Differentiate $\sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}}$ w.r.t. $x$.

$\frac{1}{2}y \left[ \frac{1}{x-3} + \frac{2x}{x^2+4} - \frac{6x+4}{3x^2+4x+5} \right]$

The correct answer is Option (3) → $\frac{1}{2}y \left[ \frac{1}{x-3} + \frac{2x}{x^2+4} - \frac{6x+4}{3x^2+4x+5} \right]$ ##

Let $y = \sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}}$

Taking logarithm on both sides, we have

$\log y = \frac{1}{2} [\log(x-3) + \log(x^2+4) - \log(3x^2+4x+5)]$$

Now, differentiating both sides w.r.t. $x$, we get

$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{(x-3)} + \frac{2x}{x^2+4} - \frac{6x+4}{3x^2+4x+5} \right]$

or $\frac{dy}{dx} = \frac{y}{2} \left[ \frac{1}{(x-3)} + \frac{2x}{x^2+4} - \frac{6x+4}{3x^2+4x+5} \right]$

$= \frac{1}{2} \sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}} \left[ \frac{1}{(x-3)} + \frac{2x}{x^2+4} - \frac{6x+4}{3x^2+4x+5} \right]$