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The first derivative of $$x^{x}$$ is
$$x log x$$
$$x x^{x-1}$$
$$x^{x}(1-log x)$$
$$x^{x}(1+log x)$$
Let us assume $$T=x^x$$
Taking log on both sides:
$$\log T=x\log x$$
$$\dfrac{1}{T}\left(\dfrac{dT}{dx}\right)=x\cdot\dfrac{1}{x}+\log x$$
$$\dfrac{dT}{dx}=x^x\left(1+\log x\right)$$
=> Option D is the correct answer.
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