Differentiate $\sqrt{\frac{(x-3)(x^{2}+4)}{3x^{2}+4x+5}}$ with respect to $x$.

Let $y = \sqrt{\frac{(x-3)(x^{2}+4)}{3x^{2}+4x+5}}$.
Taking the natural logarithm on both sides,we get:
$\log y = \frac{1}{2} [\log(x-3) + \log(x^{2}+4) - \log(3x^{2}+4x+5)]$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{d}{dx}(\log(x-3)) + \frac{d}{dx}(\log(x^{2}+4)) - \frac{d}{dx}(\log(3x^{2}+4x+5)) \right]$.
Applying the chain rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x-3} + \frac{2x}{x^{2}+4} - \frac{6x+4}{3x^{2}+4x+5} \right]$.
Therefore,the derivative is:
$\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x-3)(x^{2}+4)}{3x^{2}+4x+5}} \left[ \frac{1}{x-3} + \frac{2x}{x^{2}+4} - \frac{6x+4}{3x^{2}+4x+5} \right]$.