$$$2^{n}$$$ 的二阶导数
您的输入
求$$$\frac{d^{2}}{dn^{2}} \left(2^{n}\right)$$$。
解答
求一阶导数 $$$\frac{d}{dn} \left(2^{n}\right)$$$
应用指数法则 $$$\frac{d}{dn} \left(m^{n}\right) = m^{n} \ln\left(m\right)$$$，其中 $$$m = 2$$$：
$${\color{red}\left(\frac{d}{dn} \left(2^{n}\right)\right)} = {\color{red}\left(2^{n} \ln\left(2\right)\right)}$$因此，$$$\frac{d}{dn} \left(2^{n}\right) = 2^{n} \ln\left(2\right)$$$。
接下来，$$$\frac{d^{2}}{dn^{2}} \left(2^{n}\right) = \frac{d}{dn} \left(2^{n} \ln\left(2\right)\right)$$$
对 $$$c = \ln\left(2\right)$$$ 和 $$$f{\left(n \right)} = 2^{n}$$$ 应用常数倍法则 $$$\frac{d}{dn} \left(c f{\left(n \right)}\right) = c \frac{d}{dn} \left(f{\left(n \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dn} \left(2^{n} \ln\left(2\right)\right)\right)} = {\color{red}\left(\ln\left(2\right) \frac{d}{dn} \left(2^{n}\right)\right)}$$应用指数法则 $$$\frac{d}{dn} \left(m^{n}\right) = m^{n} \ln\left(m\right)$$$，其中 $$$m = 2$$$：
$$\ln\left(2\right) {\color{red}\left(\frac{d}{dn} \left(2^{n}\right)\right)} = \ln\left(2\right) {\color{red}\left(2^{n} \ln\left(2\right)\right)}$$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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