Derivative of $$$\frac{1}{\ln\left(x\right)}$$$
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Solution
The function $$$\frac{1}{\ln\left(x\right)}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \frac{1}{u}$$$ and $$$g{\left(x \right)} = \ln\left(x\right)$$$.
Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(\frac{1}{\ln\left(x\right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\frac{1}{u}\right) \frac{d}{dx} \left(\ln\left(x\right)\right)\right)}$$Apply the power rule $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ with $$$n = -1$$$:
$${\color{red}\left(\frac{d}{du} \left(\frac{1}{u}\right)\right)} \frac{d}{dx} \left(\ln\left(x\right)\right) = {\color{red}\left(- \frac{1}{u^{2}}\right)} \frac{d}{dx} \left(\ln\left(x\right)\right)$$Return to the old variable:
$$- \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{{\color{red}\left(u\right)}^{2}} = - \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{{\color{red}\left(\ln\left(x\right)\right)}^{2}}$$The derivative of the natural logarithm is $$$\frac{d}{dx} \left(\ln\left(x\right)\right) = \frac{1}{x}$$$:
$$- \frac{{\color{red}\left(\frac{d}{dx} \left(\ln\left(x\right)\right)\right)}}{\ln^{2}\left(x\right)} = - \frac{{\color{red}\left(\frac{1}{x}\right)}}{\ln^{2}\left(x\right)}$$Thus, $$$\frac{d}{dx} \left(\frac{1}{\ln\left(x\right)}\right) = - \frac{1}{x \ln^{2}\left(x\right)}$$$.
Answer
$$$\frac{d}{dx} \left(\frac{1}{\ln\left(x\right)}\right) = - \frac{1}{x \ln^{2}\left(x\right)}$$$A