Question

求下列函数的二阶导数？

Answer 1

y=x^2/√(1+x),
y'=2x/√(1+x)-(1/2)x^2/(1+x)^(3/2)
=[4x(1+x)-x^2]/[2(1+x)^(3/2)]
=(4x+3x^2)/[2(1+x)^(3/2)],
y''=(4+6x)/[2(1+x)^(3/2)]-(3/2)(4x+3x^2)/[2(1+x)^(5/2)]
=[2(4+6x)(1+x)-3(4x+3x^2)]/[4(1+x)^(5/2)]
=(8+8x+3x^2)/[4(1+x)^(5/2)].

Answer 2

x^2
=(1+x)^2 - 2x -1
=(1+x)^2 - 2(1+x) + 1
y
=x^2/√(1+x)
=(1+x)^(3/2) -2√(1+x) + 1/√(1+x)
y'
=(3/2)(1+x)^(1/2) - (1+x)^(-1/2) - (1/2)(1+x)^(-3/2)
y''
=(3/4)(1+x)^(-1/2) +(1/2)(1+x)^(-3/2) + (3/4) (1+x)^(-5/2)