1004. Calculate approximately:
1) $(0.96)^{2}(1.02)^{3}$;
2) $(1.02)^{3.05}$.
Solution. 1) $(0.96)^{2}(1.02)^{3}$ is a particular value of the function $f(x, y)=x^{2} y^{3}$. Using formula (3), we get

$$
(x+\Delta x)^{2}(y+\Delta y)^{3} \approx x^{2} y^{3}+2 x y^{3} \Delta x+3 x^{2} y^{2} \Delta y
$$

Let $x=1, y=1$, then $\Delta x=-0.04$, since $1+\Delta x=0.96$, $\Delta y=0.02$, since $1+\Delta y=1.02$.

Therefore, $\quad(0.96)^{2}(1.02)^{3} \approx 1^{2} \cdot 1^{3}+2 \cdot 1 \cdot 1^{3} \cdot(-0.04)+$ $+3 \cdot 1^{2} \cdot 1^{2} \cdot 0.02=1-0.08+0.06=0.98$.
2) $(1.02)^{3.05}$ is a particular value of the function $z=x^{y}$. Using formula (3), we get

$(x+\Delta x)^{y+\Delta y} \approx x^{y}+\left(x^{y}\right)_{x}^{\prime} \Delta x+\left(x^{y}\right)_{y}^{\prime} \Delta y=x^{y}+y x^{y-1} \Delta x+x^{y} \ln x \Delta y$. Let $x=1, y=3$, then $\Delta x=0.02, \Delta y=0.05$.

Therefore, $(1.02)^{3.05} \approx 1^{3}+3 \cdot 1^{3-1} \cdot 0.02+1^{3} \ln 1 \cdot 0.05=$ $=1+0.06+0=1.06$.

Find the total differentials of the following functions: