Example 3.16. Find $u_{x x}^{\prime \prime}, u_{y y}^{\prime \prime}, u_{x y}^{\prime \prime}, u_{y x}^{\prime \prime}, u_{x y z}^{\prime \prime \prime}, u_{x y^{2} z}^{I V}$, if $u=e^{x y z}$.
Solution. According to the definition of higher-order partial derivatives (3.2), it is necessary to first find the corresponding partial derivatives of a lower order. In this case, it is necessary to find the first-order partial derivatives

$$
u_{x}^{\prime}, u_{y}^{\prime}, u_{z}^{\prime} \Rightarrow u_{x}^{\prime}=e^{x y z} y z ; u_{y}^{\prime}=e^{x y z} z x ; u_{z}^{\prime}=e^{x y z} x y
$$

Next, in accordance with (3.2)

$$
\begin{aligned}
& u_{x x}^{\prime \prime}=\frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial \dot{x}}\right)=\frac{\partial}{\partial x}\left(e^{x y z} y z\right)= \\
& \quad=y z\left(e^{x y z}\right)_{x}^{\prime}=y z e^{x y z}(x y z)_{x}^{\prime}=(y z)^{2} e^{x y z} \\
& u_{y y}^{\prime \prime}=\frac{\partial^{2} u}{\partial y^{2}}=\frac{\partial}{\partial y}\left(\frac{\partial u}{\partial y}\right)=\frac{\partial}{\partial y}\left(e^{x y z} x z\right)= \\
& \quad=x z\left(e^{x y z}\right)_{y}^{\prime}=x z e^{x y z}(x y z)_{y}^{\prime}=(x z)^{2} e^{x y z}
\end{aligned}
$$

Similarly,

$$
\begin{gathered}
u_{x y}^{\prime \prime}=\frac{\partial^{2} u}{\partial y \partial x}=\left(u_{x}^{\prime}\right)_{y}^{\prime}=\frac{\partial}{\partial y}\left(\frac{\partial u}{\partial x}\right)=\frac{\partial}{\partial y}\left(\epsilon^{x y z} y z\right)= \\
=y z e^{x y z}(x y z)_{y}^{\prime}+e^{x y z}(y z)_{y}^{\prime}= \\
=e^{x y z} x y z^{2}+e^{x y z} z=e^{x y z}\left(x y z^{2}+z\right) \\
u_{y x}^{\prime \prime}=\frac{\partial^{2} u}{\partial x \partial y}=\left(u_{y}^{\prime}\right)_{x}^{\prime}=\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial y}\right)=\frac{\partial}{\partial x}\left(e^{x y z} x z\right)= \\
=x z e^{x y z}(x y z)_{x}^{\prime}+e^{x y z}(z z)_{x}^{\prime}= \\
=e^{x y z} x y z^{2}+e^{x y z} z=e^{x y z}\left(x y z^{2}+z\right)
\end{gathered}
$$

Now let's find the third-order partial derivative $u_{x y z}^{\prime \prime \prime}$. According to (3.2), we have

$$
\begin{gathered}
u_{x y z}^{\prime \prime \prime}=\frac{\partial}{\partial z}\left(\frac{\partial^{2} u}{\partial y \partial x}\right)=\frac{\partial}{\partial z}\left(u_{x y}^{\prime \prime}\right)=\frac{\partial}{\partial z}\left(e^{x y z}\left(x y z^{2}+z\right)\right)= \\
=e^{x y z}(x y z)_{z}^{\prime}\left(x y z^{2}+z\right)+e^{x y z}\left(x y z^{2}+z\right)_{z}^{\prime}= \\
=e^{x y z} x y\left(x y z^{2}+z\right)+e^{x y z}(2 x y z+1)= \\
=e^{x y z}\left(x^{2} y^{2} z^{2}+x y z+2 x y z+1\right)=e^{x y z}\left(x^{2} y^{2} z^{2}+3 x y z+1\right)
\end{gathered}
$$

The fourth-order partial derivative $u_{x y^{2} z}^{I V}$ will be found as the partial derivative with respect to the variable $z$ of the third-order partial derivative $u_{x y^{2}}^{\prime \prime \prime}$. In turn, the specified third-order derivative is found using the formula $u_{x y^{2}}^{\prime \prime \prime}=$ $=\left(u_{x y}^{\prime \prime}\right)_{y}^{\prime}$, that is, as the partial derivative with respect to the variable $y$ of the second-order partial derivative $u_{x y}^{\prime \prime}$. Thus, in accordance with the above, we have

$$
\begin{gathered}
u_{x y^{2} z}^{I V}=\left(\left(u_{x y}^{\prime \prime}\right)_{y}^{\prime}\right)_{z}^{\prime}=\frac{\partial}{\partial z} \frac{\partial}{\partial y}\left(\frac{\partial^{2} u}{\partial y \partial x}\right)=\frac{\partial}{\partial z} \frac{\partial}{\partial y}\left(e^{x y z}\left(x y z^{2}+z\right)\right)= \\
=\frac{\partial}{\partial z}\left(z\left(\left(e^{x y z}\right)_{y}^{\prime}(x y z+1)+e^{x y z}(x y z+1)_{y}^{\prime}\right)\right)= \\
=\frac{\partial}{\partial z}\left(z\left(e^{x y z} x z,(x y z+1)+e^{x y z} x z\right)\right)=\frac{\partial}{\partial z}\left(z e^{x y z}\left(x^{2} y z^{2}+2 x z\right)\right)= \\
=e^{x y z}\left(x^{2} y z^{2}+2 x z\right)+ \\
\quad+z\left(\left(e^{x y z}\right)_{z}^{\prime}\left(x^{2} y z^{2}+2 x z\right)+e^{x y z}\left(x^{2} y z^{2}+2 x z\right)_{z}^{\prime}\right)= \\
=e^{x y z}\left(x^{2} y z^{2}+2 x z\right)+ \\
\quad+z\left(e^{x y z} x y\left(x^{2} y z^{2}+2 x z\right)+e^{x y z}\left(x^{2} y z+2 x\right)\right)= \\
=e^{x y z}\left(x^{2} y z^{2}+2 x z+x^{3} y^{2} z^{3}+2 x^{2} y z^{2}+2 x^{2} y z^{2}+2 x z\right)= \\
\quad=e^{x y z}\left(x^{3} y^{2} z^{3}+5 x^{2} y z^{2}+4 x z\right)
\end{gathered}
$$