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

Let \(u = x^x\) and \(v = a^x\), so \(y = u + v\) and \(\dfrac{dy}{dx} = \dfrac{du}{dx} + \dfrac{dv}{dx}\).

\(u = x^x \Rightarrow \log u = x\log x\). Differentiating w.r.t. \(x\): \(\dfrac{1}{u}\dfrac{du}{dx} = x \cdot \dfrac{1}{x} + \log x = 1 + \log x\), so \(\dfrac{du}{dx} = x^x(1 + \log x)\).

\(v = a^x \Rightarrow \dfrac{dv}{dx} = a^x \log a\).

\(\therefore \dfrac{dy}{dx} = x^x(1 + \log x) + a^x \log a\).

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