x^2/a^2+y^2/b^2=1
两边同时对x求导,得
2x/a²+2yy'/b²=0
y'=-b²x/(a²y)
y''=-b²/a²[(y-xy')/y²]
=-b²/a²[(y-x[-b²x/(a²y)])/y²]
=-b²/a²[(y+[b²x²/(a²y)])/y²]
=-b²/a^4[(a²y²+b²x²)/y³]
x^2/a^2+y^2/b^2=1
两边同时对x求导,得
2x/a²+2yy'/b²=0
y'=-b²x/(a²y)
y''=-b²/a²[(y-xy')/y²]
=-b²/a²[(y-x[-b²x/(a²y)])/y²]
=-b²/a²[(y+[b²x²/(a²y)])/y²]
=-b²/a^4[(a²y²+b²x²)/y³]