Solution 09

 Given: y = sin⁻¹(2x√(1-x²))

Using the chain rule with d/dx[sin⁻¹(u)] = 1/√(1-u²) · du/dx

Let u = 2x√(1-x²)

Step 1: Find du/dx using the product rule. u = 2x · √(1-x²)

du/dx = 2x · d/dx[√(1-x²)] + √(1-x²) · d/dx[2x]

For d/dx[√(1-x²)] = d/dx[(1-x²)^(1/2)] = (1/2)(1-x²)^(-1/2) · (-2x) = -x/√(1-x²)

So: du/dx = 2x · (-x/√(1-x²)) + √(1-x²) · 2 du/dx = -2x²/√(1-x²) + 2√(1-x²) du/dx = (-2x² + 2(1-x²))/√(1-x²) du/dx = (-2x² + 2 - 2x²)/√(1-x²) du/dx = (2 - 4x²)/√(1-x²)

Step 2: Find 1-u². u² = (2x√(1-x²))² = 4x²(1-x²) = 4x² - 4x⁴

1-u² = 1 - (4x² - 4x⁴) = 1 - 4x² + 4x⁴ = (1-2x²)²

Therefore: √(1-u²) = |1-2x²|

Step 3: Apply the chain rule. dy/dx = 1/√(1-u²) · du/dx dy/dx = 1/|1-2x²| · (2-4x²)/√(1-x²) dy/dx = (2-4x²)/(|1-2x²|√(1-x²)) dy/dx = 2(1-2x²)/(|1-2x²|√(1-x²))

Since we typically assume the domain where the derivative exists, and noting that 1-2x² can be positive or negative:

dy/dx = 2/√(1-x²) when 1-2x² > 0 (i.e., when |x| < 1/√2)

dy/dx = -2/√(1-x²) when 1-2x² < 0 (i.e., when |x| > 1/√2)

The most common simplified form is: dy/dx = ±2/√(1-x²)

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