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

상수배 법칙 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$을 $$$c = \frac{\sqrt{6}}{3}$$$와 $$$f{\left(t \right)} = \cos{\left(t + \frac{\pi}{4} \right)}$$$에 적용합니다:

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

함수 $$$\cos{\left(t + \frac{\pi}{4} \right)}$$$는 두 함수 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$와 $$$g{\left(t \right)} = t + \frac{\pi}{4}$$$의 합성함수 $$$f{\left(g{\left(t \right)} \right)}$$$이다.

연쇄법칙 $$$\frac{d}{dt} \left(f{\left(g{\left(t \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dt} \left(g{\left(t \right)}\right)$$$을(를) 적용하십시오:

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

코사인의 도함수는 $$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$입니다:

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

역치환:

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

합/차의 도함수는 도함수들의 합/차이다:

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

멱법칙 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$을 $$$n = 1$$$에 대해 적용하면, 즉 $$$\frac{d}{dt} \left(t\right) = 1$$$:

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

상수의 도함수는 $$$0$$$입니다:

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

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