Question

x^y=y^x,求dy/dx，以及二阶导数.(用隐函数的求导公式解答)

0.0

Answer 1

对原式两边微分
yx^(y-1)dx+x^ylnxdy=y^xlnydx+xy^(x-1)dy
(x^ylnx-xy^(x-1))dy=(y^xlny-yx^(y-1))dx
两边除以(x^ylnx-xy^(x-1))dx得
y'=dy/dx=(y^xlny-yx^(y-1))/(x^ylnx-xy^(x-1))
对上面第2式两边微分
d(x^ylnx-xy^(x-1))dy+(x^ylnx-xy^(x-1))ddy=d(y^xlny-yx^(y-1))dx
(yx^(y-1)lnxdx+x^y(lnx)^2dy+x^y/xdx-y^(x-1)dx-xy^(x-1)lnydx-x(x-1)y^(x-2)dy)dy+(x^ylnx-xy^(x-1))ddy
=(y^x(lny)^2dx+xy^(x-1)lnydy+y^x1/ydy-x^(y-1)dy-y(y-1)x^(y-2)dx-yx^(y-1)lnxdy)dx
两边除以dx^2
(yx^(y-1)lnx+x^y(lnx)^2y'+x^y/x-y^(x-1)-xy^(x-1)lny-x(x-1)y^(x-2)y')y'+(x^ylnx-xy^(x-1))y''
=(y^x(lny)^2+xy^(x-1)lnyy'+y^x1/yy'-x^(y-1)y'-y(y-1)x^(y-2)-yx^(y-1)lnxy')
得2阶导数
y''=((y^x(lny)^2+xy^(x-1)lnyy'+y^x1/yy'-x^(y-1)y'-y(y-1)x^(y-2)-yx^(y-1)lnxy')-((yx^(y-1)lnx+x^y(lnx)^2y'+x^y/x-y^(x-1)-xy^(x-1)lny-x(x-1)y^(x-2)y')y'))/(x^ylnx-xy^(x-1))
晕死了，默哀5分钟
要是再把上面的第3式一阶导数y'代进去
就会得到史上
最颠三倒四
最啰嗦
最花里胡哨
最长的x、y字符串
装进骨灰盒了
不用再默哀

Answer 2

ylnx=xlny
y'lnx+y/x=lny+xy'/y
移项，整理得：
dy/dx=(lny-y/x)/(lnx-x/y)
y"=[(lny-y/x)'(lnx-x/y)-(lny-y/x)(lnx-x/y)']/(lnx-x/y)²
以下略