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प्रश्न
If y = `sqrt(tan x + sqrt(tanx + sqrt(tanx + .... + ∞)`, then show that `dy/dx = (sec^2x)/(2y - 1)`.
Find `dy/dx` at x = 0.
उत्तर
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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