如果是(根号x)/(x+100)呢
利用商法则
(f/g)'=(f'g-g'f)/g^2
=[(x^(1/2))'*(x+100)-(x^(1/2))*(x+100)']/(x+100)^2
=[1/2x^(-1/2)(x+100)-x^(1/2)*1]/(x+100)^2
=x^(-1/2)(50-1/2*x)/(x+100)^2
如果是根号[x/(x+100)]呢
利用链式加商法则
=1/2*[x/(x+100)]^(-1/2) * [x/(x+100)]'
=1/2*[x/(x+100)]^(-1/2) * [1-100/(x+100)]'
=1/2*[x/(x+100)]^(-1/2) *[100/(x+100)^2]