Question

求曲线x³ y³-xy=7在店(1,2)处的切线方程和法线方程

Answer 1

解

Answer 2

答：f(x)=1/x²求导：f'(x)=-2/x³点M（1,1）在f(x)上x=1时,f'(1)=-2切线斜率k=f'(1)=-2法线斜率k=-1/f'(1)=1/2所以：切线为y-1=-2(x-1),y=-2x+3法线为y-1=(1/2)(x-1),y=(1/2)x+1/2

Answer 3

x^3+ y^3-xy=7
3x^2 + 3y^2.dy/dx - ( y + x.dy/dx ) = 0
(3y^2 -x ) dy/dx = y - 3x^2
dy/dx = (y - 3x^2)/(3y^2 -x)
dy/dx |(1,2) = ( 2 -3)/(12 - 1) = -1/11
(1,2)处的切线方程
y-2 = -(1/11)(x-1)
(1,2)处的法线方程
y-2 = 11(x-1)

Answer 4

Answer 5

先求导算出斜率