If `y=(cosx)^(cosx)^(((cosx)dotoo))` , prove that `(dy)/(dx)=-(y^2tanx)/((1-ylogcosx))`

Explanation

Given that,
`y=(cosx)^(cosx^(cosx)dotoo)`
`=>y=(cosx)^y`
`=>logy=log(cosx)^y`
`=>logy=y log(cosx)`
`=>d/(dx)logy=d/(dx)(y log(cosx))`
`=>1/y (dy)/(dx)=(dy)/(dx)*log(cosx)+yd/(dx)log(cosx)`
`=>(dy)/(dx)[1/y-log(cosx)]=y*1/cosx(-sinx)`
`=>(dy)/(dx)[(1-ylog(cosx))/y]=-ytanx`
`=>(dy)/(dx)=(-y^2tanx)/(1-ylog(cosx))`