Example 1. Find the derivative of the scalar field

$$
u=x y z
$$

at the point $M_{0}(1,-1,1)$ in the direction from point $M_{0}$ to point $M_{1}(2,3,1)$.
Solution. We find the direction cosines of the vector $\overrightarrow{M_{0} M_{1}}=\{1,4,0\}$, the length of which is $\left|\overrightarrow{M_{0} M}\right|=\sqrt{17}$. We have

$$
\cos \alpha=\frac{1}{\sqrt{17}}, \quad \cos \beta=\frac{4}{\sqrt{17}}, \quad \cos \gamma=0
$$

The values of the partial derivatives of the function $u=x y z$ at the point $M_{0}(1,-1,1)$ are

$$
\left.\frac{\partial u}{\partial x}\right|_{\mathcal{K}_{0}}=-1,\left.\quad \frac{\partial u}{\partial y}\right|_{\mathcal{M}_{0}}=1,\left.\quad \frac{\partial u}{\partial z}\right|_{M_{0}}=-1
$$

Using formula (1), we obtain

$$
\left.\frac{\partial u}{\partial t}\right|_{M_{0}}=-\frac{1}{\sqrt{17}}+\frac{4}{\sqrt{17}}-1 \cdot 0 \neq \frac{3}{\sqrt{17}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_267d965536514099194eg-13.jpg?height=64&width=749&top_left_y=986&top_left_x=915)

![](https://cdn.mathpix.com/cropped/2024_05_22_267d965536514099194eg-13.jpg?height=35&width=224&top_left_y=1041&top_left_x=917)

Example 2. Calculate the derivative of the scalar field

$$
u=\operatorname{arctg} x y
$$

along this curve (in the direction of increasing abscissa).

Solution. The direction I of the parabola $y=a^{2}$ at the point $M_{0}(1,1)$ is considered to be the tangent to the parabola at this point (Fig. 12).

Let the tangent I to the curve at the point $M_{0}$ form an angle $\alpha$ with the axis $O_{2}$. We have

$$
y^{\prime}=2 x ; \quad \operatorname{ts} \alpha=\left.y^{\prime}\right|_{z=1}=2
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_267d965536514099194eg-14.jpg?height=265&width=305&top_left_y=179&top_left_x=104)

Fig. 12 from which the direction cosines of the tangent are

$$
\begin{aligned}
& \cos \alpha=\frac{1}{\sqrt{1+\operatorname{tg}^{2} \alpha}}=\frac{1}{\sqrt{5}} \\
& \cos \beta=\sin \alpha=\frac{\lg \alpha}{\sqrt{1+\lg ^{2} \alpha}}=\frac{2}{\sqrt{5}}
\end{aligned}
$$

The values of the partial derivatives of the given function $u(x, y)$ at the point $M_{0}(1,1)$ are:

$$
\left.\frac{\partial u}{\partial x}\right|_{M_{0}}=\left.\frac{y}{1+x^{2} y^{2}}\right|_{M_{0}}=\frac{1}{2},
$$

$$
\left.\frac{\partial u}{\partial y}\right|_{M_{0}}=\left.\frac{x}{1+x^{2} y^{2}}\right|_{M_{0}}=\frac{1}{2}
$$

we obtain

Substituting the found values into formula (2),

$$
\frac{\Delta u}{\Delta l}=\frac{1}{2} \cdot \frac{1}{\sqrt{5}}+\frac{1}{2} \cdot \frac{2}{\sqrt{5}}=\frac{3}{2 \sqrt{5}}
$$