解答求一阶导数 $$$\frac{d}{dx} \left(\sin{\left(5 x \right)}\right)$$$函数$$$\sin{\left(5 x \right)}$$$是两个函数$$$f{\left(u \right)} = \sin{\left(u \right)}$$$和$$$g{\left(x \right)} = 5 x$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。
应用链式法则 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dx} \left(\sin{\left(5 x \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\sin{\left(u \right)}\right) \frac{d}{dx} \left(5 x\right)\right)}$$正弦函数的导数为 $$$\frac{d}{du} \left(\sin{\left(u \right)}\right) = \cos{\left(u \right)}$$$:
$${\color{red}\left(\frac{d}{du} \left(\sin{\left(u \right)}\right)\right)} \frac{d}{dx} \left(5 x\right) = {\color{red}\left(\cos{\left(u \right)}\right)} \frac{d}{dx} \left(5 x\right)$$返回到原变量：
$$\cos{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(5 x\right) = \cos{\left({\color{red}\left(5 x\right)} \right)} \frac{d}{dx} \left(5 x\right)$$对 $$$c = 5$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：
$$\cos{\left(5 x \right)} {\color{red}\left(\frac{d}{dx} \left(5 x\right)\right)} = \cos{\left(5 x \right)} {\color{red}\left(5 \frac{d}{dx} \left(x\right)\right)}$$应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dx} \left(x\right) = 1$$$：
$$5 \cos{\left(5 x \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = 5 \cos{\left(5 x \right)} {\color{red}\left(1\right)}$$因此，$$$\frac{d}{dx} \left(\sin{\left(5 x \right)}\right) = 5 \cos{\left(5 x \right)}$$$。
接下来，$$$\frac{d^{2}}{dx^{2}} \left(\sin{\left(5 x \right)}\right) = \frac{d}{dx} \left(5 \cos{\left(5 x \right)}\right)$$$对 $$$c = 5$$$ 和 $$$f{\left(x \right)} = \cos{\left(5 x \right)}$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dx} \left(5 \cos{\left(5 x \right)}\right)\right)} = {\color{red}\left(5 \frac{d}{dx} \left(\cos{\left(5 x \right)}\right)\right)}$$函数$$$\cos{\left(5 x \right)}$$$是两个函数$$$f{\left(u \right)} = \cos{\left(u \right)}$$$和$$$g{\left(x \right)} = 5 x$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。
应用链式法则 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：
$$5 {\color{red}\left(\frac{d}{dx} \left(\cos{\left(5 x \right)}\right)\right)} = 5 {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right) \frac{d}{dx} \left(5 x\right)\right)}$$余弦函数的导数是$$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$：
$$5 {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right)\right)} \frac{d}{dx} \left(5 x\right) = 5 {\color{red}\left(- \sin{\left(u \right)}\right)} \frac{d}{dx} \left(5 x\right)$$返回到原变量：
$$- 5 \sin{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(5 x\right) = - 5 \sin{\left({\color{red}\left(5 x\right)} \right)} \frac{d}{dx} \left(5 x\right)$$对 $$$c = 5$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：
$$- 5 \sin{\left(5 x \right)} {\color{red}\left(\frac{d}{dx} \left(5 x\right)\right)} = - 5 \sin{\left(5 x \right)} {\color{red}\left(5 \frac{d}{dx} \left(x\right)\right)}$$应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dx} \left(x\right) = 1$$$：
$$- 25 \sin{\left(5 x \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = - 25 \sin{\left(5 x \right)} {\color{red}\left(1\right)}$$因此，$$$\frac{d}{dx} \left(5 \cos{\left(5 x \right)}\right) = - 25 \sin{\left(5 x \right)}$$$。
因此，$$$\frac{d^{2}}{dx^{2}} \left(\sin{\left(5 x \right)}\right) = - 25 \sin{\left(5 x \right)}$$$。