Question

设y=ex+x6+e,求曲线上点(1,2e+1)处的切线方程

Answer 1

y=e^x+x^6+e
y'=e^x +6x^5
y'(1) = e +6
曲线上点(1,2e+1)处的切线方程
y-(2e+1) = y'(1) .(x-1)
y-(2e+1) = (e+6)(x-1)
(e+6)x -y +e -5=0