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If y = tanx+tanx+tanx+....+ ∞, then show that dydx=sec2x2y-1. Find dydx at x = 0. - Mathematics and Statistics

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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.

Sum
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Solution

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

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