已知x^y-y^x=2确定函数y=fx,求dy/dx

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已知x^y-y^x=2确定函数y=fx,求dy/dx

已知x^y-y^x=2确定函数y=fx,求dy/dx
已知x^y-y^x=2确定函数y=fx,求dy/dx

已知x^y-y^x=2确定函数y=fx,求dy/dx
x^y-y^x=2
u=x^y
lnu=ylnx
u'/u=y'lnx+y/x
u'=(y'lnx+y/x)x^y

v=y^x
lnv=xlny
v'/v=lny+xy'/y
v'=(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=0
x^yy'lnx+yx^(y-1)-y^xlny-y^(x-1)y'=0
y'=-[yx^(y-1)-y^xlny]/[x^ylnx-y^(x-1)]

dy/dx=(y^x lny - y x^(y-1))/(x^y lnx - x y^(x-1))

已知x^y-y^x=2确定函数y=f(x),求dy/dx
令u=x^y;取对数得lnu=ylnx;取导数得(1/u)(du/dx)=lnx(dy/dx)+y/x;
故du/dx=(x^y)′=u[(lnx)y′+y/x]=(x^y)[(lnx)y′+y/x]................(1);
再令v=y^x;取对数得lnv=xlny;取导数得(1/v)(dv/dx)=...

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已知x^y-y^x=2确定函数y=f(x),求dy/dx
令u=x^y;取对数得lnu=ylnx;取导数得(1/u)(du/dx)=lnx(dy/dx)+y/x;
故du/dx=(x^y)′=u[(lnx)y′+y/x]=(x^y)[(lnx)y′+y/x]................(1);
再令v=y^x;取对数得lnv=xlny;取导数得(1/v)(dv/dx)=lny+xy′/y;
故dv/dx=(y^x)′=v[lny+xy′/y]=(y^x)[lny+xy′/y]......................(2);
利用(1)和(2)的结果,将原方程x^y-y^x=2的两边对x取导数得:
(x^y)′-(y^x)′=0,将(1)和(2)代入得:
(x^y)[(lnx)y′+y/x]-(y^x)[lny+xy′/y]=0 ,整理,将y′作为公因子提出得:
[(x^y)lnx-(y^x)(x/y)]y′=(y^x)lny-(x^y)(y/x)
∴y′=dy/dx=[(y^x)lny-(x^y)(y/x)]/[(x^y)lnx-(y^x)(x/y)]=[x(lny)y^(x+1)-y²x^y]/[y(lnx)x^(y+1)-x²y^x]

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