Differentiate `(cosx)^x+(sinx)^(1//x)` with respect to `x` :

Explanation

Given that,
`y=(cosx)^x+(sinx)^(1//x)`
`y=u+v`
`=>(dy)/(dx)=(du)/(dx)+(dv)/(dx).....(1)`
Now, `u= (cosx)^x`
taking log on both sides we get,
`log u= x*log(cos x)`
differentiate on both sides we get,
`=>1/u (du)/(dx)=x*1/cosx *(-sinx) +1*log(cosx)`
`=>(du)/(dx)=u[x(-sinx/cosx)+log(cosx)]`
`=>(du)/(dx)=(cosx)^x[-xtanx+log(cosx)]...(2)`



again `v=(sinx)^(1//x)`
`=>logv= 1/x log (sinx)`
`=>1/v(dv)/(dx)=1/x*cosx/sinx +log sinx (-1/x^2)`
`=>(dv)/(dx)=v[cotx/x-(log sinx)/x^2]`
`=>(dv)/(dx)=(sinx)^(1//x)[cotx/x-(log sinx)/x^2]...(3)`


from `(1),(2) ,(3)` we get,
`(dy)/(dx)=(cosx)^x[-xtanx+log(cosx)]+(sinx)^(1//x)[cotx/x-(log sinx)/x^2]`