Question

If \( y = \tan x + \tan x + \tan x + \dots \infty \), then show that \( \frac{dy}{dx} = \frac{2 \sec^2 x}{2 - y} \). Find \( \frac{dy}{dx} \) at \( x = 0 \).

Hint:

Verify infinite series definitions; for derivatives, test consistency at given points.

Answer 1

The series \( y = \tan x + \tan x + \dots \) is infinite, implying \( y = \tan x + y \).
\[ y = \tan x + y \quad \Rightarrow \quad \tan x = 0, \text{ which is inconsistent unless } y \text{ is redefined.} \] Assume a geometric series or misinterpretation; let’s try \( y = \tan x \). Then: \[ \frac{dy}{dx} = \sec^2 x. \] The given equation \( \frac{dy}{dx} = \frac{2 \sec^2 x}{2 - y} \) suggests a different function. Assume the problem intends a recursive or specific form. Let’s derive for a general \( y \). Differentiate \( y = \tan x + y \): \[ \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}. \] This is inconsistent, so assume a finite sum or typo. For \( y = \tan x \), check at \( x = 0 \): \[ y = \tan 0 = 0, \quad \frac{dy}{dx} = \sec^2 0 = 1. \] Try the given derivative: \[ \frac{2 \sec^2 x}{2 - y} \text{ at } x = 0, y = 0: \quad \frac{2 \cdot 1}{2 - 0} = 1. \] The series definition seems incorrect; assuming \( y = \tan x \), the result holds at \( x = 0 \).
Answer: \( \frac{dy}{dx} \big|_{x=0} = 1 \).