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)`

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