某二元二次方程式:Y^2+X^2-10X-10Y+41=0
當X=2時 斜率為無限大
求X=3時斜率多少??
d/dX(Y^2+X^2-10X-10Y+41)=0
2Y*dY/dX+2X-10-10*dY/dX=0
(2Y-10)*dY/dX=10-2X
dY/dX=(10-2X)/(2Y-10)
dY/dX=(5-X)/(Y-5)
當X=3,
Y^2+3^2-10(3)-10Y+41=0
Y^2-10Y+20=0
Y=5+sqrt(5) or 5-sqrt(5)
dY/dX=(5-3)/(5+sqrt(5)-5) or (5-3)/(5-sqrt(5)-5)
=2/sqrt(5) or -2/sqrt(5)
=2sqrt(5)/5 or -2sqrt(5)/5