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