First, consider the derivative with respect to #x# of #y=f(x)^g(x)#.
Follow these steps:
#y=f(x)^g(x)#
#ln(y)=g(x)ln(f(x))#
Using implicit differentiation,
#(dy/dx)/y=(dg)/dxln(f(x))+(df)/dxg(x)/f(x)#
#dy/dx=y((dg)/dxln(f(x))+(df)/dxg(x)/f(x))#
Since #y=f(x)^g(x)#,
#dy/dx=f(x)^g(x)((dg)/dxln(f(x))+(df)/dxg(x)/f(x))#
#" "= f(x)^g(x)\ (dg)/dxln(f(x))+(df)/dxg(x)f(x)^(g(x)-1)#
Now, with this, we can proceed to solve the equation #x^y+y^x=c#.
Assume that #y# is some function of #x# and #c# is a constant. Then, using implicit differentiation,
#d/dx(x^y)+d/dx(y^x)=d/dx(c)#
Using the formula we found for the derivative of #f(x)^g(x)#, we know that
#d/dx(x^y)=x^y dy/dx ln(x)+y x^(y-1)#
#d/dx(y^x)=y^xln(y)+dy/dxxy^(x-1)#
#d/dx(c)=0#
So, the original equation becomes
#x^y dy/dx ln(x)+y x^(y-1)+y^xln(y)+dy/dxxy^(x-1)=0#
Gather the terms containing #dy/dx# on the left-hand side and the remaining terms on the right-hand side,
#x^y dy/dx ln(x)+dy/dxxy^(x-1)=-y x^(y-1)-y^xln(y)#
#dy/dx(x^y ln(x)+xy^(x-1))=-y x^(y-1)-y^xln(y)#
Therefore,
#dy/dx=-(y x^(y-1)+y^xln(y))/(x^y ln(x)+xy^(x-1))#
