1. **Problem:** Find the second derivative $y''$ of the function $$y = \sqrt{2} - e^{2x} \arcsin \sqrt{1 - e^{2x}}.$$ 2. **Step 1: Understand the function and notation.** We have $$y = \sqrt{2} - e^{2x} \arcsin \left( \sqrt{1 - e^{2x}} \right).$$ The constant term $\sqrt{2}$ disappears upon differentiation. We need to find $y'' = \frac{d^2y}{dx^2}$. 3. **Step 2: Differentiate once to find $y'$.** Use the product rule on $$- e^{2x} \arcsin \left( \sqrt{1 - e^{2x}} \right).$$ Let $$u = e^{2x}, \quad v = \arcsin \left( \sqrt{1 - e^{2x}} \right).$$ Then $$y = \sqrt{2} - u v,$$ so $$y' = - (u' v + u v').$$ 4. **Step 3: Compute $u'$ and $v'$.** $$u' = \frac{d}{dx} e^{2x} = 2 e^{2x}.$$ For $$v = \arcsin \left( \sqrt{1 - e^{2x}} \right),$$ set $$w = \sqrt{1 - e^{2x}} = (1 - e^{2x})^{1/2}.$$ Then $$v = \arcsin(w), \quad v' = \frac{1}{\sqrt{1 - w^2}} \cdot w'.$$ 5. **Step 4: Compute $w'$ and simplify inside the square root.** $$w' = \frac{1}{2} (1 - e^{2x})^{-1/2} \cdot (-2 e^{2x}) = - \frac{e^{2x}}{\sqrt{1 - e^{2x}}}.$$ Also, $$1 - w^2 = 1 - (1 - e^{2x}) = e^{2x}.$$ So $$v' = \frac{1}{\sqrt{e^{2x}}} \cdot \left(- \frac{e^{2x}}{\sqrt{1 - e^{2x}}} \right) = \frac{1}{e^{x}} \cdot \left(- \frac{e^{2x}}{\sqrt{1 - e^{2x}}} \right) = - \frac{e^{x}}{\sqrt{1 - e^{2x}}}.$$ 6. **Step 5: Substitute $u'$, $v$, $u$, and $v'$ back into $y'$.** $$y' = - \left( 2 e^{2x} \cdot \arcsin \left( \sqrt{1 - e^{2x}} \right) + e^{2x} \cdot \left(- \frac{e^{x}}{\sqrt{1 - e^{2x}}} \right) \right) = - 2 e^{2x} \arcsin \left( \sqrt{1 - e^{2x}} \right) + \frac{e^{3x}}{\sqrt{1 - e^{2x}}}.$$ 7. **Step 6: Differentiate $y'$ to find $y''$.** $$y'' = \frac{d}{dx} \left( - 2 e^{2x} \arcsin \left( \sqrt{1 - e^{2x}} \right) + \frac{e^{3x}}{\sqrt{1 - e^{2x}}} \right).$$ 8. **Step 7: Differentiate each term separately.** - For $$- 2 e^{2x} \arcsin \left( \sqrt{1 - e^{2x}} \right),$$ use product rule again: $$-2 \left( 2 e^{2x} \arcsin \left( \sqrt{1 - e^{2x}} \right) + e^{2x} \cdot v' \right) = -4 e^{2x} \arcsin \left( \sqrt{1 - e^{2x}} \right) - 2 e^{2x} v'.$$ Recall from step 5 that $$v' = - \frac{e^{x}}{\sqrt{1 - e^{2x}}}.$$ So $$- 2 e^{2x} v' = - 2 e^{2x} \left(- \frac{e^{x}}{\sqrt{1 - e^{2x}}} \right) = \frac{2 e^{3x}}{\sqrt{1 - e^{2x}}}.$$ - For $$\frac{e^{3x}}{\sqrt{1 - e^{2x}}},$$ use quotient or product rule: Rewrite as $$e^{3x} (1 - e^{2x})^{-1/2}.$$ Derivative is $$3 e^{3x} (1 - e^{2x})^{-1/2} + e^{3x} \cdot \left(- \frac{1}{2} \right) (1 - e^{2x})^{-3/2} \cdot (-2 e^{2x} \cdot 2)$$ (since derivative of $1 - e^{2x}$ is $-2 e^{2x} \cdot 2$). Simplify the second term: $$- \frac{1}{2} \cdot (-4 e^{2x}) = 2 e^{2x}.$$ So the derivative is $$3 e^{3x} (1 - e^{2x})^{-1/2} + 2 e^{3x} e^{2x} (1 - e^{2x})^{-3/2} = 3 e^{3x} (1 - e^{2x})^{-1/2} + 2 e^{5x} (1 - e^{2x})^{-3/2}.$$ 9. **Step 8: Combine all parts for $y''$.** $$y'' = -4 e^{2x} \arcsin \left( \sqrt{1 - e^{2x}} \right) + \frac{2 e^{3x}}{\sqrt{1 - e^{2x}}} + 3 e^{3x} (1 - e^{2x})^{-1/2} + 2 e^{5x} (1 - e^{2x})^{-3/2}.$$ 10. **Step 9: Simplify terms with common denominators.** Note that $$\frac{2 e^{3x}}{\sqrt{1 - e^{2x}}} + 3 e^{3x} (1 - e^{2x})^{-1/2} = 5 e^{3x} (1 - e^{2x})^{-1/2}.$$ 11. **Final answer:** $$\boxed{y'' = -4 e^{2x} \arcsin \left( \sqrt{1 - e^{2x}} \right) + 5 e^{3x} (1 - e^{2x})^{-1/2} + 2 e^{5x} (1 - e^{2x})^{-3/2}}.$$