Question

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

Answer 1

Given that y = `(cos x)^((cos x)^((cosx)....oo)`,

⇒ y = (cos x)^y  .....`[y = (cos x)^((cos x)^((cosx)....oo))]`

Taking log on both sides log y = y.log(cos x)

Differentiating both sides w.r.t. x

`1/y * "dy"/"dx" = y * "d"/"dx" log (cos x ) + log(cos x) * "dy"/"dx"`

⇒ `1/y * "dy"/"dx" = y * 1/cosx * "d"/"dx" (cos x) + log(cos x) * "dy"/"dx"`

⇒ `1/y * "dy"/"dx" = y* 1/cosx * (- sin x) + log(cosx) * "dy"/"dx"`

⇒ `1/y * "dy"/"dx" - log(cos x) "dy"/"dx"` = – y tan x

⇒ `[1/y - log (cosx)] "dy"/"dx"` = – y tan x

⇒ `"dy"/"dx" = (- y tanx)/(1/y - log(cosx))`

= `(y^2 tanx)/(y log cos x - 1)`

Hence, `"dy"/"dx" = (y^2 tanx)/(y log cos x - 1)`.

Hence proved.