高等数学,第2题的(1)(2)小题怎么做,需要过程,求高手帮忙
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2 (1) . u = x^2y^2/(x+y)
u'<x> = [2xy^2(x+y)-x^2y^2]/(x+y)^2 = xy^2(x+2y)/(x+y)^2
u'<y> = [2x^2y(x+y)-x^2y^2]/(x+y)^2 = x^2y(2x+y)/(x+y)^2
xu'<x> + yu'<y> = x^2y^2(3x+3y)/(x+y)^2
= 3x^2y^2/(x+y) = 3u
(2) z = ln(e^x+e^y),
z'<x> = e^x/(e^x+e^y) = 1-e^y/(e^x+e^y),
z'<y> = e^y/(e^x+e^y) = 1-e^x/(e^x+e^y),
z''<xx> = e^ye^x/(e^x+e^y)^2
z''<yy> = e^xe^y/(e^x+e^y)^2
z''<xy> = -e^xe^y/(e^x+e^y)^2
则 z''<xx> z''<yy> - (z''<xy>)^2 = 0
u'<x> = [2xy^2(x+y)-x^2y^2]/(x+y)^2 = xy^2(x+2y)/(x+y)^2
u'<y> = [2x^2y(x+y)-x^2y^2]/(x+y)^2 = x^2y(2x+y)/(x+y)^2
xu'<x> + yu'<y> = x^2y^2(3x+3y)/(x+y)^2
= 3x^2y^2/(x+y) = 3u
(2) z = ln(e^x+e^y),
z'<x> = e^x/(e^x+e^y) = 1-e^y/(e^x+e^y),
z'<y> = e^y/(e^x+e^y) = 1-e^x/(e^x+e^y),
z''<xx> = e^ye^x/(e^x+e^y)^2
z''<yy> = e^xe^y/(e^x+e^y)^2
z''<xy> = -e^xe^y/(e^x+e^y)^2
则 z''<xx> z''<yy> - (z''<xy>)^2 = 0
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