公式(x^n)'=nx^(n-1)
f(x)=v(x)+-u(x),f'(x)=v'(x)+-u'(x)
y=x^(1/2)-x^(-1/2)
y'=[x^(1/2)]'-[x^(-1/2)]'
=1/2*x^(1/2-1)-(-1/2)*x^(-1/2-1)
=x^(-1/2)/2+x^(-3/2)/2
=[x^(-1/2)+x^(-3/2)]/2
反函数自己求吧~用比例,几下就出来了