Question

x/根号下(a^2-x^2)的导数,

Answer 1

复合函数的导数：

y’=1/根号下(a^2-x^2)

+x*{[（a^2-x^2)^(-1/2)]'}

[（a^2-x^2)^(-1/2)]'

=(-1/2)*[（a^2-x^2)^(-3/2)]*[(a^2-x^2)']

=x[（a^2-x^2)^(-3/2)]

所以

y’=1/根号下(a^2-x^2)

+(x^2)*[（a^2-x^2)^(-3/2)]