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

Terapkan aturan kelipatan konstanta $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$ dengan $$$c = \frac{\sqrt{6}}{3}$$$ dan $$$f{\left(t \right)} = \cos{\left(t + \frac{\pi}{4} \right)}$$$:

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

Fungsi $$$\cos{\left(t + \frac{\pi}{4} \right)}$$$ merupakan komposisi $$$f{\left(g{\left(t \right)} \right)}$$$ dari dua fungsi $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ dan $$$g{\left(t \right)} = t + \frac{\pi}{4}$$$.

Terapkan aturan rantai $$$\frac{d}{dt} \left(f{\left(g{\left(t \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dt} \left(g{\left(t \right)}\right)$$$:

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

Turunan fungsi kosinus adalah $$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$:

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

Kembalikan ke variabel semula:

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

Turunan dari jumlah/selisih adalah jumlah/selisih dari turunan:

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

Terapkan aturan pangkat $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ dengan $$$n = 1$$$, dengan kata lain, $$$\frac{d}{dt} \left(t\right) = 1$$$:

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

Turunan dari suatu konstanta adalah $$$0$$$:

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

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