Question

y=根号下a^2-x^2 求二阶导数 要过程

Answer 1

y=√(a²-x²)，
那么
y'= (a²-x²)' / [2√(a²-x²)]
= -2x / [2√(a²-x²)]
= -x /√(a²-x²)，
所以
y"={ (-x)' * √(a²-x²) + x* [√(a²-x²)] ' }/ (a²-x²)
= [-√(a²-x²) - x²/√(a²-x²) ] / (a²-x²)
= -a² / (a²-x²)^(3/2)

Answer 2

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