$$$e^{- 8 x^{4}}$$$ 的導數
您的輸入
求$$$\frac{d}{dx} \left(e^{- 8 x^{4}}\right)$$$。
解答
函數 $$$e^{- 8 x^{4}}$$$ 是兩個函數 $$$f{\left(u \right)} = e^{u}$$$ 與 $$$g{\left(x \right)} = - 8 x^{4}$$$ 之複合 $$$f{\left(g{\left(x \right)} \right)}$$$。
應用鏈式法則 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dx} \left(e^{- 8 x^{4}}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(- 8 x^{4}\right)\right)}$$指數函數的導數為 $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$：
$${\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(- 8 x^{4}\right) = {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(- 8 x^{4}\right)$$返回原變數：
$$e^{{\color{red}\left(u\right)}} \frac{d}{dx} \left(- 8 x^{4}\right) = e^{{\color{red}\left(- 8 x^{4}\right)}} \frac{d}{dx} \left(- 8 x^{4}\right)$$套用常數倍法則 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$，使用 $$$c = -8$$$ 與 $$$f{\left(x \right)} = x^{4}$$$：
$$e^{- 8 x^{4}} {\color{red}\left(\frac{d}{dx} \left(- 8 x^{4}\right)\right)} = e^{- 8 x^{4}} {\color{red}\left(- 8 \frac{d}{dx} \left(x^{4}\right)\right)}$$套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 4$$$：
$$- 8 e^{- 8 x^{4}} {\color{red}\left(\frac{d}{dx} \left(x^{4}\right)\right)} = - 8 e^{- 8 x^{4}} {\color{red}\left(4 x^{3}\right)}$$因此，$$$\frac{d}{dx} \left(e^{- 8 x^{4}}\right) = - 32 x^{3} e^{- 8 x^{4}}$$$。
答案
$$$\frac{d}{dx} \left(e^{- 8 x^{4}}\right) = - 32 x^{3} e^{- 8 x^{4}}$$$A