815. Find the differentials of the functions:
1) $y=x^{3}$ at the point $x=0$, if $\left.\Delta x=0.3 ; 2\right) y=x^{3}-x^{2}+$ $+3 x-1$; 3) $r=\varphi^{4}+2^{\sin 3 \varphi}$; 4) $y=\ln \left(x^{2}+1\right)+\operatorname{arctg} \sqrt{x}$ at the point $x=1$, if $\Delta x=0.1$.
Solution. 1) Let's find the increment of the given function at the point $x$:

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

The first term of the function's increment contains $\Delta x$ in the first degree, i.e., it is the linear part of the function's increment with respect to $\Delta x$. Therefore, by definition, $d y=3 x^{2} \Delta x$. If $x=0, \Delta x=0.3$, then $d f(0)=0$.

In the subsequent examples, we will find the differentials of the functions much more simply, using formula (3):
2) $d y=y^{\prime} d x=\left(x^{3}-x^{2}+3 x-1\right)^{\prime} d x=\left(3 x^{2}-2 x+3\right) d x$.
3) $d r=\left(\varphi^{4}+2^{\sin 3 \varphi}\right)^{\prime} d \varphi=\left(4 \varphi^{3}+3 \cos 3 \varphi 2^{\sin 3 \varphi} \ln 2\right) d \varphi$.
4) $d y=\left(\ln \left(x^{2}+1\right)+\operatorname{arctg} \sqrt{x}\right)^{\prime} d x=\left(\frac{2 x}{x^{2}+1}+\frac{1}{2(1+x) \sqrt{x}}\right) d x$. Assuming $x=1$ and $d x=0.1$, we find $d y=\left(\frac{2}{2}+\frac{1}{2 \cdot 2}\right) 0.1=0.125$.

816. Find the differentials of the following functions:

1) $y=\sin x$ at the point $x_{0}=0$, if $\Delta x=\frac{\pi}{2}$;
2) $y=\operatorname{tg} x$ at the point $x_{0}=\frac{\pi}{4}$, if $\Delta x=0.5 ; 0.1 ; 0.01$;
3) $y=x^{2}$ at the point $x_{0}=1$, if $\Delta x=2 ; 1 ;-0.1 ;-0.01$. Find the differentials of the following functions: