If x^x = y^y, then find dy/dx
The normal method is taking log both sides which gives
x.lnx= y.lny
differentiating both sides we get
x(1/x)+1.lnx= y(1/y)+1.lny(dy/dx)
therefore, dy/dx= 1+lnx/1+lny
BUT by using partial derivative,
f(x,y)=x^x-y^y
The ans comes out to be
dy/dx= -[x^x(lnx+1)]/-[y^y(lny+1)]
Can someone plz help..
