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If x=ey+ey+ey+…∞x = e^{y + e^{y + e^{y + \ldots \infty}}}, x>0x > 0, then dydx\frac{dy}{dx} is equal to

Solution

✅ Correct Option: 3

By the recursive structure, x=ey+xx = e^{y+x}.

Taking log: ln⁡x=y+x\ln x = y + x, so y=ln⁡x−xy = \ln x - x.

Differentiating: dydx=1x−1=1−xx\frac{dy}{dx} = \frac{1}{x} - 1 = \frac{1-x}{x}.

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