Solution:

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

(i) Logarithmic differentiation

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

$\frac{1}{y} \frac{d y}{d x}=\frac{1}{x^{2}-5 x+8} \frac{d}{d x}\left(x^{2}-5 x+8\right)+\frac{1}{x^{3}+7 x+9} \frac{d}{d x}\left(x^{3}+7 x+9\right)$

$\frac{1}{y} \frac{d y}{d x}=\frac{1}{x^{2}-5 x+8}(2 x-5)+\frac{1}{x^{3}+7 x+9}\left(3 x^{2}+7\right)$

$\frac{d y}{d x}=y\left[\frac{2 x-5}{x^{2}-5 x+8}+\frac{3 x^{2}+7}{x^{3}+7 x+9}\right]$

$\frac{d y}{d x}=y\left[\frac{2 x-5}{x^{2}-5 x+8}+\frac{3 x^{2}+7}{x^{3}+7 x+9}\right]$

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

$\frac{d y}{d x}=y\left[\frac{2 x^{4}+14 x^{2}+18 x-5 x^{3}-35 x-45+3 x^{4}-15 x^{3}+24 x^{2}+7 x^{2}-35 x+56}{\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)}\right]$

$\frac{d y}{d x}=y\left[\frac{5 x^{4}-20 x^{3}+45 x^{2}-52 x+11}{\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)}\right]$

$\frac{d y}{d x}=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)\left[\frac{5 x^{4}-20 x^{3}+45 x^{2}-52 x+11}{\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)}\right] \text { [using the value of y]}$

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

As a result, the value of dy/dx is same as obtained by three different methods.