if xy _{+ y}x = _ab find dy/dx

x^y+ y^x= a^b, here ab is const .

let x^y= u ; take log on both sides == y log x = log u == y*1/x + logx * dy/dx (by products rule) = 1/u du / dx

==du/dx = x^y(y/x + logx dy/dx)

let y^x= v ; take log on both sides == x log y = logv == x*1/ydy/dx + logy*1 (by products rule) = 1/v dv/dx

== dv/dx = y^x(x/y dy/dx + logy)

== u + v = a^b

differentiating wrt x we get , du/dx + dv/dx = 0 == x^y(y/x + logx dy/dx) + y^x(x/y dy/dx + logy) = 0

== x^ylogx dy/dx + y^xx/y dy/dx = - x^y*y/x - y^xlogy

=== dy/dx = - (x^y* y/x + y^xlogy )/ x^ylogx + y^xx/y