Derivatan av $$$\frac{\sqrt{6} \cos{\left(t + \frac{\pi}{4} \right)}}{3}$$$

Tillämpa konstantfaktorregeln $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$ med $$$c = \frac{\sqrt{6}}{3}$$$ och $$$f{\left(t \right)} = \cos{\left(t + \frac{\pi}{4} \right)}$$$:

$${\color{red}\left(\frac{d}{dt} \left(\frac{\sqrt{6} \cos{\left(t + \frac{\pi}{4} \right)}}{3}\right)\right)} = {\color{red}\left(\frac{\sqrt{6}}{3} \frac{d}{dt} \left(\cos{\left(t + \frac{\pi}{4} \right)}\right)\right)}$$

Funktionen $$$\cos{\left(t + \frac{\pi}{4} \right)}$$$ är sammansättningen $$$f{\left(g{\left(t \right)} \right)}$$$ av två funktioner $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ och $$$g{\left(t \right)} = t + \frac{\pi}{4}$$$.

Tillämpa kedjeregeln $$$\frac{d}{dt} \left(f{\left(g{\left(t \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dt} \left(g{\left(t \right)}\right)$$$:

$$\frac{\sqrt{6} {\color{red}\left(\frac{d}{dt} \left(\cos{\left(t + \frac{\pi}{4} \right)}\right)\right)}}{3} = \frac{\sqrt{6} {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right) \frac{d}{dt} \left(t + \frac{\pi}{4}\right)\right)}}{3}$$

Derivatan av cosinus är $$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$:

$$\frac{\sqrt{6} {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right)\right)} \frac{d}{dt} \left(t + \frac{\pi}{4}\right)}{3} = \frac{\sqrt{6} {\color{red}\left(- \sin{\left(u \right)}\right)} \frac{d}{dt} \left(t + \frac{\pi}{4}\right)}{3}$$

Återgå till den ursprungliga variabeln:

$$- \frac{\sqrt{6} \sin{\left({\color{red}\left(u\right)} \right)} \frac{d}{dt} \left(t + \frac{\pi}{4}\right)}{3} = - \frac{\sqrt{6} \sin{\left({\color{red}\left(t + \frac{\pi}{4}\right)} \right)} \frac{d}{dt} \left(t + \frac{\pi}{4}\right)}{3}$$

Derivatan av en summa/differens är summan/differensen av derivatorna:

$$- \frac{\sqrt{6} \sin{\left(t + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{dt} \left(t + \frac{\pi}{4}\right)\right)}}{3} = - \frac{\sqrt{6} \sin{\left(t + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{dt} \left(t\right) + \frac{d}{dt} \left(\frac{\pi}{4}\right)\right)}}{3}$$

Tillämpa potensregeln $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ med $$$n = 1$$$, det vill säga $$$\frac{d}{dt} \left(t\right) = 1$$$:

$$- \frac{\sqrt{6} \left({\color{red}\left(\frac{d}{dt} \left(t\right)\right)} + \frac{d}{dt} \left(\frac{\pi}{4}\right)\right) \sin{\left(t + \frac{\pi}{4} \right)}}{3} = - \frac{\sqrt{6} \left({\color{red}\left(1\right)} + \frac{d}{dt} \left(\frac{\pi}{4}\right)\right) \sin{\left(t + \frac{\pi}{4} \right)}}{3}$$

Derivatan av en konstant är $$$0$$$:

$$- \frac{\sqrt{6} \left({\color{red}\left(\frac{d}{dt} \left(\frac{\pi}{4}\right)\right)} + 1\right) \sin{\left(t + \frac{\pi}{4} \right)}}{3} = - \frac{\sqrt{6} \left({\color{red}\left(0\right)} + 1\right) \sin{\left(t + \frac{\pi}{4} \right)}}{3}$$

Alltså, $$$\frac{d}{dt} \left(\frac{\sqrt{6} \cos{\left(t + \frac{\pi}{4} \right)}}{3}\right) = - \frac{\sqrt{6} \sin{\left(t + \frac{\pi}{4} \right)}}{3}$$$.