What is the slope of the tangent to the curve y=f(x) at x=0.5?

Explanation

Rewrite y = x(1-x^2)^{-1/2}\sin^{-1}x + \frac{1}{2}\ln(1-x^2). Differentiating the first term using the product rule: \frac{d}{dx}[x(1-x^2)^{-1/2}]\sin^{-1}x + x(1-x^2)^{-1/2}(\frac{1}{\sqrt{1-x^2}}). The derivative of x(1-x^2)^{-1/2} is (1-x^2)^{-3/2}. The derivative of the logarithmic term is \frac{1}{2}(\frac{-2x}{1-x^2}) = \frac{-x}{1-x^2}. Adding them up, y' = \frac{\sin^{-1}x}{(1-x^2)^{3/2}} + \frac{x}{1-x^2} - \frac{x}{1-x^2} = \frac{\sin^{-1}x}{(1-x^2)^{3/2}}. At x=0.5, \sin^{-1}(0.5) = \pi/6 and (1-0.5^2)^{3/2} = (3/4)^{3/2} = \frac{3\sqrt{3}}{8}. Evaluating gives y'(0.5) = \frac{\pi/6}{3\sqrt{3}/8} = \frac{4\pi\sqrt{3}}{27}.

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