If `y^x+x^y=(x+y)^(x+y)` find `dy/dx`

Explanation

`y^x+x^y = (x+y)^(x+y)`
Let `f(x) = y^x`
Taking log both sides,
`log(f(x)) = xlogy`
Now, differentiating it w.r.t. `x`,
`=>1/(f(x))f'(x) = x/ydy/dx+logy`
`=>f'(x) = f(x)x/ydy/dx+logy`
`=>f'(x) = y^x (x/ydy/dx+logy)`
Similarly, `d/dx(x^y) = x^y(y/x+dy/dxlogx)`
and `d/dx((x+y)^(x+y)) = (x+y)^(x+y)[(1+dy/dx)log(x+y)+(x+y)/(x+y)(1+dy/dx)]`
So, differentiation of given equation becomes,
`y^x (xdy/dx+logy)+x^y(y/x+dy/dxlogx) = (x+y)^(x+y)[(1+dy/dx)log(x+y)+(1+dy/dx)]`
`=>dy/dx(y/x y^x + x^y logx - (x+y)^(x+y) log(x+y) - (x+y)^(x+y)) = (x+y)^(x+y)log(x+y)+1+y^xlogy - x^y y/x`
`=>dy/dx = ((x+y)^(x+y)log(x+y)+1+y^xlogy - x^y y/x)/(y/x y^x + x^y logx - (x+y)^(x+y) log(x+y) - (x+y)^(x+y))`