Question:

<p><strong>Step 1:</strong> Find the first derivative y'.</p><p>Given: y = e^(2sin⁻¹x)</p><p>Using chain rule: y' = e^(2sin⁻¹x) · 2 · d/dx(sin⁻¹x)</p><p>y' = e^(2sin⁻¹x) · 2 · 1/√(1-x²)</p><p>y' = 2y/√(1-x²)</p><p></p><p><strong>Step 2:</strong> Find the second derivative y''.</p><p>From y' = 2y/√(1-x²), differentiate using quotient rule:</p><p>y'' = 2[y'√(1-x²) - y · (-2x)/(2√(1-x²))]/(1-x²)</p><p>y'' = 2[y'√(1-x²) + xy/√(1-x²)]/(1-x²)</p><p>y'' = 2[y'(1-x²) + xy]/[(1-x²)^(3/2)]</p><p></p><p><strong>Step 3:</strong> Substitute y' = 2y/√(1-x²) into the expression for y''.</p><p>y'' = 2[2y/√(1-x²) · (1-x²) + xy]/[(1-x²)^(3/2)]</p><p>y'' = 2[2y√(1-x²) + xy]/[(1-x²)^(3/2)]</p><p>y'' = [4y√(1-x²) + 2xy]/[(1-x²)^(3/2)]</p><p></p><p><strong>Step 4:</strong> Calculate (x²-1)y'' + xy'.</p><p>(x²-1)y'' = (x²-1) · [4y√(1-x²) + 2xy]/[(1-x²)^(3/2)]</p><p>= -(1-x²) · [4y√(1-x²) + 2xy]/[(1-x²)^(3/2)]</p><p>= -[4y√(1-x²) + 2xy]/√(1-x²)</p><p>= -[4y + 2xy/√(1-x²)]</p><p></p><p>xy' = x · 2y/√(1-x²) = 2xy/√(1-x²)</p><p></p><p><strong>Step 5:</strong> Add the terms.</p><p>(x²-1)y'' + xy' = -4y - 2xy/√(1-x²) + 2xy/√(1-x²)</p><p>= -4y</p><p></p><p><strong>Step 6:</strong> Divide by y.</p><p>[(x²-1)y'' + xy']/y = -4y/y = -4</p><p></p><p>∴ Answer: <strong>-4</strong></p>