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Answer: \(\dfrac{dy}{dx} = x^{e^x} e^x\left(\log x + \dfrac{1}{x}\right)\)

Taking logarithms, \(\log y = e^x \log x\).

Differentiating w.r.t. \(x\): \(\dfrac{1}{y}\dfrac{dy}{dx} = e^x\cdot\dfrac{1}{x} + \log x\cdot e^x\).

\(\therefore \dfrac{dy}{dx} = y\,e^x\left(\dfrac{1}{x} + \log x\right) = x^{e^x} e^x\left(\log x + \dfrac{1}{x}\right)\).

Doubtly AI · 29 Oct 2026 AI-verified · Nov 2026 Report
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