d/dx [sin^-1(x) + x√(1-x²)]= d/dx sin^-1(x) + d/dx x√(1-x²)= 1/√(1-x²) + [√(1-x²) + x*(-x)/√(1-x²)]= 1/√(1-x²) + √(1-x²) - x²/√(1-x²)= 1/√(1-x...d/dx sin^-1(x)是怎么变到1/√(1-x2) 的？反函數的導數：(sin^-1(x))' = 1/√(1-x^2)證明：設y = sin^-1(x)siny = x，兩邊取導數cosy * dy/dx = 1dy/dx = 1/cosy∵siny = x ，cosy = √(1-sin^2(y)) = √(1-x^2)∴ dy/dx = 1/√(1-x^2)