函数y=f(x)由方程y=1+xe^xy确定,求y'|x=0和y''|x=0各是多少
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y=1+x.e^(xy)
x=0,y=1
y' = [1+ x(xy' +x) ] e^(xy)
y' [1- x^2.e^(xy)] = (1+x)e^(xy)
y' = (1+x)e^(xy) /[1- x^2.e^(xy)]
y'|x=0 = 1
y' = (1+x)e^(xy) /[1- x^2.e^(xy)]
y'' ={ [1- x^2.e^(xy)] [1+ (1+x)(xy'+y) ] e^(xy) - (1+x)e^(xy).[ -2x+ x^2(xy'+y)]e^(xy) } /[1- x^2.e^(xy)]^2
y''|x=0 =(2 -0)/1 =2
x=0,y=1
y' = [1+ x(xy' +x) ] e^(xy)
y' [1- x^2.e^(xy)] = (1+x)e^(xy)
y' = (1+x)e^(xy) /[1- x^2.e^(xy)]
y'|x=0 = 1
y' = (1+x)e^(xy) /[1- x^2.e^(xy)]
y'' ={ [1- x^2.e^(xy)] [1+ (1+x)(xy'+y) ] e^(xy) - (1+x)e^(xy).[ -2x+ x^2(xy'+y)]e^(xy) } /[1- x^2.e^(xy)]^2
y''|x=0 =(2 -0)/1 =2
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