731. Calculate the approximate value:
1) $1.08^{3.96}$;
2) $\frac{\sin 1.49 \cdot \operatorname{arctg} 0.07}{2^{2.95}}$.
Solution. If it is required to calculate the value of the function $f(x, y, \ldots, t)$ at the point $M_{1}\left(x_{1}, y_{1}, \ldots, t_{1}\right)$ and if it is easier to calculate the values of this function and its partial derivatives at the point $M_{0}\left(x_{0}, y_{0}, \ldots, t_{0}\right)$, then for sufficiently small, in absolute value, differences $x_{1}-x_{0}=d x, y_{1}-y_{0}=d y, \ldots$, $t_{1}-t_{0}=d t$ one can replace the total increment of the function by its total differential:

$$
f\left(M_{1}\right)-f\left(M_{0}\right) \approx f_{x}^{\prime}\left(M_{0}\right) d x+f_{y}^{\prime}\left(M_{0}\right) d y+\ldots+f_{t}^{\prime}\left(M_{0}\right) d t
$$

and from this, find the approximate value of the desired quantity by the formula

$$
f\left(M_{1}\right) \approx f\left(M_{0}\right)+f_{x}^{\prime}\left(M_{0}\right) d x+f_{y}^{\prime}\left(M_{0}\right) d y+\ldots+f_{t}^{\prime}\left(M_{0}\right) d t
$$

1) Assuming that $1.08^{3.96}$ is a particular value of the function $f(x, y)=x^{y}$ at the point $M_{1}(1.08 ; 3.96)$ and that the auxiliary point will be $M_{0}(1 ; 4)$, we get

$$
\begin{gathered}
f\left(M_{0}\right)=1^{4}=1 ; f_{x}^{\prime}\left(M_{0}\right)=\left.y x^{y-1}\right|_{\substack{x=1 \\
y=4}}=4 ; f_{1 \prime}^{\prime}\left(M_{0}\right)=\left.x^{y} \ln x\right|_{\substack{x=1 \\
y=4}}=0 \\
d x=1.08-1=0.08 ; d y=3.96-4=-0.04
\end{gathered}
$$

Substituting into formula ( $a$ ), we find

$$
1.08^{3.96} \approx f\left(M_{0}\right)+f_{x}^{\prime}\left(M_{0}\right) d x+f_{y}^{\prime}\left(M_{0}\right) d y=1+4 \cdot 0.08=1.32
$$

2) Let $\frac{\sin 1.49 \cdot \operatorname{arctg} 0.07}{2^{2.95}}$ be a particular value of the function of three variables $\varphi(x, y, z)=2^{x} \sin y \operatorname{arctg} z$ at the point $M_{1}(-2.95 ; 1.49$; $0.07)$ and let the auxiliary point be $M_{0}\left(-3 ; \frac{\pi}{2} ; 0\right)$. Then $\quad a^{\prime} x=-2.95-(-3)=0.05 ; \quad d y=1.49-1.57=-0.08$; $d z=0.07$ $\varphi\left(M_{0}\right)=2^{-3} \sin \frac{\pi}{2} \operatorname{arctg} 0=0 ; \varphi_{x}^{\prime}\left(M_{0}\right)=\left.2^{x} \ln 2 \cdot \sin y \operatorname{arctg} z\right|_{M_{0}}=0 ;$ $\varphi_{1 /}^{\prime}\left(M_{r_{1}}\right)=2^{x} \cos y \operatorname{arctg} z\left|M_{0}=0 ; \varphi_{z}^{\prime}\left(M_{0}\right)=\frac{2^{x} \sin y}{1+z^{2}}\right|_{M_{0}}=2^{-3}$.

Substituting into formula ( $a$ ), we get

$$
\frac{\sin 1.49 \cdot \operatorname{arc} 1 \operatorname{g~} 0.07}{2^{2.95}} \approx 2^{-3} \cdot 0.07 \approx 0.01
$$

Find the total differentials of the functions: