Derivative of $$$\frac{v \left(\ln\left(b\right) + 1\right)}{\ln\left(b\right)}$$$ with respect to $$$v$$$
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Find $$$\frac{d}{dv} \left(\frac{v \left(\ln\left(b\right) + 1\right)}{\ln\left(b\right)}\right)$$$.
Solution
Apply the constant multiple rule $$$\frac{d}{dv} \left(c f{\left(v \right)}\right) = c \frac{d}{dv} \left(f{\left(v \right)}\right)$$$ with $$$c = \frac{\ln\left(b\right) + 1}{\ln\left(b\right)}$$$ and $$$f{\left(v \right)} = v$$$:
$${\color{red}\left(\frac{d}{dv} \left(\frac{v \left(\ln\left(b\right) + 1\right)}{\ln\left(b\right)}\right)\right)} = {\color{red}\left(\frac{\ln\left(b\right) + 1}{\ln\left(b\right)} \frac{d}{dv} \left(v\right)\right)}$$Apply the power rule $$$\frac{d}{dv} \left(v^{n}\right) = n v^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dv} \left(v\right) = 1$$$:
$$\frac{\left(\ln\left(b\right) + 1\right) {\color{red}\left(\frac{d}{dv} \left(v\right)\right)}}{\ln\left(b\right)} = \frac{\left(\ln\left(b\right) + 1\right) {\color{red}\left(1\right)}}{\ln\left(b\right)}$$Simplify:
$$\frac{\ln\left(b\right) + 1}{\ln\left(b\right)} = 1 + \frac{1}{\ln\left(b\right)}$$Thus, $$$\frac{d}{dv} \left(\frac{v \left(\ln\left(b\right) + 1\right)}{\ln\left(b\right)}\right) = 1 + \frac{1}{\ln\left(b\right)}$$$.
Answer
$$$\frac{d}{dv} \left(\frac{v \left(\ln\left(b\right) + 1\right)}{\ln\left(b\right)}\right) = 1 + \frac{1}{\ln\left(b\right)}$$$A