Solution:

Let’s take $y=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)$

(i) By using the product rule:

$\begin{aligned}
&\frac{d y}{d x}=\left(x^{2}-5 x+8\right) \frac{d}{d x}\left(x^{3}+7 x+9\right)+\left(x^{3}+7 x+9\right) \frac{d}{d x}\left(x^{2}-5 x+8\right) \\
&\frac{d y}{d x}=\left(x^{2}-5 x+8\right)\left(3 x^{2}+7\right)+\left(x^{3}+7 x+9\right)(2 x-5)
\end{aligned}$

$\frac{d y}{d x}=3 x^{4}+7 x^{2}-15 x^{3}-35 x+24 x^{2}+56+2 x^{4}-5 x^{3}+14 x^{2}-35 x+18 x-45$

$\frac{d y}{d x}=5 x^{4}-20 x^{3}+45 x^{2}+11$

(ii) To obtain a single polynomial expanding the product

$y=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)$

$y=x^{5}+7 x^{3}+9 x^{2}-5 x^{4}-35 x^{2}-45 x+8 x^{3}+56 x+72$

$y=x^{5}-5 x^{4}+15 x^{3}-26 x^{2}+11 x+ 72 $

$\frac{d y}{d x}=5 x^{4}-20 x^{3}+45 x^{2}-52 x+11$