Example 3.35. Find the derivative of the function $u=x^{3} y^{3} z^{3}$ at the point $M(1,1,1)$ in the direction that forms angles of $60^{\circ}, 45^{\circ}, 60^{\circ}$ with the coordinate axes.
Solution. According to (3.26) and (3.29), it is necessary to find the partial derivatives of the function $u$ at point $M$. We have

$$
\frac{\partial u}{\partial x}=3 x^{2} y^{3} z^{3}; \frac{\partial u}{\partial y}=3 x^{3} y^{2} z^{3} ; \frac{\partial u}{\partial z}=3 x^{3} y^{3} z^{2}
$$

At point $M$, the values of the derivatives are

$$
\frac{\partial u(1,1,1)}{\partial x}=3 ; \frac{\partial u(1,1,1)}{\partial y}=3 ; \frac{\partial u(1,1,1)}{\partial z}=3
$$

Substituting the found values into (3.29), we get:

$$
\begin{aligned}
\frac{\partial u(1,1,1)}{\partial \ell}=3 \cos 60^{\circ} & +3 \cos 45^{\circ}+3 \cos 60^{\circ}= \\
& =3\left(\frac{1}{2}+\frac{\sqrt{2}}{2}+\frac{1}{2}\right)=\frac{3}{2}(2+\sqrt{2})
\end{aligned}
$$

Note that the gradient of this function at point $M(1,1,1)$ is $\operatorname{grad} u(1,1,1)=3(\vec{i}+\vec{j}+\vec{k})$.