if **xy _{+} _{y}x = _{a}b 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^{y}logx dy/dx + y^{x}x/y dy/dx = - x^{y}*y/x - y^{x}logy

=== dy/dx = - (x^{y}* y/x + y^{x}logy )/ x^{y}logx + y^{x}x/y