Ableitung von $$$\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$
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Ihre Eingabe
Bestimme $$$\frac{d}{du} \left(\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}\right)$$$.
Lösung
Die Funktion $$$\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$ ist die Komposition $$$f{\left(g{\left(u \right)} \right)}$$$ der beiden Funktionen $$$f{\left(v \right)} = \tan{\left(v \right)}$$$ und $$$g{\left(u \right)} = \frac{u}{2} + \frac{\pi}{4}$$$.
Wende die Kettenregel $$$\frac{d}{du} \left(f{\left(g{\left(u \right)} \right)}\right) = \frac{d}{dv} \left(f{\left(v \right)}\right) \frac{d}{du} \left(g{\left(u \right)}\right)$$$ an:
$${\color{red}\left(\frac{d}{du} \left(\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}\right)\right)} = {\color{red}\left(\frac{d}{dv} \left(\tan{\left(v \right)}\right) \frac{d}{du} \left(\frac{u}{2} + \frac{\pi}{4}\right)\right)}$$Die Ableitung des Tangens ist $$$\frac{d}{dv} \left(\tan{\left(v \right)}\right) = \sec^{2}{\left(v \right)}$$$:
$${\color{red}\left(\frac{d}{dv} \left(\tan{\left(v \right)}\right)\right)} \frac{d}{du} \left(\frac{u}{2} + \frac{\pi}{4}\right) = {\color{red}\left(\sec^{2}{\left(v \right)}\right)} \frac{d}{du} \left(\frac{u}{2} + \frac{\pi}{4}\right)$$Zurück zur ursprünglichen Variable:
$$\sec^{2}{\left({\color{red}\left(v\right)} \right)} \frac{d}{du} \left(\frac{u}{2} + \frac{\pi}{4}\right) = \sec^{2}{\left({\color{red}\left(\frac{u}{2} + \frac{\pi}{4}\right)} \right)} \frac{d}{du} \left(\frac{u}{2} + \frac{\pi}{4}\right)$$Die Ableitung einer Summe/Differenz ist die Summe/Differenz der Ableitungen:
$$\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{du} \left(\frac{u}{2} + \frac{\pi}{4}\right)\right)} = \sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{du} \left(\frac{u}{2}\right) + \frac{d}{du} \left(\frac{\pi}{4}\right)\right)}$$Die Ableitung einer Konstante ist $$$0$$$:
$$\left({\color{red}\left(\frac{d}{du} \left(\frac{\pi}{4}\right)\right)} + \frac{d}{du} \left(\frac{u}{2}\right)\right) \sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} = \left({\color{red}\left(0\right)} + \frac{d}{du} \left(\frac{u}{2}\right)\right) \sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$Wende die Konstantenfaktorregel $$$\frac{d}{du} \left(c f{\left(u \right)}\right) = c \frac{d}{du} \left(f{\left(u \right)}\right)$$$ mit $$$c = \frac{1}{2}$$$ und $$$f{\left(u \right)} = u$$$ an:
$$\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{du} \left(\frac{u}{2}\right)\right)} = \sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{\frac{d}{du} \left(u\right)}{2}\right)}$$Wenden Sie die Potenzregel $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ mit $$$n = 1$$$ an, mit anderen Worten, $$$\frac{d}{du} \left(u\right) = 1$$$:
$$\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{du} \left(u\right)\right)}}{2} = \frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} {\color{red}\left(1\right)}}{2}$$Vereinfachen:
$$\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2} = \frac{1}{1 - \sin{\left(u \right)}}$$Somit gilt $$$\frac{d}{du} \left(\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}\right) = \frac{1}{1 - \sin{\left(u \right)}}$$$.
Antwort
$$$\frac{d}{du} \left(\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}\right) = \frac{1}{1 - \sin{\left(u \right)}}$$$A