Question

求函数的二阶导数 y=arcsinx / 根号(1-x^2)

Answer 1

y=arcsinx/√(1-x^2)

y'=[(arcsinx)'√(1-x^2)-arcsinx*(√(1-x^2))']/(1-x^2)

=[1+arcsinx* x/√(1-x^2)]/(1-x^2)

=1/(1-x^2)+xarcsinx *(1-x^2)^(-3/2)

y"=2x/(1-x^2)+(arcsinx+x/√(1-x^2))*(1-x^2)^(-3/2)+xarcsinx*(-3/2)*(1-x^2)^(-5/2)* (-2x)

=2x/(1-x^2)+[arcsinx+x/√(1-x^2)]*(1-x^2)^(-3/2)+6x^2arcsinx*(1-x^2)^(-5/2)