For the following three (03) items: Let y = f(x) = x ^ -1 x 1 - x^2...

Given $y = f(x) = \dfrac{x\sin^{-1} x}{\sqrt{1 - x^2}} + \ln\sqrt{1 - x^2}$

We can rewrite the function as:
$y = \dfrac{x\sin^{-1} x}{\sqrt{1 - x^2}} + \dfrac{1}{2}\ln(1 - x^2)$

Differentiating with respect to $x$ using the product rule and chain rule:
$\dfrac{dy}{dx} = \dfrac{d}{dx}\left(\dfrac{x}{\sqrt{1 - x^2}}\right)\sin^{-1} x + \dfrac{x}{\sqrt{1 - x^2}}\dfrac{d}{dx}(\sin^{-1} x) + \dfrac{1}{2}\dfrac{1}{1 - x^2}(-2x)$

$\dfrac{dy}{dx} = \left[\dfrac{1\cdot\sqrt{1 - x^2} - x\cdot\left(\dfrac{-x}{\sqrt{1 - x^2}}\right)}{1 - x^2}\right]\sin^{-1} x + \dfrac{x}{\sqrt{1 - x^2}}\cdot\dfrac{1}{\sqrt{1 - x^2}} - \dfrac{x}{1 - x^2}$

$\dfrac{dy}{dx} = \left[\dfrac{1 - x^2 + x^2}{(1 - x^2)^{3/2}}\right]\sin^{-1} x + \dfrac{x}{1 - x^2} - \dfrac{x}{1 - x^2}$

$\dfrac{dy}{dx} = \dfrac{\sin^{-1} x}{(1 - x^2)^{3/2}}$

To find the slope of the tangent at $x = 0.5 = \dfrac{1}{2}$, we substitute this value into the derivative:
$\dfrac{dy}{dx}\bigg|_{x=1/2} = \dfrac{\sin^{-1}(1/2)}{(1 - (1/2)^2)^{3/2}}$

$\dfrac{dy}{dx}\bigg|_{x=1/2} = \dfrac{\pi/6}{(3/4)^{3/2}} = \dfrac{\pi/6}{3\sqrt{3}/8}$

$\dfrac{dy}{dx}\bigg|_{x=1/2} = \dfrac{8\pi}{18\sqrt{3}} = \dfrac{4\pi}{9\sqrt{3}}$

Rationalizing the denominator by multiplying the numerator and denominator by $\sqrt{3}$:
$\dfrac{dy}{dx}\bigg|_{x=1/2} = \dfrac{4\pi\sqrt{3}}{27}$