Find y' for y = arcsin(2 sin x)
Find $y'$ for $y = \arcsin(2\sin x)$.
Explanation of Method
An arcsine of a stretched sine, with a restricted domain.
Step 1: Apply the chain rule
With $u = 2\sin x$ and $u' = 2\cos x$: $y' = \frac{u'}{\sqrt{1-u^{2}}} = \frac{2\cos x}{\sqrt{1-4\sin^{2}x}}$.
Step 2: Verification
\[x = 0:\quad y = \arcsin 0 = 0,\quad y' = \frac{2}{\sqrt{1}} = 2; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx 2.0 \;\;\checkmark\]
\[x = \tfrac{\pi}{12}:\quad \sin x \approx 0.25882 \;\Longrightarrow\; y' = \frac{2\cos(\pi/12)}{\sqrt{1-4(0.25882)^{2}}} \approx 2.25789 \;\;\checkmark\]
Step 3: Domain
The arcsine needs $|2\sin x| \leq 1$, that is $|\sin x| \leq \tfrac12$, which holds on the intervals $\left[k\pi-\tfrac{\pi}{6},\,k\pi+\tfrac{\pi}{6}\right]$. The derivative excludes the endpoints, where $1-4\sin^{2}x = 0$ and the graph has vertical tangents.
Result
Final answer:
\[\boxed{\,y' = \dfrac{2\cos x}{\sqrt{1-4\sin^{2}x}}, \qquad |\sin x| < \tfrac12\,}\]
Remark
Remark. Unlike $\arcsin(\sin x)$, which is piecewise linear with slope $\pm 1$, the factor $2$ breaks the exact cancellation: the function exists only on a disconnected union of intervals, and on each of them it runs from $-\tfrac{\pi}{2}$ to $\tfrac{\pi}{2}$ with infinite slope at both ends.