Example 3.36. Find the directional derivative of the function $u=x^{2}+y^{3}+$ $+z^{3}$ at the point $A(1,2,3)$ in the direction leading from this point to the point $B(13,5,7)$.
Solution. Unlike the previous problem, the direction cosines of the unit vector of the given direction are not specified here. They can be easily found if the coordinates of the vector are known: $\overrightarrow{A B}: \overrightarrow{A B}=12 \vec{i}+3 \vec{j}+4 \vec{k}$, then

$$
\begin{gathered}
\cos \alpha=\frac{12}{\sqrt{144+9+16}}=\frac{12}{13}, \cos \beta=\frac{3}{13}, \cos \gamma=\frac{4}{13} \\
\vec{\ell}=\frac{12}{13} \vec{i}+\frac{3}{13} \vec{j}+\frac{4}{13} \vec{k}
\end{gathered}
$$

Now let's find the first-order partial derivatives:

$$
\begin{gathered}
\frac{\partial u(1,2,3)}{\partial x}=3 x^{2}=3 ; \frac{\partial u(1,2,3)}{\partial y}=3 y^{2}=12 \\
\frac{\partial u(1,2.3)}{\partial z}=3 z^{2}=27
\end{gathered}
$$

And the directional derivative of $\vec{\ell}$ at point $A$:

$$
\frac{\partial u(1,2,3)}{\partial \ell}=3 \frac{12}{13}+12 \frac{3}{13}+27 \frac{4}{13}=\frac{36+36+108}{13}=\frac{180}{13}
$$