Find y' for y = arcsin sqrt(1-x²)

Find $y'$ for $y = \arcsin\sqrt{1-x^{2}}$.

Step 1: Set up the two layers

Let $u = \sqrt{1-x^{2}}$, so that $y = \arcsin u$ and

\[ \frac{dy}{du} = \frac{1}{\sqrt{1-u^{2}}}, \qquad \frac{du}{dx} = \frac{-2x}{2\sqrt{1-x^{2}}} = \frac{-x}{\sqrt{1-x^{2}}}. \]

Step 2: Compute $1-u^{2}$

\[ 1-u^{2} = 1-\bigl(1-x^{2}\bigr) = x^{2}, \qquad\text{so}\qquad \sqrt{1-u^{2}} = \sqrt{x^{2}} = |x|. \]

The absolute value is the whole point of this problem: $\sqrt{x^{2}}$ is $|x|$, not $x$.

Step 3: Multiply

\[ y' = \frac{1}{|x|}\cdot\frac{-x}{\sqrt{1-x^{2}}} = -\frac{x}{|x|}\cdot\frac{1}{\sqrt{1-x^{2}}} = -\frac{\operatorname{sgn}(x)}{\sqrt{1-x^{2}}}. \]

Written out by cases: $y' = -\frac{1}{\sqrt{1-x^{2}}}\ \ (x \gt 0), \qquad y' = +\frac{1}{\sqrt{1-x^{2}}}\ \ (x \lt 0)$.

Step 4: Verification

On $[-1,1]$ the function is simply $\arcsin\sqrt{1-x^{2}} = \arccos|x|$, whose derivative is

\[\bigl(\arccos|x|\bigr)' = -\frac{1}{\sqrt{1-x^{2}}}\cdot\operatorname{sgn}(x) \;\;\checkmark\]

\[x = \tfrac12:\quad \arcsin\sqrt{0.75} = \frac{\pi}{3} = \arccos\tfrac12 \;\;\checkmark;\qquad y' = -\frac{1}{\sqrt{0.75}} \approx -1.15470 \;\;\checkmark\]