设函数y=f(x)由方程e^(2x+y)—cos(xy)=e—1确定,则曲线f(x)在点(0,1)处的法线方程为

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设函数y=f(x)由方程e^(2x+y)—cos(xy)=e—1确定,则曲线f(x)在点(0,1)处的法线方程为
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设函数y=f(x)由方程e^(2x+y)—cos(xy)=e—1确定,则曲线f(x)在点(0,1)处的法线方程为
设函数y=f(x)由方程e^(2x+y)—cos(xy)=e—1确定,则曲线f(x)在点(0,1)处的法线方程为

设函数y=f(x)由方程e^(2x+y)—cos(xy)=e—1确定,则曲线f(x)在点(0,1)处的法线方程为
显然f(x)过点(0,1)
将y视为x的函数,求微分
(2+y')e^(2x+y)+sin(xy)*(y+xy')=0
代入x=0时y=1,代入,有:
y'=-2
所以法线方程为y-1=1/2*(x-0)
即x-2y+2=0

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