a domain bounded by a single closed contour) and satisfy the condition

$$
\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}
$$

Finding the function $f(x, y)$ from its total differential consists of three stages.

The first stage involves determining the domain where condition (3.11) is satisfied.

The second stage consists of integrating the following obvious equalities: $\frac{\partial f}{\partial x}=P(x, y)$ and $\frac{\partial f}{\partial y}=Q(x, y)$, which leads to the expressions:

$$
f(x, y)=\int P(x, y) d x+U(y) \text { and } f(x, y)=\int Q(x, y) d x+V(x)
$$

The third stage involves comparing these two expressions and adding to the known terms of the first expression the missing terms, depending on $y$ from the second expression. It is also possible to use the second expression first; the missing terms, depending on $x$ from the first expression, should be added to it.