`y = x^(cosx)`
taking log to the base 'e' both sides we get
`lny = (cosx)*lnx`
Thus, `(1/y)*dy/dx = -sinx*(lnx) + (cosx/x)`
`or, dy/dx = y*[(cosx/x) - sinx*(lnx)]`
`or, dy/dx = x^(cosx)*[(cosx/x) - sinx*(logx)]`
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