Funktion $$$\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$ derivaatta

Funktio $$$\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$ on kahden funktion $$$f{\left(v \right)} = \tan{\left(v \right)}$$$ ja $$$g{\left(u \right)} = \frac{u}{2} + \frac{\pi}{4}$$$ yhdistelmä $$$f{\left(g{\left(u \right)} \right)}$$$.

Sovella ketjusääntöä $$$\frac{d}{du} \left(f{\left(g{\left(u \right)} \right)}\right) = \frac{d}{dv} \left(f{\left(v \right)}\right) \frac{d}{du} \left(g{\left(u \right)}\right)$$$:

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

Tangenttifunktion derivaatta on $$$\frac{d}{dv} \left(\tan{\left(v \right)}\right) = \sec^{2}{\left(v \right)}$$$:

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

Palaa alkuperäiseen muuttujaan:

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

Summan/erotuksen derivaatta on derivaattojen summa/erotus:

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

Vakion derivaatta on $$$0$$$:

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

Sovella vakion kerroinsääntöä $$$\frac{d}{du} \left(c f{\left(u \right)}\right) = c \frac{d}{du} \left(f{\left(u \right)}\right)$$$ käyttäen $$$c = \frac{1}{2}$$$ ja $$$f{\left(u \right)} = u$$$:

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

Sovella potenssisääntöä $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ käyttäen $$$n = 1$$$, toisin sanoen, $$$\frac{d}{du} \left(u\right) = 1$$$:

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

Sievennä:

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

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