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