If `y = (cos x)^((cosx)^((cos x)...oo)), " then " dy/dx ` is equal to

Options

A. `(y tan x)/(y log cos x-1)`
B. `(y^(2) tan x)/(y log cos x-1)`
C. `(y tan x)/(1+y log cos x)`
D. None of these

Answer

B

Explanation

Given that , `y = (cosx)^((cos x )^((cos x )^(...oo))) rArr y = (cos x)^(y)`
On taking log on both sides , we get
` log y = y log (cos x) `
Now , differentiating both sides w.r.t.x, we get
`(1)/(y) .(dy)/(dx) = y .(1)/(cos x ) (- sin x ) + log (cos x ).(dy)/(dx)`
` rArr (dy)/(dx)= y{- y tanx + log cos x . (dy)/(dx)}`
`rArr(1 - y log cos x )(dy)/(dx) = - y^(2) tan x `
` rArr (dy)/(dx) = (y^(2) tanx)/((y log cos x - 1))`