Question

高数求切线和法线find the equations of tangent line and normal line o

Answer 1

x^2+y^3-3*x*y=0

(x^2+y^3-3*x*y)'=0'

2*x+2*y^2*y'-(3*y'*x+3*y)=0

y'=(2*x-3*y)/(3*x-2*y^2)

隐函数求导

注：(y^3)'=y'*y*y+y*y'*y+y*y*y'

y是一个函数,比如f(x).

而x是自变量

(x^3)'=3*x^2

代入x,y即可

法线,据互相垂直的直线斜率k1*k2=-1