935. Find the derivative of the function $u=x y+y z+1$ in the direction of the vector $\bar{l}\{12 ;-3 ;-4\}$ at any point and at points $A(0 ;-2 ;-1)$ and $B(3 ; 3 ; 5)$.
Solution. Let's find the partial derivatives of the function and the direction cosines of the vector $\bar{l}$:

$$
\begin{gathered}
u_{x}^{\prime}=y ; u_{y}^{\prime}=x+z ; u_{z}^{\prime}=y \\
\cos \alpha=\frac{12}{13} ; \cos \beta=-\frac{3}{13} ; \quad \cos \gamma=-\frac{4}{13}
\end{gathered}
$$

Substituting into formula (a), we find the derivative of the function $u$ in the direction of $\bar{l}$ at any point:

$$
u_{l}^{\prime}=\frac{12}{13} y-\frac{3}{13}(x+z)-\frac{4}{13} y=\frac{8 y-3(x+z)}{13}
$$

Substituting the coordinates of points $A$ and $B$, we get $u_{l}^{\prime}(A)=-1$; $u_{l}^{\prime}(B)=0$.