Derivative of $$$\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$

The function $$$\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$ is the composition $$$f{\left(g{\left(u \right)} \right)}$$$ of two functions $$$f{\left(v \right)} = \tan{\left(v \right)}$$$ and $$$g{\left(u \right)} = \frac{u}{2} + \frac{\pi}{4}$$$.

Apply the chain rule $$$\frac{d}{du} \left(f{\left(g{\left(u \right)} \right)}\right) = \frac{d}{dv} \left(f{\left(v \right)}\right) \frac{d}{du} \left(g{\left(u \right)}\right)$$$:

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

The derivative of the tangent is $$$\frac{d}{dv} \left(\tan{\left(v \right)}\right) = \sec^{2}{\left(v \right)}$$$:

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

Return to the old variable:

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

The derivative of a sum/difference is the sum/difference of derivatives:

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

The derivative of a constant is $$$0$$$:

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

Apply the constant multiple rule $$$\frac{d}{du} \left(c f{\left(u \right)}\right) = c \frac{d}{du} \left(f{\left(u \right)}\right)$$$ with $$$c = \frac{1}{2}$$$ and $$$f{\left(u \right)} = u$$$:

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

Apply the power rule $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{du} \left(u\right) = 1$$$:

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

Simplify:

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

Thus, $$$\frac{d}{du} \left(\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}\right) = \frac{1}{1 - \sin{\left(u \right)}}$$$.