The question asks me to find the arc length of $$y= (x-x^2)^{1/2} + \sin^{-1}(x^{1/2})$$ I know I need to take the derivative: $$\frac{1-2x}{2(x-x^2)^{1/2}} + \frac{1}{(1-x)^{1/2}}$$ I've tried manipulating it to simplify it, since I need some equation that is easy to square, but I've gotten nowhere. I tried getting a common denominator but that didn't really help. I also tried rationalizing the denominator but got stuck as well.
Please help!
Ok, sorry! The derivative of $$\sin^{-1}(x^{1/2})$$ should be $$\frac{1}{2(x)^{1/2}(1-x)^{1/2}}$$ That makes more sense!
The second expression in your derivative (the derivative of the arcsin term) is incorrect as you need to use the chain rule.
$$\frac{d}{dx}\sin^{-1}(\sqrt{x})=\frac{d}{d(\sqrt{x})}\sin^{-1}(\sqrt{x})\frac{d}{dx}(\sqrt{x})=\frac{1}{\sqrt{1-x}}\frac{1}{2\sqrt{x}}$$
So the differential should be
$$\frac{dy}{dx}=\frac{1-2x}{2(x-x^2)^{1/2}}+\frac{1}{2x^{1/2}(1-x)^{1/2}}\\=\frac{1-2x}{2x^{1/2}(1-x)^{1/2}}+\frac{1}{2x^{1/2}(1-x)^{1/2}}\\=\frac{2(1-x)}{2x^{1/2}(1-x)^{1/2}}=\sqrt{\frac{1-x}{x}}$$
I will leave it to you to do the rest.