Example 3. Given the function $z=\arcsin \frac{y}{x^{2}}$, point $A(-2,-1)$, and vector $\vec{a}=3 \vec{i}-4 \vec{j}$. Find:

1) $\overrightarrow{\operatorname{grad}} z$ at point $A$;
2) the derivative at point $A$ in the direction of vector $\vec{a}$.
Solution. 1) To find the coordinates of the vector $\overrightarrow{\operatorname{grad}} z$ (see point $3^{\circ}$), we will find the partial derivatives

$$
z_{x}^{\prime}=\frac{1}{\sqrt{1-\left(\frac{y}{x^{2}}\right)^{2}}}\left(-\frac{y}{x^{4}}\right) \cdot 2 x=-\frac{2 y}{x \sqrt{x^{4}-y^{2}}} \quad \text { and } \quad z_{y}^{\prime}=\frac{1}{\sqrt{x^{4}-y^{2}}}
$$

Their values at point $A(-2,-1)$ are as follows:

$$
z_{x}^{\prime}(-2,-1)=-\frac{1}{\sqrt{15}}, \quad z_{y}^{\prime}(-2,-1) \fallingdotseq \frac{1}{\sqrt{15}}
$$

Thus,

$$
\overrightarrow{\operatorname{grad}} z=-\frac{1}{\sqrt{14}} \vec{i}+\frac{1}{\sqrt{15}} \vec{j}
$$

2) Find the direction cosines of the vector $\vec{a}=3 i-4 j$:

$$
\cos \alpha=\frac{3}{\sqrt{9+16}}=\frac{3}{5}, \cos \beta=-\frac{4}{5}=\sin \alpha
$$

Then (see point $2^{\circ}$)

$$
\begin{aligned}
\frac{\partial z}{\partial e}=z_{x}^{\prime}(-2,-1) \cos \alpha+z_{y}^{\prime}(-2, & -1) \sin \alpha= \\
& =-\frac{1}{\sqrt{15}} \cdot \frac{3}{5}+\frac{1}{\sqrt{15}}\left(-\frac{4}{5}\right)=-\frac{7}{5 \sqrt{15}}
\end{aligned}
$$

The function decreases in the direction of vector $\vec{a}$.

## Exercises

1. Find the derivative of the function $z=x^{2}+y^{2}-3 x+2 y$ at point $M_{0}(0,0)$ in the direction of point $M_{1}(3,4)$.
2. Find the derivative of the function $z=\frac{1}{\sqrt{x^{2}+y^{2}}}$ at point $M_{0}(1,-1)$ in the direction of vector $\vec{a}=\{-\sqrt{2}, \sqrt{2}\}$.
3. Find the gradient of the scalar field $z=\frac{x}{x^{2}+y^{2}}$ at point $M_{0}(1,2)$.
4. Find the maximum rate of increase of the scalar field $z=\ln \left(x^{2}+4 y^{2}\right)$ at point $M_{0}(6,4)$.
5. Find the gradient of the function $u=\frac{1}{r^{3}}$, where $r=\sqrt{x^{2}+y^{2}}$.
6. Find the direction of maximum growth of the function $z=3 x^{2}+x y-2 y^{2}$ at point $(2,1)$.

## Answers

1. $\quad-\frac{1}{5} . \quad 2 . \quad \frac{1}{2}$. 3. $\quad \overrightarrow{\operatorname{grad}} z=\left(\frac{3}{5} ;-\frac{4}{5}\right)=\frac{3}{5} \vec{i}-\frac{4}{5} \vec{j} . \quad$ 4. $\quad \frac{\sqrt{73}}{25}$. 5. $-\frac{3(x \vec{i}+y \vec{j})}{r^{5}}$. 6. $\vec{l}=(13,-2)$.

## § 7. Taylor's Formula for Functions of Two Variables

Let $z=f(x, y)$ be a function of two variables $x$ and $y$, continuous along with its partial derivatives up to the $(n+1)$-th order inclusive in a neighborhood of a given point $M_{0}(a, b)$. Then, similarly to the case of functions of one variable (see Chapter VII § 12), the function of two variables can be represented as a polynomial of degree $n$ in $x-a$ and $y-b$ and some remainder term:

$$
f(x, y)=f(a, b)+\frac{1}{1!} d f(a, b)+\frac{1}{2!} d^{2} f(a, b)+\cdots+\frac{1}{n!} d^{n} f(a, b)+R_{n}
$$

Here, it can be assumed that $d x=x-a, d y=y-b$.

The polynomial appearing in this formula is called the Taylor polynomial of the function $f(x, y)$. It represents an approximate value of the given function in the neighborhood of the point $M_{0}(a, b)$.

## Examples with Solutions