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求全微分z=e^[sin(x+y)]*y^x
人气:195 ℃ 时间:2020-05-25 08:08:42
解答
dz=e^[sin(x+y)]*cos(x+y)*(dx+dy)*y^x+e^[sin(x+y)]*y^x[(lny)dx+(x/y)dy]
={e^[sin(x+y)]*cos(x+y)*y^x+e^[sin(x+y)]*y^x(lny)}dx+{e^[sin(x+y)]*cos(x+y)*y^x+e^[sin(x+y)]*y^x(x/y)}dy
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