丙肝hcvrna103e+07
@昌薇1988:请帮忙看看结果是否有问题 - 肝病科 - 复禾健康问答
邬盼13479106337…… [答案] 复合函数求导 因d/dx(e^y)=e^y*y' d/dx(xy)=y+xy' d/dx(e)=0 故 e^y+xy-e=0的导数dy/dx 有 e^y*y'+y+xy'-0=0 y'=-y/(e^y+x)
邬盼13479106337…… [答案] 复合函数求导 因d/dx(e^y)=e^y*y' d/dx(xy)=y+xy' d/dx(e)=0 故 e^y+xy-e=0的导数dy/dx 有 e^y*y'+y+xy'-0=0 y'=-y/(e^y+x)