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已知x^y-y^x=2确定函数y=fx,求dy/dx
人气:136 ℃ 时间:2020-06-08 00:39:44
解答
x^y-y^x=2u=x^ylnu=ylnxu'/u=y'lnx+y/xu'=(y'lnx+y/x)x^y v=y^xlnv=xlnyv'/v=lny+xy'/yv'=(lny+xy'/y)y^x (x^y-y^x)'=0(y'lnx+y/x)x^y-(lny+xy'/y)y^x=0(y'lnx+y/x)x^y-(lny+xy'/y)y^x=0x^yy'lnx+yx^(y-1)-y^xlny-y^...
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