Question

1.求方程ye^x+lny=1的微分dy 2.求曲线e^y-xy=e在点（0,1）处的切线方程。

请各位帮我解答下，步骤详细点，感谢给位了

Answer 1

Answer 2

(1)F＝ye^x+lny-1.
Fx＝ye^x,Fy＝e^x+1/y.
dy/dx＝-Fx/Fy＝-ye^x/(e^x+1/y)
∴dy＝[-y²e^x/(ye^x+1)]dx.

(2)F＝e^y-xy-e
Fx＝-y,Fy＝e^y-x.
dy/dx＝-Fx/Fy＝y/(e^y-x)
∴k＝1/(e¹-0)＝1/e.
故切线为：y-1＝(1/e)(x-0),
即x-ey+e＝0。