Differentiate $(x^{2}-5x+8)(x^{3}+7x+9)$ using logarithmic differentiation.

Let $y = (x^{2}-5x+8)(x^{3}+7x+9)$.
Taking logarithm on both sides,we obtain:
$\log y = \log(x^{2}-5x+8) + \log(x^{3}+7x+9)$.
Differentiating both sides with respect to $x$,we obtain:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \log(x^{2}-5x+8) + \frac{d}{dx} \log(x^{3}+7x+9)$.
Using the chain rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x^{2}-5x+8} \cdot (2x-5) + \frac{1}{x^{3}+7x+9} \cdot (3x^{2}+7)$.
Therefore:
$\frac{dy}{dx} = y \left[ \frac{2x-5}{x^{2}-5x+8} + \frac{3x^{2}+7}{x^{3}+7x+9} \right]$.
Substituting $y$ back:
$\frac{dy}{dx} = (x^{2}-5x+8)(x^{3}+7x+9) \left[ \frac{2x-5}{x^{2}-5x+8} + \frac{3x^{2}+7}{x^{3}+7x+9} \right]$.
Expanding the terms:
$\frac{dy}{dx} = (2x-5)(x^{3}+7x+9) + (3x^{2}+7)(x^{2}-5x+8)$.
$\frac{dy}{dx} = (2x^{4} + 14x^{2} + 18x - 5x^{3} - 35x - 45) + (3x^{4} - 15x^{3} + 24x^{2} + 7x^{2} - 35x + 56)$.
Combining like terms:
$\frac{dy}{dx} = 5x^{4} - 20x^{3} + 45x^{2} - 52x + 11$.