623. Find $f^{\prime}(0), f^{\prime}(1), f^{\prime}(2)$, if $f(x)=2 x^{3}+4 x-5$.
Solution. First, we find the derivative in general form. By rule II, and then I, we will have

$$
f^{\prime}(x)=\left(2 x^{3}\right)^{\prime}+(4 x)^{\prime}-(5)^{\prime}=2\left(x^{3}\right)^{\prime}+4(x)^{\prime}-(5)^{\prime}
$$

Using the derivative formulas (1), (2), (3), we finally get

$$
f^{\prime}(x)=6 x^{2}+4
$$

Now we find the values of the derivative at the specified points:

$$
f^{\prime}(0)=4 ; \quad f^{\prime}(1)=10 ; \quad f^{\prime}(2)=28
$$