fgo金闪闪技能:x/(x^2+4)的导数怎么求??
来源:百度文库 编辑:查人人中国名人网 时间:2024/07/03 09:28:06
(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