求z=(1+xy)^y在(1,1)处的偏导数
人气:423 ℃ 时间:2020-06-28 21:08:38
解答
z = (1+xy)^y
∂z/∂x = y²(1+xy)^(y-1)
lnz = yln(1+xy)
∂z/∂y /z = ln(1+xy) + xy/(1+xy)
∂z/∂y = [ln(1+xy) + xy/(1+xy)] (1+xy)^y
∂z/∂x (1,1) = 1²(1+1*1)^(1-1)=1
∂z/∂y (1,1) = [ln(1+1*1)+1*1/(1+1*1)](1+1*1)^1 = 2(ln2 + ½) = 1+ ln4
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