Use logarithmic differentiation to find the de \( y=\sqrt{\left(x^{2}+3\right)(x-3)^{2}} \) \( \frac{d y}{d x}=\square \)

Let \[ y = \sqrt{(x^2+3)(x-3)^2}. \] Taking the natural logarithm of both sides gives \[ \ln y = \ln \left( \sqrt{(x^2+3)(x-3)^2} \right) = \frac{1}{2}\ln\left[(x^2+3)(x-3)^2\right]. \] Using the logarithm property \(\ln(ab)=\ln a+\ln b\), we have \[ \ln y = \frac{1}{2}\left[\ln(x^2+3) + \ln\left((x-3)^2\right)\right]. \] Since \(\ln\left((x-3)^2\right)=2\ln|x-3|\), the equation simplifies to \[ \ln y = \frac{1}{2}\ln(x^2+3) + \ln|x-3|. \] Differentiate both sides with respect to \(x\). The derivative of the left-hand side is \[ \frac{1}{y}\frac{dy}{dx}. \] For the right-hand side, differentiate term by term: \[ \frac{d}{dx}\left[\frac{1}{2}\ln(x^2+3)\right] = \frac{1}{2}\cdot \frac{1}{x^2+3}\cdot 2x = \frac{x}{x^2+3}, \] \[ \frac{d}{dx}\left[\ln|x-3|\right] = \frac{1}{x-3}. \] So, we obtain \[ \frac{1}{y}\frac{dy}{dx} = \frac{x}{x^2+3} + \frac{1}{x-3}. \] Multiplying both sides by \(y\) yields \[ \frac{dy}{dx} = y\left(\frac{x}{x^2+3} + \frac{1}{x-3}\right). \] Finally, substituting back \( y=\sqrt{(x^2+3)(x-3)^2} \) gives \[ \frac{dy}{dx} = \sqrt{(x^2+3)(x-3)^2}\left(\frac{x}{x^2+3} + \frac{1}{x-3}\right). \]