JEE MainMathematicsModerateDifferentiation

If $y = \left( x^{x} \right)^{x},$ then $\frac{dy}{dx} =$

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y = \left( x^{x} \right)^{x}

\therefore logy = xlogx^{x} = x^{2}logx

\therefore\frac{1}{y}\frac{dy}{dx} = 2xlogx + x^{2} \cdot \frac{1}{x} = x(1 + 2logx)

\therefore\frac{dy}{dx} = x \cdot y(1 + 2logx)

Source: Grass-G / M-12 Mathematics Paper 2 Nov.docx

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