100分求一道数学题,Find an equation for the tangent line to the curve x^2y + xy^3 = 2 at the point (1; 1).

100分求一道数学题,Find an equation for the tangent line to the curve x^2y + xy^3 = 2 at the point (1; 1).
100分求一道数学题,
Find an equation for the tangent line to the curve x^2y + xy^3 = 2 at the point (1; 1).

100分求一道数学题,Find an equation for the tangent line to the curve x^2y + xy^3 = 2 at the point (1; 1).

题目意思是,求曲线 x^2y + xy^3 = 2在点(1,1)处的切线方程.
对方程两边取全导,
d(x^2y)+d(xy^3)=2,
2xydx+x^2dy+y^3dx+3xy^2dy=0,
整理得 dy/dx=-(2xy+y^3)/(x^2+3xy^2)
那么曲线在(1,1)处的斜率为y'=-(2+1)/(1+3)=-3/4,
那么切线方程为 y=-3/4*(x-1)+1=-3/4*x+7/4.
O(∩_∩)O~

解法2两边求导
12x+3y+3xy'+4yy'+17y'=0
y'=-(12x+3y)/(3x+4y+17)
则过点(-1,0)的切线斜率为
k=-[12*(-1)+3*0]/[3*(-1)+4*0+17]
=12/14
=6/7
所以切线方程为y=(6/7)(x+1)
即6x-7y+6=0
垂线的斜率为k'=-1/k=-7...

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解法2两边求导
12x+3y+3xy'+4yy'+17y'=0
y'=-(12x+3y)/(3x+4y+17)
则过点(-1,0)的切线斜率为
k=-[12*(-1)+3*0]/[3*(-1)+4*0+17]
=12/14
=6/7
所以切线方程为y=(6/7)(x+1)
即6x-7y+6=0
垂线的斜率为k'=-1/k=-7/6
方程为y=-7/6(x+1)
即7x+6y+7=0

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