Question

If y = `sqrt(tan x + sqrt(tanx + sqrt(tanx + .... +  ∞)`, then show that `dy/dx = (sec^2x)/(2y - 1)`.

Find `dy/dx` at x = 0.

Answer 1

Given that : y = `sqrt(tan x + sqrt(tanx + sqrt(tanx + ...  ∞)`    ...(I)

Squaring both sides, we get

y² = `tanx + sqrt(tanx + sqrt(tanx + ...  ∞)`, which is same as 

y² = `tanx + sqrt(tanx + sqrt(tanx + sqrt(tanx + ...  ∞)`

y² = tan x + y    ...[From (I)]

Differentiate w. r. t. x

`d/dx (y^2) = d/dx (tan x) + dy/dx`

`2y dy/dx - dy/dx` = sec² x

`(2y - 1) dy/dx` = sec² x

∴ `dy/dx = (sec^2x)/(2y - 1)`

x = 0, y = 0

`dy/dx = 1/(0 - 1)` = – 1