$$$\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$$의 도함수

함수 $$$\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$$는 두 함수 $$$f{\left(u \right)} = \tan{\left(u \right)}$$$와 $$$g{\left(\theta \right)} = \frac{\theta}{2} + \frac{\pi}{4}$$$의 합성함수 $$$f{\left(g{\left(\theta \right)} \right)}$$$이다.

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

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

탄젠트 함수의 도함수는 $$$\frac{d}{du} \left(\tan{\left(u \right)}\right) = \sec^{2}{\left(u \right)}$$$:

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

역치환:

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

합/차의 도함수는 도함수들의 합/차이다:

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

상수배 법칙 $$$\frac{d}{d\theta} \left(c f{\left(\theta \right)}\right) = c \frac{d}{d\theta} \left(f{\left(\theta \right)}\right)$$$을 $$$c = \frac{1}{2}$$$와 $$$f{\left(\theta \right)} = \theta$$$에 적용합니다:

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

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

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

상수의 도함수는 $$$0$$$입니다:

$$\left({\color{red}\left(\frac{d}{d\theta} \left(\frac{\pi}{4}\right)\right)} + \frac{1}{2}\right) \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)} = \left({\color{red}\left(0\right)} + \frac{1}{2}\right) \sec^{2}{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}$$

간단히 하시오:

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

따라서, $$$\frac{d}{d\theta} \left(\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}\right) = \frac{1}{1 - \sin{\left(\theta \right)}}$$$.