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$$$\frac{1}{\ln\left(x\right)}$$$的导数
您的输入
求$$$\frac{d}{dx} \left(\frac{1}{\ln\left(x\right)}\right)$$$。
解答
函数$$$\frac{1}{\ln\left(x\right)}$$$是两个函数$$$f{\left(u \right)} = \frac{1}{u}$$$和$$$g{\left(x \right)} = \ln\left(x\right)$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。
应用链式法则 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dx} \left(\frac{1}{\ln\left(x\right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\frac{1}{u}\right) \frac{d}{dx} \left(\ln\left(x\right)\right)\right)}$$应用幂次法则 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$，其中 $$$n = -1$$$:
$${\color{red}\left(\frac{d}{du} \left(\frac{1}{u}\right)\right)} \frac{d}{dx} \left(\ln\left(x\right)\right) = {\color{red}\left(- \frac{1}{u^{2}}\right)} \frac{d}{dx} \left(\ln\left(x\right)\right)$$返回到原变量：
$$- \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{{\color{red}\left(u\right)}^{2}} = - \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{{\color{red}\left(\ln\left(x\right)\right)}^{2}}$$自然对数的导数为 $$$\frac{d}{dx} \left(\ln\left(x\right)\right) = \frac{1}{x}$$$：
$$- \frac{{\color{red}\left(\frac{d}{dx} \left(\ln\left(x\right)\right)\right)}}{\ln^{2}\left(x\right)} = - \frac{{\color{red}\left(\frac{1}{x}\right)}}{\ln^{2}\left(x\right)}$$因此，$$$\frac{d}{dx} \left(\frac{1}{\ln\left(x\right)}\right) = - \frac{1}{x \ln^{2}\left(x\right)}$$$。
答案
$$$\frac{d}{dx} \left(\frac{1}{\ln\left(x\right)}\right) = - \frac{1}{x \ln^{2}\left(x\right)}$$$A