设y=√x²-1,则y''= 求二阶导数!

问题描述:

设y=√x²-1,则y''= 求二阶导数!

y'=x/√x²-1
y''=[√(x²-1)-x²/√(x²-1)]/(x²-1)
=-1/(x²-1)^(3/2)

y'=1/(2√(x^2-1))*2x=x/√(x^2-1)
y''=(√(x^2-1)-x*1/(2√(x^2-1))*2x)/(x^2-1)=-1/(x^2-1)^(3/2)