|
|
|
|
|
|
|
|
|
|
|
|
$$$e^{4 t}$$$ 的導數
您的輸入
求$$$\frac{d}{dt} \left(e^{4 t}\right)$$$。
解答
函數 $$$e^{4 t}$$$ 是兩個函數 $$$f{\left(u \right)} = e^{u}$$$ 與 $$$g{\left(t \right)} = 4 t$$$ 之複合 $$$f{\left(g{\left(t \right)} \right)}$$$。
應用鏈式法則 $$$\frac{d}{dt} \left(f{\left(g{\left(t \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dt} \left(g{\left(t \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dt} \left(e^{4 t}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dt} \left(4 t\right)\right)}$$指數函數的導數為 $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$：
$${\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dt} \left(4 t\right) = {\color{red}\left(e^{u}\right)} \frac{d}{dt} \left(4 t\right)$$返回原變數：
$$e^{{\color{red}\left(u\right)}} \frac{d}{dt} \left(4 t\right) = e^{{\color{red}\left(4 t\right)}} \frac{d}{dt} \left(4 t\right)$$套用常數倍法則 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$，使用 $$$c = 4$$$ 與 $$$f{\left(t \right)} = t$$$：
$$e^{4 t} {\color{red}\left(\frac{d}{dt} \left(4 t\right)\right)} = e^{4 t} {\color{red}\left(4 \frac{d}{dt} \left(t\right)\right)}$$套用冪次法則 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$，取 $$$n = 1$$$，也就是 $$$\frac{d}{dt} \left(t\right) = 1$$$：
$$4 e^{4 t} {\color{red}\left(\frac{d}{dt} \left(t\right)\right)} = 4 e^{4 t} {\color{red}\left(1\right)}$$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