哪些方法能治疗白点病:x/(x^2+4)的导数怎么求??
来源:百度文库 编辑:高考问答 时间:2024/05/05 16:58:07
(a/b)'=(a'*b-a*b')/(b)^2
x/(x^2+4)=[x'*(x^2+4)-(x^2+4)'*x]/[(x^2+4)^2]=[(x^2+4)-2x*x]/[(x^2+4)]^2=-(x^2-4)/[(x^2)+4]^2
x/(x^2+4)=[x'*(x^2+4)-(x^2+4)'*x]/[(x^2+4)^2]=[(x^2+4)-2x*x]/[(x^2+4)]^2=-(x^2-4)/[(x^2)+4]^2