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高等数学 一道求全微分的题 求详细解题过程 在线等速度采纳
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z = arctan(y/x)
z'<x> = (-y/x^2)/[1+(y/x)^2] = -y/(x^2+y^2)
z'<y> = (1/x)/[1+(y/x)^2] = x/(x^2+y^2)
dz = (-ydx+xdy)/(x^2+y^2)
z'<x> = (-y/x^2)/[1+(y/x)^2] = -y/(x^2+y^2)
z'<y> = (1/x)/[1+(y/x)^2] = x/(x^2+y^2)
dz = (-ydx+xdy)/(x^2+y^2)
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