Find y' for y = arcsin sqrt(1-4x)

Find $y'$ for $y = \arcsin\sqrt{1-4x}$.

Step 1: Set up the two layers

Let $u = \sqrt{1-4x} = (1-4x)^{1/2}$, so that $y = \arcsin u$ and

\[ \frac{dy}{du} = \frac{1}{\sqrt{1-u^{2}}}, \qquad \frac{du}{dx} = \frac{-4}{2\sqrt{1-4x}} = \frac{-2}{\sqrt{1-4x}}. \]

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

This is where the problem is engineered to simplify: squaring undoes the root exactly.

\[ 1-u^{2} = 1-(1-4x) = 4x. \]

Step 3: Multiply the factors

\[ y' = \frac{1}{\sqrt{4x}}\cdot\frac{-2}{\sqrt{1-4x}} = \frac{-2}{2\sqrt{x}\sqrt{1-4x}} = \frac{-1}{\sqrt{x(1-4x)}}. \]

Step 4: Verification

\[x = \tfrac18:\quad y' = -\frac{1}{\sqrt{\tfrac18\cdot\tfrac12}} = -\frac{1}{\sqrt{1/16}} = -4\]

\[\frac{y(0.125+10^{-6})-y(0.125-10^{-6})}{2\times 10^{-6}} \approx -4.0 \;\;\checkmark\]

Step 5: Domain

The function requires $1-4x \geq 0$ and $\sqrt{1-4x} \leq 1$, i.e. $0 \leq x \leq \tfrac14$. The derivative excludes both endpoints, where the graph has vertical tangents. The derivative is negative throughout: as $x$ grows the radicand shrinks, so the arcsine decreases.