Find dx/dy for the function y=x^cos(x)

Answer 1

recall that if
y = a^v, y' = ln(a) a^v v'
y = u^n, y' = n u^(n-1) u'

So, interestingly enough, if

y = u^v, y' = vu^(v-1) u' + ln(u) u^v v'

so, here we have

y' = cosx x^(cosx-1) + ln(cosx) x^cosx (-sinx)
= x^cosx (cosx/x - ln(cosx) sinx)