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$$$\cos^{2}{\left(x \right)}$$$ 的二階導數
您的輸入
求$$$\frac{d^{2}}{dx^{2}} \left(\cos^{2}{\left(x \right)}\right)$$$。
解答
求第一階導數 $$$\frac{d}{dx} \left(\cos^{2}{\left(x \right)}\right)$$$
函數 $$$\cos^{2}{\left(x \right)}$$$ 是兩個函數 $$$f{\left(u \right)} = u^{2}$$$ 與 $$$g{\left(x \right)} = \cos{\left(x \right)}$$$ 之複合 $$$f{\left(g{\left(x \right)} \right)}$$$。
應用鏈式法則 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dx} \left(\cos^{2}{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(\cos{\left(x \right)}\right)\right)}$$套用冪次法則 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$，取 $$$n = 2$$$：
$${\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(\cos{\left(x \right)}\right) = {\color{red}\left(2 u\right)} \frac{d}{dx} \left(\cos{\left(x \right)}\right)$$返回原變數：
$$2 {\color{red}\left(u\right)} \frac{d}{dx} \left(\cos{\left(x \right)}\right) = 2 {\color{red}\left(\cos{\left(x \right)}\right)} \frac{d}{dx} \left(\cos{\left(x \right)}\right)$$餘弦函數的導數為 $$$\frac{d}{dx} \left(\cos{\left(x \right)}\right) = - \sin{\left(x \right)}$$$：
$$2 \cos{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\cos{\left(x \right)}\right)\right)} = 2 \cos{\left(x \right)} {\color{red}\left(- \sin{\left(x \right)}\right)}$$化簡：
$$- 2 \sin{\left(x \right)} \cos{\left(x \right)} = - \sin{\left(2 x \right)}$$因此，$$$\frac{d}{dx} \left(\cos^{2}{\left(x \right)}\right) = - \sin{\left(2 x \right)}$$$。
接下來，$$$\frac{d^{2}}{dx^{2}} \left(\cos^{2}{\left(x \right)}\right) = \frac{d}{dx} \left(- \sin{\left(2 x \right)}\right)$$$
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 $$$c = -1$$$ 與 $$$f{\left(x \right)} = \sin{\left(2 x \right)}$$$：
$${\color{red}\left(\frac{d}{dx} \left(- \sin{\left(2 x \right)}\right)\right)} = {\color{red}\left(- \frac{d}{dx} \left(\sin{\left(2 x \right)}\right)\right)}$$函數 $$$\sin{\left(2 x \right)}$$$ 是兩個函數 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ 與 $$$g{\left(x \right)} = 2 x$$$ 之複合 $$$f{\left(g{\left(x \right)} \right)}$$$。
應用鏈式法則 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：
$$- {\color{red}\left(\frac{d}{dx} \left(\sin{\left(2 x \right)}\right)\right)} = - {\color{red}\left(\frac{d}{du} \left(\sin{\left(u \right)}\right) \frac{d}{dx} \left(2 x\right)\right)}$$正弦函數的導數為$$$\frac{d}{du} \left(\sin{\left(u \right)}\right) = \cos{\left(u \right)}$$$：
$$- {\color{red}\left(\frac{d}{du} \left(\sin{\left(u \right)}\right)\right)} \frac{d}{dx} \left(2 x\right) = - {\color{red}\left(\cos{\left(u \right)}\right)} \frac{d}{dx} \left(2 x\right)$$返回原變數：
$$- \cos{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(2 x\right) = - \cos{\left({\color{red}\left(2 x\right)} \right)} \frac{d}{dx} \left(2 x\right)$$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 $$$c = 2$$$ 與 $$$f{\left(x \right)} = x$$$：
$$- \cos{\left(2 x \right)} {\color{red}\left(\frac{d}{dx} \left(2 x\right)\right)} = - \cos{\left(2 x \right)} {\color{red}\left(2 \frac{d}{dx} \left(x\right)\right)}$$套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是 $$$\frac{d}{dx} \left(x\right) = 1$$$：
$$- 2 \cos{\left(2 x \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = - 2 \cos{\left(2 x \right)} {\color{red}\left(1\right)}$$因此，$$$\frac{d}{dx} \left(- \sin{\left(2 x \right)}\right) = - 2 \cos{\left(2 x \right)}$$$。
因此，$$$\frac{d^{2}}{dx^{2}} \left(\cos^{2}{\left(x \right)}\right) = - 2 \cos{\left(2 x \right)}$$$。
答案
$$$\frac{d^{2}}{dx^{2}} \left(\cos^{2}{\left(x \right)}\right) = - 2 \cos{\left(2 x \right)}$$$A