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If y = (cosx)^((cosx)^((cosx)^("....."oo...

If `y = (cosx)^((cosx)^((cosx)^("....."oo)))` , then show that `(dy)/(dx) =(y^(2)tanx)/(ylogcosx-1)`

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We have, `y = (cosx)^((cosx)^((cosx)^("....."oo)))`
`rArr y = (cosx)^(y)`
`:. log y = log(cosx)^(y)`
`rArr logy = y logcosx`
On differentiating w.r.t. we get
`1/y.(dy)/(dx)= y.(d)/(dx)log cosx+logcosx.(dy)/(dx)`
`rArr 1/y.(dy)/(dx) = (y)/(cosx).(d)/(dx) cosx+logcosx.(dy)/(dx)`
`rArr (dy)/(dx)[1/y-logcosx] = (-ysinx)/(cosx)=-ytanx`
`:. (dy)/(dx) = (-y^(2)tanx)/((1-ylogcosx))`
` = (y^(2)tanx)/(ylogcosx-1)`
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