Find y' for y = sqrt(1+ arcsin x)
Find $y'$ for $y = \sqrt{1+\arcsin x}$.
Explanation of Method
A square root wrapped around an inverse sine.
Step 1: Apply the chain rule
Write $y = u^{1/2}$ with $u = 1+\arcsin x$, so that
\[ u' = \frac{1}{\sqrt{1-x^{2}}}, \qquad y' = \frac{u'}{2\sqrt{u}} = \frac{1}{2\sqrt{1+\arcsin x}}\cdot\frac{1}{\sqrt{1-x^{2}}}. \]
Step 2: Combine the radicals
\[ y' = \frac{1}{2\sqrt{\bigl(1-x^{2}\bigr)\bigl(1+\arcsin x\bigr)}}. \]
Step 3: Verification
\[x = 0:\quad y = 1,\quad y' = \frac{1}{2\sqrt{1\cdot 1}} = \frac12; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx 0.5 \;\;\checkmark\]
\[x = \tfrac12:\quad y' = \frac{1}{2\sqrt{0.75\bigl(1+\tfrac{\pi}{6}\bigr)}} \approx 0.46774 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{1}{2\sqrt{\bigl(1-x^{2}\bigr)\bigl(1+\arcsin x\bigr)}}\,}\]
Remark
Remark. Two conditions shape the domain. The radicand needs $1+\arcsin x \geq 0$, i.e. $\arcsin x \geq -1$, which holds for $x \geq \sin(-1) \approx -0.8415$; and the derivative additionally needs $|x| \lt 1$. So the formula is valid on $\bigl(\sin(-1),\,1\bigr)$.