求2y-x=(x-y)ln(x-y)在点(2,1)处的切线方程和法线方程

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求2y-x=(x-y)ln(x-y)在点(2,1)处的切线方程和法线方程
求2y-x=(x-y)ln(x-y)在点(2,1)处的切线方程和法线方程

求2y-x=(x-y)ln(x-y)在点(2,1)处的切线方程和法线方程
两边对x求导:
2y'-1=(1-y')ln(x-y)+(x-y)/(x-y)*(1-y')
2y'-1=(1-y')ln(x-y)+1-y'
3y'+y'ln(x-y)=ln(x-y)+2
y'=[ln(x-y)+2]/[3+ln(x-y)]
代入点(2,1),得y'=2/3
因此切线方程为:y=2/3*(x-2)+1=2x/3-1/3
法线方程为:y=-3/2*(x-2)+1=-3x/2+4