Example 3.29. Find $\frac{\partial^{2} z}{\partial y \partial x}$ if $Z=f\left(x^{2}+y^{2}, x^{2} y^{2}\right)$.
Solution. The desired partial derivative will be found as the partial derivative with respect to the variable of the partial derivative $\frac{\partial z}{\partial x}$, which has already been found in problem $3.28(2)$. We have

$$
\begin{gathered}
\frac{\partial^{2} z}{\partial y \partial x}=\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial u} 2 x+\frac{\partial f}{\partial v} 2 x y^{2}\right)+\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial u}\right) 2 x+ \\
+\frac{\partial f}{\partial u} \frac{\partial}{\partial y}(2 y)+\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial v}\right) 2 x y^{2}+\frac{\partial f}{\partial v} \frac{\partial}{\partial y}\left(2 x y^{2}\right)= \\
=\frac{\partial^{2} f}{\partial u^{2}} 4 x^{2}+\frac{\partial^{2} f}{\partial v \partial u} 2 x 2 x y^{2}+\frac{\partial^{2} f}{\partial u \partial v} 2 x 2 x y^{2}+\frac{\partial^{2} f}{\partial v^{2}} 4 x^{2} y^{4}+\frac{\partial f}{\partial v} 4 x y= \\
=\frac{\partial^{2} f}{\partial u^{2}} 4 x^{2}+8 x^{2} y^{2} \frac{\partial^{2} f}{\partial u \partial v}+\frac{\partial^{2} f}{\partial v^{2}} 4 x^{2} y^{4}+\frac{\partial f}{\partial v} 4 x y .
\end{gathered}
$$

Here we have taken into account that $\frac{\partial}{\partial y}(2 x)=0$, since $x$ and $y$ are independent variables, and also that for a twice differentiable function, the second-order partial derivatives are continuous, and therefore in the case of calculating mixed derivatives $\frac{\partial^{2} f}{\partial u \partial v}$ and $\frac{\partial^{2} f}{\partial v \partial u}$, their magnitude does not depend on the order of differentiation.

Using the apparatus of differential calculus of functions of several variables, various expressions containing partial derivatives can be transformed into new variables, which is particularly widely used in solving various problems of mathematical physics.