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$$$x \ln\left(x\right)$$$的导数
您的输入
求$$$\frac{d}{dx} \left(x \ln\left(x\right)\right)$$$。
解答
对 $$$f{\left(x \right)} = x$$$ 和 $$$g{\left(x \right)} = \ln\left(x\right)$$$ 应用乘积法则 $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(x \ln\left(x\right)\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x\right) \ln\left(x\right) + x \frac{d}{dx} \left(\ln\left(x\right)\right)\right)}$$自然对数的导数为 $$$\frac{d}{dx} \left(\ln\left(x\right)\right) = \frac{1}{x}$$$：
$$x {\color{red}\left(\frac{d}{dx} \left(\ln\left(x\right)\right)\right)} + \ln\left(x\right) \frac{d}{dx} \left(x\right) = x {\color{red}\left(\frac{1}{x}\right)} + \ln\left(x\right) \frac{d}{dx} \left(x\right)$$应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dx} \left(x\right) = 1$$$：
$$\ln\left(x\right) {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + 1 = \ln\left(x\right) {\color{red}\left(1\right)} + 1$$因此，$$$\frac{d}{dx} \left(x \ln\left(x\right)\right) = \ln\left(x\right) + 1$$$。
答案
$$$\frac{d}{dx} \left(x \ln\left(x\right)\right) = \ln\left(x\right) + 1$$$A
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