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