Derivatan av $$$\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$$
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Din inmatning
Bestäm $$$\frac{d}{d\theta} \left(\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}\right)$$$.
Lösning
Funktionen $$$\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$$ är sammansättningen $$$f{\left(g{\left(\theta \right)} \right)}$$$ av två funktioner $$$f{\left(u \right)} = \tan{\left(u \right)}$$$ och $$$g{\left(\theta \right)} = \frac{\theta}{2} + \frac{\pi}{4}$$$.
Tillämpa kedjeregeln $$$\frac{d}{d\theta} \left(f{\left(g{\left(\theta \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{d\theta} \left(g{\left(\theta \right)}\right)$$$:
$${\color{red}\left(\frac{d}{d\theta} \left(\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\tan{\left(u \right)}\right) \frac{d}{d\theta} \left(\frac{\theta}{2} + \frac{\pi}{4}\right)\right)}$$Derivatan av tangens är $$$\frac{d}{du} \left(\tan{\left(u \right)}\right) = \sec^{2}{\left(u \right)}$$$:
$${\color{red}\left(\frac{d}{du} \left(\tan{\left(u \right)}\right)\right)} \frac{d}{d\theta} \left(\frac{\theta}{2} + \frac{\pi}{4}\right) = {\color{red}\left(\sec^{2}{\left(u \right)}\right)} \frac{d}{d\theta} \left(\frac{\theta}{2} + \frac{\pi}{4}\right)$$Återgå till den ursprungliga variabeln:
$$\sec^{2}{\left({\color{red}\left(u\right)} \right)} \frac{d}{d\theta} \left(\frac{\theta}{2} + \frac{\pi}{4}\right) = \sec^{2}{\left({\color{red}\left(\frac{\theta}{2} + \frac{\pi}{4}\right)} \right)} \frac{d}{d\theta} \left(\frac{\theta}{2} + \frac{\pi}{4}\right)$$Derivatan av en summa/differens är summan/differensen av derivatorna:
$$\sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{d\theta} \left(\frac{\theta}{2} + \frac{\pi}{4}\right)\right)} = \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{d\theta} \left(\frac{\theta}{2}\right) + \frac{d}{d\theta} \left(\frac{\pi}{4}\right)\right)}$$Tillämpa konstantfaktorregeln $$$\frac{d}{d\theta} \left(c f{\left(\theta \right)}\right) = c \frac{d}{d\theta} \left(f{\left(\theta \right)}\right)$$$ med $$$c = \frac{1}{2}$$$ och $$$f{\left(\theta \right)} = \theta$$$:
$$\left({\color{red}\left(\frac{d}{d\theta} \left(\frac{\theta}{2}\right)\right)} + \frac{d}{d\theta} \left(\frac{\pi}{4}\right)\right) \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)} = \left({\color{red}\left(\frac{\frac{d}{d\theta} \left(\theta\right)}{2}\right)} + \frac{d}{d\theta} \left(\frac{\pi}{4}\right)\right) \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$Tillämpa potensregeln $$$\frac{d}{d\theta} \left(\theta^{n}\right) = n \theta^{n - 1}$$$ med $$$n = 1$$$, det vill säga $$$\frac{d}{d\theta} \left(\theta\right) = 1$$$:
$$\left(\frac{{\color{red}\left(\frac{d}{d\theta} \left(\theta\right)\right)}}{2} + \frac{d}{d\theta} \left(\frac{\pi}{4}\right)\right) \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)} = \left(\frac{{\color{red}\left(1\right)}}{2} + \frac{d}{d\theta} \left(\frac{\pi}{4}\right)\right) \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$Derivatan av en konstant är $$$0$$$:
$$\left({\color{red}\left(\frac{d}{d\theta} \left(\frac{\pi}{4}\right)\right)} + \frac{1}{2}\right) \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)} = \left({\color{red}\left(0\right)} + \frac{1}{2}\right) \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$Förenkla:
$$\frac{\sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}}{2} = \frac{1}{1 - \sin{\left(\theta \right)}}$$Alltså, $$$\frac{d}{d\theta} \left(\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}\right) = \frac{1}{1 - \sin{\left(\theta \right)}}$$$.
Svar
$$$\frac{d}{d\theta} \left(\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}\right) = \frac{1}{1 - \sin{\left(\theta \right)}}$$$A