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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Differentiation

NTA Abhyas 2020 · Mathematics · Differentiation

Let y = x^ x^ x ∞ , then (dy)/(dx) is equal to (given x > 0 )

Let y=xxx…∞y = x^{x^{x \ldots \infty}}, then dydx\dfrac{dy}{dx} is equal to (given x>0x > 0 )

Options

  1. Ayxy−1yx^{y - 1}
  2. By2x(1−ylnx)\dfrac{y^{2}}{x ( 1 - ylnx )}
  3. Cyx(1+ylnx)\dfrac{y}{x ( 1 + ylnx )}
  4. DNone of these

Correct answer

B. y2x(1−ylnx)\dfrac{y^{2}}{x ( 1 - ylnx )}

Step-by-step solution

Given, y=xxx...∞y = x^{x^{x...\infty}} then y=xyy = x^{y}

Taking lnln on both the sides, we get

ln⁡y=ylnx\ln y = ylnx

⇒1y.dydx=y.1x+dydx.ln⁡x⇒ \dfrac{1}{y}. \dfrac{dy}{dx} = y. \dfrac{1}{x} + \dfrac{dy}{dx}. \ln x

⇒dydx[1y−lnx]=yx⇒ \dfrac{dy}{dx}\left[\dfrac{1}{y} - ln x\right] = \dfrac{y}{x}

⇒dydx=y2x(1−ylnx)⇒ \dfrac{dy}{dx} = \dfrac{y^{2}}{x ( 1 - ylnx )}
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