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Question: If \(y = x^{x^{x.....\infty}}\), then \(x(1 - y\log_{e}x)\frac{dy}{dx}\) is...

If y=xxx.....∞y = x^{x^{x.....\infty}}, then x(1−ylog⁡ex)dydxx(1 - y\log_{e}x)\frac{dy}{dx} is

A

x2x^{2}

B

y2y^{2}

C

xy2xy^{2}

D

None of these

Answer

y2y^{2}

Explanation

Solution

y=xxx......∞y = x^{x^{x......\infty}} ⇒ y=xyy = x^{y} ⇒ log⁡ey=ylog⁡ex\log_{e}y = y\log_{e}x

⇒ 1y⋅dydx=yx+log⁡exdydx\frac{1}{y} \cdot \frac{dy}{dx} = \frac{y}{x} + \log_{e}x\frac{dy}{dx} ⇒ (1y−log⁡ex)dydx=yx\left( \frac{1}{y} - \log_{e}x \right)\frac{dy}{dx} = \frac{y}{x}

⇒ x(1−ylog⁡ex)dydx=y2x(1 - y\log_{e}x)\frac{dy}{dx} = y^{2}