Funktion $$$\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$$ derivaatta
Aiheeseen liittyvät laskurit: Logaritmisen derivoinnin laskin, Vaiheittainen implisiittisen derivoinnin laskin
Syötteesi
Määritä $$$\frac{d}{d\theta} \left(\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}\right)$$$.
Ratkaisu
Funktio $$$\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$$ on kahden funktion $$$f{\left(u \right)} = \tan{\left(u \right)}$$$ ja $$$g{\left(\theta \right)} = \frac{\theta}{2} + \frac{\pi}{4}$$$ yhdistelmä $$$f{\left(g{\left(\theta \right)} \right)}$$$.
Sovella ketjusääntöä $$$\frac{d}{d\theta} \left(f{\left(g{\left(\theta \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{d\theta} \left(g{\left(\theta \right)}\right)$$$:
$${\color{red}\left(\frac{d}{d\theta} \left(\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\tan{\left(u \right)}\right) \frac{d}{d\theta} \left(\frac{\theta}{2} + \frac{\pi}{4}\right)\right)}$$Tangenttifunktion derivaatta on $$$\frac{d}{du} \left(\tan{\left(u \right)}\right) = \sec^{2}{\left(u \right)}$$$:
$${\color{red}\left(\frac{d}{du} \left(\tan{\left(u \right)}\right)\right)} \frac{d}{d\theta} \left(\frac{\theta}{2} + \frac{\pi}{4}\right) = {\color{red}\left(\sec^{2}{\left(u \right)}\right)} \frac{d}{d\theta} \left(\frac{\theta}{2} + \frac{\pi}{4}\right)$$Palaa alkuperäiseen muuttujaan:
$$\sec^{2}{\left({\color{red}\left(u\right)} \right)} \frac{d}{d\theta} \left(\frac{\theta}{2} + \frac{\pi}{4}\right) = \sec^{2}{\left({\color{red}\left(\frac{\theta}{2} + \frac{\pi}{4}\right)} \right)} \frac{d}{d\theta} \left(\frac{\theta}{2} + \frac{\pi}{4}\right)$$Summan/erotuksen derivaatta on derivaattojen summa/erotus:
$$\sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{d\theta} \left(\frac{\theta}{2} + \frac{\pi}{4}\right)\right)} = \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{d\theta} \left(\frac{\theta}{2}\right) + \frac{d}{d\theta} \left(\frac{\pi}{4}\right)\right)}$$Sovella vakion kerroinsääntöä $$$\frac{d}{d\theta} \left(c f{\left(\theta \right)}\right) = c \frac{d}{d\theta} \left(f{\left(\theta \right)}\right)$$$ käyttäen $$$c = \frac{1}{2}$$$ ja $$$f{\left(\theta \right)} = \theta$$$:
$$\left({\color{red}\left(\frac{d}{d\theta} \left(\frac{\theta}{2}\right)\right)} + \frac{d}{d\theta} \left(\frac{\pi}{4}\right)\right) \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)} = \left({\color{red}\left(\frac{\frac{d}{d\theta} \left(\theta\right)}{2}\right)} + \frac{d}{d\theta} \left(\frac{\pi}{4}\right)\right) \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$Sovella potenssisääntöä $$$\frac{d}{d\theta} \left(\theta^{n}\right) = n \theta^{n - 1}$$$ käyttäen $$$n = 1$$$, toisin sanoen, $$$\frac{d}{d\theta} \left(\theta\right) = 1$$$:
$$\left(\frac{{\color{red}\left(\frac{d}{d\theta} \left(\theta\right)\right)}}{2} + \frac{d}{d\theta} \left(\frac{\pi}{4}\right)\right) \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)} = \left(\frac{{\color{red}\left(1\right)}}{2} + \frac{d}{d\theta} \left(\frac{\pi}{4}\right)\right) \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$Vakion derivaatta on $$$0$$$:
$$\left({\color{red}\left(\frac{d}{d\theta} \left(\frac{\pi}{4}\right)\right)} + \frac{1}{2}\right) \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)} = \left({\color{red}\left(0\right)} + \frac{1}{2}\right) \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$Sievennä:
$$\frac{\sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}}{2} = \frac{1}{1 - \sin{\left(\theta \right)}}$$Näin ollen, $$$\frac{d}{d\theta} \left(\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}\right) = \frac{1}{1 - \sin{\left(\theta \right)}}$$$.
Vastaus
$$$\frac{d}{d\theta} \left(\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}\right) = \frac{1}{1 - \sin{\left(\theta \right)}}$$$A