Find the derivative of y = ((x (x))/(sqrt(1-x2))).

The property $\ln\left(\frac{A}{B}\right) = \ln(A) - \ln(B)$ and $\ln(AB) = \ln(A) + \ln(B)$ will be used.

$$y = \ln(x \sin(x)) - \ln(\sqrt{1-x^2})$$ $$y = \ln(x) + \ln(\sin(x)) - \ln((1-x^2)^{1/2})$$ $$y = \ln(x) + \ln(\sin(x)) - \frac{1}{2}\ln(1-x^2)$$

Recall the differentiation rules:
$\frac{d}{dx}(\ln(u)) = \frac{1}{u} \frac{du}{dx}$
$\frac{d}{dx}(\sin(x)) = \cos(x)$
$\frac{d}{dx}(x) = 1$
$\frac{d}{dx}(1-x^2) = -2x$

Differentiate $\ln(x)$:

$$\frac{d}{dx}(\ln(x)) = \frac{1}{x}$$

Differentiate $\ln(\sin(x))$:

$$\frac{d}{dx}(\ln(\sin(x))) = \frac{1}{\sin(x)} \cdot \cos(x) = \cot(x)$$

Differentiate $-\frac{1}{2}\ln(1-x^2)$:

$$\frac{d}{dx}\left(-\frac{1}{2}\ln(1-x^2)\right) = -\frac{1}{2} \cdot \frac{1}{1-x^2} \cdot (-2x) = \frac{x}{1-x^2}$$

$$\frac{dy}{dx} = \frac{1}{x} + \cot(x) + \frac{x}{1-x^2}$$

The final answer is $\frac{1}{x} + \cot(x) + \frac{x}{1-x^2}$.