See this is a very un often kind of a equation. Let u(x)=sinx -cosx. Thus the equation provided reduces to u^u. Let the function f(x)=u^u. Now taking log in both sides we get,

log f(x)= u log(u)

Differentiating both side we get: 1/f(x) * f'(x)= u' logu+u'/u => f'(x)=f(x)[u' logu+u'/u] where u' is the first order derivative of u and f'(x) is the first order derivative of f(x) with respect to x.

Now u'=cosx+sinx Thus, f'(x)=(sinx-cosx)^(sinx-cosx)[(cosx+sinx) log (sinx-cosx)+(sin+cosx)/(sinx-cosx)].

This can be one of the solution of the derivative of the given function.

I hope this answer helps. All the very best for your future endeavors!