794. Find the derivatives of the specified order of the given functions:
1) $y=x^{3}+2 x^{2}-4 x+1 ; y^{\prime \prime \prime}=$ ?
2) $y=\ln x ; y^{(4)}=$ ?
3) $s=t^{2}-t+1 ; s^{\prime \prime}(0)=$ ?
4) $f(x)=\sin 2 x ; f^{(5)}\left(\frac{\pi}{2}\right)=$ ?
Solution. 1) Find the first derivative

$$
y^{\prime}=3 x^{2}+4 x-4
$$

Now find the second derivative

$$
y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(3 x^{2}+4 x-4\right)^{\prime}=6 x+4
$$

Differentiate one more time to get the third-order derivative

$$
y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=(6 x+4)^{\prime}=6
$$

2) Similarly, find the first-order derivative, and then the higher-order derivatives:

$$
\begin{gathered}
y^{\prime}=\frac{1}{x} ; \quad y^{\prime \prime}=\left(\frac{1}{x}\right)^{\prime}=\left(x^{-1}\right)^{\prime}=-x^{-2} ; \\
y^{\prime \prime \prime}=\left(-x^{-2}\right)^{\prime}=-\left(x^{-2}\right)^{\prime}=2 x^{-3} ; y^{(4)}=\left(2 x^{-3}\right)^{\prime}= \\
=2\left(x^{-3}\right)^{\prime}=-6 x^{-4}=-\frac{6}{x^{4}} .
\end{gathered}
$$

3) $s^{\prime}=2 t-1 ; s^{\prime \prime}=2$, therefore, $s^{\prime \prime}(0)=2$, since the second derivative of the given function $s^{\prime \prime}(t)=2$ for any $t$.
4) $f^{\prime}(x)=2 \cos 2 x ; \quad f^{\prime \prime}(x)=-4 \sin 2 x ; \quad f^{\prime \prime \prime}(x)=-8 \cos 2 x ;$ $f^{(4)}(x)=16 \sin 2 x ; f^{(5)}(x)=32 \cos 2 x$.

At $x=\frac{\pi}{2}$, find

$$
f^{(5)}\left(\frac{\pi}{2}\right)=32 \cos \pi=-32
$$