Example 1. Derive the Taylor formula for $n=2$ for the function $f(x, y)=x^{y}$ in the neighborhood of the point $M_{0}(1,1)$ and approximate $1.1^{1.02}$.
Solution. We have

$$
f(x, y)=f(1,1)+d f(1,1)+\frac{1}{2} d^{2} f(1,1)+R_{2}
$$

First, we find the partial derivatives of the first two orders:

$$
\begin{gathered}
f_{x}^{\prime}=y \cdot x^{y-1} ; \quad f_{y}^{\prime}=x^{y} \cdot \ln x \\
f_{x^{2}}^{\prime \prime}=y(y-1) x^{y-2} ; f_{x y}^{\prime \prime}=x^{y-1}+x^{y-1} \cdot y \ln x ; f_{y^{2}}^{\prime \prime}=x^{y}(\ln x)^{2}
\end{gathered}
$$

Now we calculate the values of the obtained partial derivatives at the given point $(1,1): f_{x}^{\prime}(1,1)=1 ; f_{y}^{\prime}(1,1)=0 ; f_{x^{2}}^{\prime \prime}=0 ; f_{x y}^{\prime \prime}(1,1)=1 ; f_{y^{2}}^{\prime \prime}(1,1)=0$ and $f(1,1)=1$.

The corresponding Taylor formula is

$$
x^{y}=1+1 \cdot(x-1)+1 \cdot(x-1)(y-1)+R_{2}
$$

i.e.

$$
x^{y} \approx 1+(x-1)+(x-1)(y-1)
$$

Substituting $x=1.1, y=1.02, x-1=0.1, y-1=0.02$, we get $1.1^{1.02} \approx 1+0.1+0.1 \cdot 0.02=1.102$.

## Exercises

Construct the Taylor polynomial of degree $n$ for the given function $f(x, y)$ in the neighborhood of the given point $M_{0}(1-4)$.

1. $f(x, y)=\sqrt{1-x^{2}-y^{2}}, n=2, M_{0}(0,0)$.
2. $f(x, y)=\sin x \cdot \sin y, n=2, M_{0}\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$.
3. $f(x, y)=e^{x+y}, n \in N, M_{0}(0,0)$.
4. $f(x, y)=-x^{2}+2 x y+3 y^{2}-6 x-2 y-4, n=2, M_{0}(-2,1)$.
5. Using Taylor formulas up to terms of the second order, approximate:
1) $0.95^{2.01}$; 2) $\sqrt[3]{8.07} \cdot \sqrt{8.98}$

## Answers

1. $1-\frac{1}{2}\left(x^{2}+y^{2}\right)$. 2. $\frac{1}{2}+\frac{1}{2}\left(x-\frac{\pi}{4}\right)+\frac{1}{2}\left(y-\frac{\pi}{4}\right)-\frac{1}{4}\left[\left(x-\frac{\pi}{4}\right)^{2}-\right.$ $\left.-2\left(x-\frac{\pi}{4}\right)\left(y-\frac{\pi}{4}\right)+\left(y-\frac{\pi}{4}\right)^{2}\right]$. 3. $1+(x+y)+\frac{1}{2!}(x+y)^{2}+$ $+\cdots+\frac{1}{n!}(x+y)^{n}$. 4. $1+(x+2)^{2}+2(x+2)(y-1)+3(y-1)^{2}$. 5. 1) 0.902 . 2) 5.998 .

## § 8. Extremum of a Function of Two Variables

$1^{\circ}$. The function $z=f(x, y)$ has a maximum (minimum) at the point $\left(x_{0}, y_{0}\right)$, if $f\left(x_{0}, y_{0}\right)>f(x, y)\left(f\left(x_{0}, y_{0}\right)<f(x, y)\right)$ for all points $(x, y)$ in some neighborhood of $\left(x_{0}, y_{0}\right)$.
$2^{\circ}$. Let $A=f_{x x}^{\prime \prime}\left(x_{0}, y_{0}\right), B=f_{x y}^{\prime \prime}\left(x_{0}, y_{0}\right), C=f_{y y}^{\prime \prime}\left(x_{0}, y_{0}\right)$. Then:
1) If $A C-B^{2}<0$, then $\left(x_{0}, y_{0}\right)$ is a saddle point.
2) If $A C-B^{2}>0$ and $A<0(C<0)$, then $\left(x_{0}, y_{0}\right)$ is a point of maximum.
3) If $A C-B^{2}>0$ and $A>0(C>0)$, then $\left(x_{0}, y_{0}\right)$ is a point of minimum.
$3^{\circ}$. If $A C-B^{2}=0$, then the test is inconclusive, and further investigation is required.

## Examples with Solutions