$$$\tan{\left(2 x + \frac{\pi}{4} \right)}$$$的导数

函数$$$\tan{\left(2 x + \frac{\pi}{4} \right)}$$$是两个函数$$$f{\left(u \right)} = \tan{\left(u \right)}$$$和$$$g{\left(x \right)} = 2 x + \frac{\pi}{4}$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。

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

$${\color{red}\left(\frac{d}{dx} \left(\tan{\left(2 x + \frac{\pi}{4} \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\tan{\left(u \right)}\right) \frac{d}{dx} \left(2 x + \frac{\pi}{4}\right)\right)}$$

正切函数的导数为$$$\frac{d}{du} \left(\tan{\left(u \right)}\right) = \sec^{2}{\left(u \right)}$$$:

$${\color{red}\left(\frac{d}{du} \left(\tan{\left(u \right)}\right)\right)} \frac{d}{dx} \left(2 x + \frac{\pi}{4}\right) = {\color{red}\left(\sec^{2}{\left(u \right)}\right)} \frac{d}{dx} \left(2 x + \frac{\pi}{4}\right)$$

返回到原变量：

$$\sec^{2}{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(2 x + \frac{\pi}{4}\right) = \sec^{2}{\left({\color{red}\left(2 x + \frac{\pi}{4}\right)} \right)} \frac{d}{dx} \left(2 x + \frac{\pi}{4}\right)$$

和/差的导数等于导数的和/差：

$$\sec^{2}{\left(2 x + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{dx} \left(2 x + \frac{\pi}{4}\right)\right)} = \sec^{2}{\left(2 x + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{dx} \left(2 x\right) + \frac{d}{dx} \left(\frac{\pi}{4}\right)\right)}$$

对 $$$c = 2$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：

$$\left({\color{red}\left(\frac{d}{dx} \left(2 x\right)\right)} + \frac{d}{dx} \left(\frac{\pi}{4}\right)\right) \sec^{2}{\left(2 x + \frac{\pi}{4} \right)} = \left({\color{red}\left(2 \frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(\frac{\pi}{4}\right)\right) \sec^{2}{\left(2 x + \frac{\pi}{4} \right)}$$

常数的导数是$$$0$$$:

$$\left({\color{red}\left(\frac{d}{dx} \left(\frac{\pi}{4}\right)\right)} + 2 \frac{d}{dx} \left(x\right)\right) \sec^{2}{\left(2 x + \frac{\pi}{4} \right)} = \left({\color{red}\left(0\right)} + 2 \frac{d}{dx} \left(x\right)\right) \sec^{2}{\left(2 x + \frac{\pi}{4} \right)}$$

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

$$2 \sec^{2}{\left(2 x + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = 2 \sec^{2}{\left(2 x + \frac{\pi}{4} \right)} {\color{red}\left(1\right)}$$

化简：

$$2 \sec^{2}{\left(2 x + \frac{\pi}{4} \right)} = \frac{4}{1 - \sin{\left(4 x \right)}}$$

因此，$$$\frac{d}{dx} \left(\tan{\left(2 x + \frac{\pi}{4} \right)}\right) = \frac{4}{1 - \sin{\left(4 x \right)}}$$$。