$$$2^{n}$$$ 的二階導數
您的輸入
求$$$\frac{d^{2}}{dn^{2}} \left(2^{n}\right)$$$。
解答
求第一階導數 $$$\frac{d}{dn} \left(2^{n}\right)$$$
套用指數法則 $$$\frac{d}{dn} \left(m^{n}\right) = m^{n} \ln\left(m\right)$$$，令 $$$m = 2$$$：
$${\color{red}\left(\frac{d}{dn} \left(2^{n}\right)\right)} = {\color{red}\left(2^{n} \ln\left(2\right)\right)}$$因此，$$$\frac{d}{dn} \left(2^{n}\right) = 2^{n} \ln\left(2\right)$$$。
接下來，$$$\frac{d^{2}}{dn^{2}} \left(2^{n}\right) = \frac{d}{dn} \left(2^{n} \ln\left(2\right)\right)$$$
套用常數倍法則 $$$\frac{d}{dn} \left(c f{\left(n \right)}\right) = c \frac{d}{dn} \left(f{\left(n \right)}\right)$$$，使用 $$$c = \ln\left(2\right)$$$ 與 $$$f{\left(n \right)} = 2^{n}$$$：
$${\color{red}\left(\frac{d}{dn} \left(2^{n} \ln\left(2\right)\right)\right)} = {\color{red}\left(\ln\left(2\right) \frac{d}{dn} \left(2^{n}\right)\right)}$$套用指數法則 $$$\frac{d}{dn} \left(m^{n}\right) = m^{n} \ln\left(m\right)$$$，令 $$$m = 2$$$：
$$\ln\left(2\right) {\color{red}\left(\frac{d}{dn} \left(2^{n}\right)\right)} = \ln\left(2\right) {\color{red}\left(2^{n} \ln\left(2\right)\right)}$$因此，$$$\frac{d}{dn} \left(2^{n} \ln\left(2\right)\right) = 2^{n} \ln^{2}\left(2\right)$$$。
因此，$$$\frac{d^{2}}{dn^{2}} \left(2^{n}\right) = 2^{n} \ln^{2}\left(2\right)$$$。
答案
$$$\frac{d^{2}}{dn^{2}} \left(2^{n}\right) = 2^{n} \ln^{2}\left(2\right)$$$A
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