Question

求复合函数y=（x^2）/（根号(1+x^2)）的导数

Answer 1

答案：x(2 + x²)/(1 + x²)^(3/2)

d/dx √(1 + x²) = 2x/2√(1 + x²) = x/√(1 + x²)
y = x²/√(1 + x²)
dy/dx = [√(1 + x²) d/dx (x²) - x² d/dx √(1 + x²)]/(1 + x²) <=商法则(u/v)' = (vu' - uv')/v²
= [√(1 + x²) • 2x - x² • x/√(1 + x²)]/(1 + x²)
= [(1 + x²) • 2x - x³]/(1 + x²)^(3/2)
= x(2 + x²)/(1 + x²)^(3/2)

Answer 2

y'=[2x(根号(1+x^2)-x^2*(根号(1+x^2)）)']/（根号(1+x^2)）^2
=[2x(根号(1+x^2)-x^2*(1/2)((1+x^2)^(-1/2)]/(1+x^2)
=[2x(1+x^2)-x^2/2](1+x^2)^(-1/2)*(1+x^2)^(-1)
=[2x+3x^2/2](1+x^2)^(-3/2)
=[2x+3x^2/2](1+x^2)^(-3/2)=