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