Second derivative of $$$2^{n}$$$
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Find $$$\frac{d^{2}}{dn^{2}} \left(2^{n}\right)$$$.
Solution
Find the first derivative $$$\frac{d}{dn} \left(2^{n}\right)$$$
Apply the exponential rule $$$\frac{d}{dn} \left(m^{n}\right) = m^{n} \ln\left(m\right)$$$ with $$$m = 2$$$:
$${\color{red}\left(\frac{d}{dn} \left(2^{n}\right)\right)} = {\color{red}\left(2^{n} \ln\left(2\right)\right)}$$Thus, $$$\frac{d}{dn} \left(2^{n}\right) = 2^{n} \ln\left(2\right)$$$.
Next, $$$\frac{d^{2}}{dn^{2}} \left(2^{n}\right) = \frac{d}{dn} \left(2^{n} \ln\left(2\right)\right)$$$
Apply the constant multiple rule $$$\frac{d}{dn} \left(c f{\left(n \right)}\right) = c \frac{d}{dn} \left(f{\left(n \right)}\right)$$$ with $$$c = \ln\left(2\right)$$$ and $$$f{\left(n \right)} = 2^{n}$$$:
$${\color{red}\left(\frac{d}{dn} \left(2^{n} \ln\left(2\right)\right)\right)} = {\color{red}\left(\ln\left(2\right) \frac{d}{dn} \left(2^{n}\right)\right)}$$Apply the exponential rule $$$\frac{d}{dn} \left(m^{n}\right) = m^{n} \ln\left(m\right)$$$ with $$$m = 2$$$:
$$\ln\left(2\right) {\color{red}\left(\frac{d}{dn} \left(2^{n}\right)\right)} = \ln\left(2\right) {\color{red}\left(2^{n} \ln\left(2\right)\right)}$$Thus, $$$\frac{d}{dn} \left(2^{n} \ln\left(2\right)\right) = 2^{n} \ln^{2}\left(2\right)$$$.
Therefore, $$$\frac{d^{2}}{dn^{2}} \left(2^{n}\right) = 2^{n} \ln^{2}\left(2\right)$$$.
Answer
$$$\frac{d^{2}}{dn^{2}} \left(2^{n}\right) = 2^{n} \ln^{2}\left(2\right)$$$A