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Differential Calculus

Question : 20 of 69

Marks: +1, -0

y=x^{x^{x}}

\; \frac{d y}{d x}

Solution:

Concept:

Differentiate using logarithmic differentiation, considering the tower exponent.

Explanation:

Let

y = x^{x^x}

Take natural log:

\ln y = x^x \ln x

Let

u = x^x

, then

\ln u = x \ln x

, so

\frac{du}{dx} = u(1+\ln x)

Differentiate

\ln y = u \ln x

\frac{1}{y}\frac{dy}{dx} = u' \ln x + u \cdot \frac{1}{x}

Substitute

u'

\frac{dy}{dx} = y\left[ u(1+\ln x)\ln x + \frac{u}{x} \right] = y u \left[ \ln x + (\ln x)^2 + \frac{1}{x} \right]

Since

u = \frac{\ln y}{\ln x}

, we get

\frac{dy}{dx} = y \ln y \left(1+\ln x + \frac{1}{x\ln x}\right)

This does not simplify to any of the given forms.

Answer:

None of the options match.

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