$$$\frac{\sqrt{6} \cos{\left(t + \frac{\pi}{4} \right)}}{3}$$$の導関数

定数倍の法則 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$ を $$$c = \frac{\sqrt{6}}{3}$$$ と $$$f{\left(t \right)} = \cos{\left(t + \frac{\pi}{4} \right)}$$$ に対して適用します:

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

関数$$$\cos{\left(t + \frac{\pi}{4} \right)}$$$は、2つの関数$$$f{\left(u \right)} = \cos{\left(u \right)}$$$と$$$g{\left(t \right)} = t + \frac{\pi}{4}$$$の合成$$$f{\left(g{\left(t \right)} \right)}$$$である。

連鎖律 $$$\frac{d}{dt} \left(f{\left(g{\left(t \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dt} \left(g{\left(t \right)}\right)$$$ を適用する:

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

余弦関数の導関数は$$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$:

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

元の変数に戻す:

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

和/差の導関数は、導関数の和/差である:

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

$$$n = 1$$$ を用いて冪法則 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ を適用すると、すなわち $$$\frac{d}{dt} \left(t\right) = 1$$$:

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

定数の導数は$$$0$$$です:

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

したがって、$$$\frac{d}{dt} \left(\frac{\sqrt{6} \cos{\left(t + \frac{\pi}{4} \right)}}{3}\right) = - \frac{\sqrt{6} \sin{\left(t + \frac{\pi}{4} \right)}}{3}$$$。