$$$\cos{\left(b - x \right)}$$$ 关于 $$$x$$$ 的导数

函数$$$\cos{\left(b - x \right)}$$$是两个函数$$$f{\left(u \right)} = \cos{\left(u \right)}$$$和$$$g{\left(x \right)} = b - x$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。

应用链式法则 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：

$${\color{red}\left(\frac{d}{dx} \left(\cos{\left(b - x \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right) \frac{d}{dx} \left(b - x\right)\right)}$$

余弦函数的导数是$$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$：

$${\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right)\right)} \frac{d}{dx} \left(b - x\right) = {\color{red}\left(- \sin{\left(u \right)}\right)} \frac{d}{dx} \left(b - x\right)$$

返回到原变量：

$$- \sin{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(b - x\right) = - \sin{\left({\color{red}\left(b - x\right)} \right)} \frac{d}{dx} \left(b - x\right)$$

和/差的导数等于导数的和/差：

$$- \sin{\left(b - x \right)} {\color{red}\left(\frac{d}{dx} \left(b - x\right)\right)} = - \sin{\left(b - x \right)} {\color{red}\left(\frac{db}{dx} - \frac{d}{dx} \left(x\right)\right)}$$

常数的导数是$$$0$$$:

$$- \left({\color{red}\left(\frac{db}{dx}\right)} - \frac{d}{dx} \left(x\right)\right) \sin{\left(b - x \right)} = - \left({\color{red}\left(0\right)} - \frac{d}{dx} \left(x\right)\right) \sin{\left(b - x \right)}$$

应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dx} \left(x\right) = 1$$$：

$$\sin{\left(b - x \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = \sin{\left(b - x \right)} {\color{red}\left(1\right)}$$

因此，$$$\frac{d}{dx} \left(\cos{\left(b - x \right)}\right) = \sin{\left(b - x \right)}$$$。