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Question: If \(x = e^{y + e^{y + ........\text{to }\infty}}\), then \(\frac{dy}{dx}\) is...

If x=ey+ey+........to ∞x = e^{y + e^{y + ........\text{to }\infty}}, then dydx\frac{dy}{dx} is

A

1+xx\frac{1 + x}{x}

B

1x\frac{1}{x}

C

1−xx\frac{1 - x}{x}

D

x1+x\frac{x}{1 + x}

Answer

1−xx\frac{1 - x}{x}

Explanation

Solution

x=ey+xx = e^{y + x}

Taking log both sides, log⁡x=(y+x)log⁡e=y+x\log x = (y + x)\log e = y + x

⇒ y+x=log⁡xy + x = \log x ⇒ dydx+1=1x\frac{dy}{dx} + 1 = \frac{1}{x} ⇒ dydx=1x−1=1−xx\frac{dy}{dx} = \frac{1}{x} - 1 = \frac{1 - x}{x}