Ableitung von $$$\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$$

Die Funktion $$$\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$$ ist die Komposition $$$f{\left(g{\left(\theta \right)} \right)}$$$ der beiden Funktionen $$$f{\left(u \right)} = \tan{\left(u \right)}$$$ und $$$g{\left(\theta \right)} = \frac{\theta}{2} + \frac{\pi}{4}$$$.

Wende die Kettenregel $$$\frac{d}{d\theta} \left(f{\left(g{\left(\theta \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{d\theta} \left(g{\left(\theta \right)}\right)$$$ an:

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

Die Ableitung des Tangens ist $$$\frac{d}{du} \left(\tan{\left(u \right)}\right) = \sec^{2}{\left(u \right)}$$$:

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

Zurück zur ursprünglichen Variable:

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

Die Ableitung einer Summe/Differenz ist die Summe/Differenz der Ableitungen:

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

Wende die Konstantenfaktorregel $$$\frac{d}{d\theta} \left(c f{\left(\theta \right)}\right) = c \frac{d}{d\theta} \left(f{\left(\theta \right)}\right)$$$ mit $$$c = \frac{1}{2}$$$ und $$$f{\left(\theta \right)} = \theta$$$ an:

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

Wenden Sie die Potenzregel $$$\frac{d}{d\theta} \left(\theta^{n}\right) = n \theta^{n - 1}$$$ mit $$$n = 1$$$ an, mit anderen Worten, $$$\frac{d}{d\theta} \left(\theta\right) = 1$$$:

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

Die Ableitung einer Konstante ist $$$0$$$:

$$\left({\color{red}\left(\frac{d}{d\theta} \left(\frac{\pi}{4}\right)\right)} + \frac{1}{2}\right) \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)} = \left({\color{red}\left(0\right)} + \frac{1}{2}\right) \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$

Vereinfachen:

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

Somit gilt $$$\frac{d}{d\theta} \left(\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}\right) = \frac{1}{1 - \sin{\left(\theta \right)}}$$$.