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$$$e^{4 t}$$$的导数
您的输入
求$$$\frac{d}{dt} \left(e^{4 t}\right)$$$。
解答
函数$$$e^{4 t}$$$是两个函数$$$f{\left(u \right)} = e^{u}$$$和$$$g{\left(t \right)} = 4 t$$$的复合$$$f{\left(g{\left(t \right)} \right)}$$$。
应用链式法则 $$$\frac{d}{dt} \left(f{\left(g{\left(t \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dt} \left(g{\left(t \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dt} \left(e^{4 t}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dt} \left(4 t\right)\right)}$$指数函数的导数为 $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$：
$${\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dt} \left(4 t\right) = {\color{red}\left(e^{u}\right)} \frac{d}{dt} \left(4 t\right)$$返回到原变量：
$$e^{{\color{red}\left(u\right)}} \frac{d}{dt} \left(4 t\right) = e^{{\color{red}\left(4 t\right)}} \frac{d}{dt} \left(4 t\right)$$对 $$$c = 4$$$ 和 $$$f{\left(t \right)} = t$$$ 应用常数倍法则 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$：
$$e^{4 t} {\color{red}\left(\frac{d}{dt} \left(4 t\right)\right)} = e^{4 t} {\color{red}\left(4 \frac{d}{dt} \left(t\right)\right)}$$应用幂法则 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dt} \left(t\right) = 1$$$：
$$4 e^{4 t} {\color{red}\left(\frac{d}{dt} \left(t\right)\right)} = 4 e^{4 t} {\color{red}\left(1\right)}$$因此，$$$\frac{d}{dt} \left(e^{4 t}\right) = 4 e^{4 t}$$$。
答案
$$$\frac{d}{dt} \left(e^{4 t}\right) = 4 e^{4 t}$$$A