Example 3.19. Find $d^{2} f$ :

$$
\text { 1) } f(x, y)=x^{6} y^{8}, \text { 2) } f(x, y)=x^{6}+y^{8}
$$
Solution. 1) To compute using formula (3.5), we need to find the necessary second-order partial derivatives. We have:

$$
\begin{gathered}
\frac{\partial f}{\partial x}=6 x^{5} y^{8} ; \frac{\partial f}{\partial y}=8 x^{6} y^{7} ; \frac{\partial^{2} f}{\partial x^{2}}=\left(6 x^{5} y^{8}\right)_{x}^{\prime}=30 x^{4} y^{8} \\
\frac{\partial^{2} f}{\partial y \partial x}=\left(6 x^{5} y^{8}\right)_{y}^{\prime}=48 x^{5} y^{7} ; \frac{\partial^{2} f}{\partial y^{2}}=\left(8 x^{6} y^{7}\right)_{y}^{\prime}=56 x^{6} y^{6} \\
\frac{\partial^{2} f}{\partial x \partial y}=\left(8 x^{6} y^{7}\right)_{y}^{\prime}=48 x^{5} y^{7}
\end{gathered}
$$

Substituting the obtained expressions for the second-order partial derivatives into (3.5), we get

$$
d^{2} f=d^{2}\left(x^{6} y^{8}\right)=30 x^{4} y^{8} d x^{2}+96 x^{5} y^{7} d x d y+56 x^{6} y^{6} d y^{2}
$$

2) We find sequentially

$$
\begin{aligned}
& \frac{\partial f}{\partial x}=6 x^{5} ; \frac{\partial f}{\partial y}=8 y^{7} ; \frac{\partial^{2} f}{\partial x^{2}}=30 x^{4} \\
& \frac{\partial^{2} f}{\partial y \partial x}=0 ; \frac{\partial^{2} f}{\partial y^{2}}=56 y^{6} ; \frac{\partial^{2} f}{\partial x \partial y}=0
\end{aligned}
$$

Thus, $d^{2} f=d^{2}\left(x^{6}+y^{8}\right)=30 x^{4} d x^{2}+56 y^{6} d y^{2}$.