934. Find the derivative of the function $u=\sqrt{x^{2}+y^{2}}$ at the point $A(3 ; 4):$
1) in the direction of the bisector of the first coordinate angle
2) in the direction of the radius vector of the point $A$;
3) in the direction of the vector $\bar{q}\{4 ;-3\}$.
Solution. We find the partial derivatives of the function $u$ and calculate their values at point $A$:

$$
\begin{gathered}
\frac{\partial u}{\partial x}=\frac{x}{\sqrt{x^{2}+y^{2}}} ;\left.\frac{\partial u}{\partial x}\right|_{A}=\frac{3}{5} ; \\
\frac{\partial u}{\partial y}=\frac{y}{\sqrt{x^{2}+y^{2}}} ;\left.\frac{\partial u}{\partial y}\right|_{A}=\frac{4}{5} \\
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\end{gathered}
$$

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

$$
\left.\frac{\partial u}{\partial l}\right|_{A}=\frac{3}{5} \cos \alpha+\frac{4}{5} \cos \beta
$$

We then find the cosines of the angles $\alpha$ and $\beta$ formed by the given direction of differentiation with the coordinate axes, and the derivative of the function $u$ in the given direction:

1) For the bisector of the first coordinate angle: $\alpha=\beta=45^{\circ}$, $\cos \alpha=\cos \beta=\frac{\sqrt{2}}{2}$

$$
\left.\frac{\partial u}{\partial l_{1}}\right|_{A}=\frac{3}{5} \cdot \frac{\sqrt{2}}{2}+\frac{4}{5} \cdot \frac{\sqrt{2}}{2}=\frac{7 \sqrt{2}}{10}
$$

2) For the vector $\overline{O A}\{3 ; 4\}: \quad \cos \alpha=\frac{3}{5}, \cos \beta=\frac{4}{5}$; $\left.\frac{\partial u}{\partial l_{2}}\right|_{A}=\frac{3^{2}}{5^{2}}+\frac{4^{2}}{5^{2}}=1$.
3) For the vector $\bar{q}\{4 ;-3\}: \cos \alpha=\frac{4}{5}, \cos \beta=-\frac{3}{5} ;\left.\frac{\partial u}{\partial q}\right|_{A}=0$.