$$$\frac{\left(x + 1\right)^{6}}{\left(x^{2} + 8\right)^{6}}$$$의 도함수

$$$H{\left(x \right)} = \frac{\left(x + 1\right)^{6}}{\left(x^{2} + 8\right)^{6}}$$$라고 하자.

양변에 로그를 취합니다: $$$\ln\left(H{\left(x \right)}\right) = \ln\left(\frac{\left(x + 1\right)^{6}}{\left(x^{2} + 8\right)^{6}}\right)$$$.

로그의 성질을 이용하여 우변을 다시 쓰십시오: $$$\ln\left(H{\left(x \right)}\right) = 6 \ln\left(x + 1\right) - 6 \ln\left(x^{2} + 8\right)$$$

방정식의 양변을 각각 미분하시오: $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{d}{dx} \left(6 \ln\left(x + 1\right) - 6 \ln\left(x^{2} + 8\right)\right)$$$

방정식의 좌변을 미분하세요.

함수 $$$\ln\left(H{\left(x \right)}\right)$$$는 두 함수 $$$f{\left(u \right)} = \ln\left(u\right)$$$와 $$$g{\left(x \right)} = H{\left(x \right)}$$$의 합성함수 $$$f{\left(g{\left(x \right)} \right)}$$$이다.

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

$${\color{red}\left(\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(H{\left(x \right)}\right)\right)}$$

자연로그 함수의 도함수는 $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(H{\left(x \right)}\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(H{\left(x \right)}\right)$$

역치환:

$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{{\color{red}\left(u\right)}} = \frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{{\color{red}\left(H{\left(x \right)}\right)}}$$

따라서, $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}}$$$.

방정식의 우변을 미분하시오.

합/차의 도함수는 도함수들의 합/차이다:

$${\color{red}\left(\frac{d}{dx} \left(6 \ln\left(x + 1\right) - 6 \ln\left(x^{2} + 8\right)\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{d}{dx} \left(6 \ln\left(x^{2} + 8\right)\right)\right)}$$

상수배 법칙 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$을 $$$c = 6$$$와 $$$f{\left(x \right)} = \ln\left(x^{2} + 8\right)$$$에 적용합니다:

$$- {\color{red}\left(\frac{d}{dx} \left(6 \ln\left(x^{2} + 8\right)\right)\right)} + \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) = - {\color{red}\left(6 \frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right)\right)} + \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right)$$

함수 $$$\ln\left(x^{2} + 8\right)$$$는 두 함수 $$$f{\left(u \right)} = \ln\left(u\right)$$$와 $$$g{\left(x \right)} = x^{2} + 8$$$의 합성함수 $$$f{\left(g{\left(x \right)} \right)}$$$이다.

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

$$- 6 {\color{red}\left(\frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right)\right)} + \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) = - 6 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x^{2} + 8\right)\right)} + \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right)$$

자연로그 함수의 도함수는 $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

$$- 6 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x^{2} + 8\right) + \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) = - 6 {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x^{2} + 8\right) + \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right)$$

역치환:

$$\frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{6 \frac{d}{dx} \left(x^{2} + 8\right)}{{\color{red}\left(u\right)}} = \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{6 \frac{d}{dx} \left(x^{2} + 8\right)}{{\color{red}\left(x^{2} + 8\right)}}$$

합/차의 도함수는 도함수들의 합/차이다:

$$\frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{6 {\color{red}\left(\frac{d}{dx} \left(x^{2} + 8\right)\right)}}{x^{2} + 8} = \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{6 {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) + \frac{d}{dx} \left(8\right)\right)}}{x^{2} + 8}$$

상수의 도함수는 $$$0$$$입니다:

$$\frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{6 \left({\color{red}\left(\frac{d}{dx} \left(8\right)\right)} + \frac{d}{dx} \left(x^{2}\right)\right)}{x^{2} + 8} = \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{6 \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{2}\right)\right)}{x^{2} + 8}$$

거듭제곱법칙 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$을 $$$n = 2$$$에 적용합니다:

$$\frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{6 {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)}}{x^{2} + 8} = \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{6 {\color{red}\left(2 x\right)}}{x^{2} + 8}$$

상수배 법칙 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$을 $$$c = 6$$$와 $$$f{\left(x \right)} = \ln\left(x + 1\right)$$$에 적용합니다:

$$- \frac{12 x}{x^{2} + 8} + {\color{red}\left(\frac{d}{dx} \left(6 \ln\left(x + 1\right)\right)\right)} = - \frac{12 x}{x^{2} + 8} + {\color{red}\left(6 \frac{d}{dx} \left(\ln\left(x + 1\right)\right)\right)}$$

함수 $$$\ln\left(x + 1\right)$$$는 두 함수 $$$f{\left(u \right)} = \ln\left(u\right)$$$와 $$$g{\left(x \right)} = x + 1$$$의 합성함수 $$$f{\left(g{\left(x \right)} \right)}$$$이다.

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

$$- \frac{12 x}{x^{2} + 8} + 6 {\color{red}\left(\frac{d}{dx} \left(\ln\left(x + 1\right)\right)\right)} = - \frac{12 x}{x^{2} + 8} + 6 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x + 1\right)\right)}$$

자연로그 함수의 도함수는 $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

$$- \frac{12 x}{x^{2} + 8} + 6 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x + 1\right) = - \frac{12 x}{x^{2} + 8} + 6 {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x + 1\right)$$

역치환:

$$- \frac{12 x}{x^{2} + 8} + \frac{6 \frac{d}{dx} \left(x + 1\right)}{{\color{red}\left(u\right)}} = - \frac{12 x}{x^{2} + 8} + \frac{6 \frac{d}{dx} \left(x + 1\right)}{{\color{red}\left(x + 1\right)}}$$

합/차의 도함수는 도함수들의 합/차이다:

$$- \frac{12 x}{x^{2} + 8} + \frac{6 {\color{red}\left(\frac{d}{dx} \left(x + 1\right)\right)}}{x + 1} = - \frac{12 x}{x^{2} + 8} + \frac{6 {\color{red}\left(\frac{d}{dx} \left(x\right) + \frac{d}{dx} \left(1\right)\right)}}{x + 1}$$

상수의 도함수는 $$$0$$$입니다:

$$- \frac{12 x}{x^{2} + 8} + \frac{6 \left({\color{red}\left(\frac{d}{dx} \left(1\right)\right)} + \frac{d}{dx} \left(x\right)\right)}{x + 1} = - \frac{12 x}{x^{2} + 8} + \frac{6 \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x\right)\right)}{x + 1}$$

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

$$- \frac{12 x}{x^{2} + 8} + \frac{6 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{x + 1} = - \frac{12 x}{x^{2} + 8} + \frac{6 {\color{red}\left(1\right)}}{x + 1}$$

따라서, $$$\frac{d}{dx} \left(6 \ln\left(x + 1\right) - 6 \ln\left(x^{2} + 8\right)\right) = - \frac{12 x}{x^{2} + 8} + \frac{6}{x + 1}$$$.

따라서 $$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}} = - \frac{12 x}{x^{2} + 8} + \frac{6}{x + 1}$$$.

따라서 $$$\frac{d}{dx} \left(H{\left(x \right)}\right) = \left(- \frac{12 x}{x^{2} + 8} + \frac{6}{x + 1}\right) H{\left(x \right)} = - \frac{6 \left(x - 2\right) \left(x + 1\right)^{5} \left(x + 4\right)}{\left(x^{2} + 8\right)^{7}}.$$$