Contact us
Choose a category below so we can get back to you as quickly as possible.
由題意得,y′=ex+xex,∴在x=1處的切線的斜率是2e,且切點座標是(1,e),則在x=1處的切線方程是:y-e=2e(x-1),即2ex-y-e=0,故答案為:2ex-y-e=0.
We and our partners use cookies and other technologies to analyze traffic and optimize your experience. View more info and control your cookies settings at any time in our Cookies Policy.