导数题 2

对 $x>0$，令 $y=x^x$。取对数得 $\ln y=x\ln x$。

两边对 $x$ 求导：

$$\frac{y'}{y}=\ln x+1.$$

因此

$$\boxed{\frac{d}{dx}x^x=x^x(\ln x+1)}.$$

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英文解析

For $x>0$, let $y=x^x$. Taking the logarithm gives $\ln y=x\ln x$.

Differentiating both sides with respect to $x$:

$$\frac{y'}{y}=\ln x+1.$$

Therefore

$$\boxed{\frac{d}{dx}x^x=x^x(\ln x+1)}.$$