$$$\left(x^{3} + 4\right)^{4} \left(x^{5} + 2\right)^{2}$$$의 도함수

$$$H{\left(x \right)} = \left(x^{3} + 4\right)^{4} \left(x^{5} + 2\right)^{2}$$$라고 하자.

양변에 로그를 취합니다: $$$\ln\left(H{\left(x \right)}\right) = \ln\left(\left(x^{3} + 4\right)^{4} \left(x^{5} + 2\right)^{2}\right)$$$.

로그의 성질을 이용하여 우변을 다시 쓰십시오: $$$\ln\left(H{\left(x \right)}\right) = 4 \ln\left(x^{3} + 4\right) + 2 \ln\left(x^{5} + 2\right)$$$

방정식의 양변을 각각 미분하시오: $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{d}{dx} \left(4 \ln\left(x^{3} + 4\right) + 2 \ln\left(x^{5} + 2\right)\right)$$$

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:

$${\color{red}\left(\frac{d}{dx} \left(4 \ln\left(x^{3} + 4\right) + 2 \ln\left(x^{5} + 2\right)\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(4 \ln\left(x^{3} + 4\right)\right) + \frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right)\right)}$$

상수배 법칙 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$을 $$$c = 4$$$와 $$$f{\left(x \right)} = \ln\left(x^{3} + 4\right)$$$에 적용합니다:

$${\color{red}\left(\frac{d}{dx} \left(4 \ln\left(x^{3} + 4\right)\right)\right)} + \frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right) = {\color{red}\left(4 \frac{d}{dx} \left(\ln\left(x^{3} + 4\right)\right)\right)} + \frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right)$$

상수배 법칙 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$을 $$$c = 2$$$와 $$$f{\left(x \right)} = \ln\left(x^{5} + 2\right)$$$에 적용합니다:

$${\color{red}\left(\frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right)\right)} + 4 \frac{d}{dx} \left(\ln\left(x^{3} + 4\right)\right) = {\color{red}\left(2 \frac{d}{dx} \left(\ln\left(x^{5} + 2\right)\right)\right)} + 4 \frac{d}{dx} \left(\ln\left(x^{3} + 4\right)\right)$$

함수 $$$\ln\left(x^{3} + 4\right)$$$는 두 함수 $$$f{\left(u \right)} = \ln\left(u\right)$$$와 $$$g{\left(x \right)} = x^{3} + 4$$$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:

$$4 {\color{red}\left(\frac{d}{dx} \left(\ln\left(x^{3} + 4\right)\right)\right)} + 2 \frac{d}{dx} \left(\ln\left(x^{5} + 2\right)\right) = 4 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x^{3} + 4\right)\right)} + 2 \frac{d}{dx} \left(\ln\left(x^{5} + 2\right)\right)$$

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

$$4 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x^{3} + 4\right) + 2 \frac{d}{dx} \left(\ln\left(x^{5} + 2\right)\right) = 4 {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x^{3} + 4\right) + 2 \frac{d}{dx} \left(\ln\left(x^{5} + 2\right)\right)$$

역치환:

$$2 \frac{d}{dx} \left(\ln\left(x^{5} + 2\right)\right) + \frac{4 \frac{d}{dx} \left(x^{3} + 4\right)}{{\color{red}\left(u\right)}} = 2 \frac{d}{dx} \left(\ln\left(x^{5} + 2\right)\right) + \frac{4 \frac{d}{dx} \left(x^{3} + 4\right)}{{\color{red}\left(x^{3} + 4\right)}}$$

합/차의 도함수는 도함수들의 합/차이다:

$$2 \frac{d}{dx} \left(\ln\left(x^{5} + 2\right)\right) + \frac{4 {\color{red}\left(\frac{d}{dx} \left(x^{3} + 4\right)\right)}}{x^{3} + 4} = 2 \frac{d}{dx} \left(\ln\left(x^{5} + 2\right)\right) + \frac{4 {\color{red}\left(\frac{d}{dx} \left(x^{3}\right) + \frac{d}{dx} \left(4\right)\right)}}{x^{3} + 4}$$

상수의 도함수는 $$$0$$$입니다:

$$2 \frac{d}{dx} \left(\ln\left(x^{5} + 2\right)\right) + \frac{4 \left({\color{red}\left(\frac{d}{dx} \left(4\right)\right)} + \frac{d}{dx} \left(x^{3}\right)\right)}{x^{3} + 4} = 2 \frac{d}{dx} \left(\ln\left(x^{5} + 2\right)\right) + \frac{4 \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{3}\right)\right)}{x^{3} + 4}$$

거듭제곱법칙 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$을 $$$n = 3$$$에 적용합니다:

$$2 \frac{d}{dx} \left(\ln\left(x^{5} + 2\right)\right) + \frac{4 {\color{red}\left(\frac{d}{dx} \left(x^{3}\right)\right)}}{x^{3} + 4} = 2 \frac{d}{dx} \left(\ln\left(x^{5} + 2\right)\right) + \frac{4 {\color{red}\left(3 x^{2}\right)}}{x^{3} + 4}$$

함수 $$$\ln\left(x^{5} + 2\right)$$$는 두 함수 $$$f{\left(u \right)} = \ln\left(u\right)$$$와 $$$g{\left(x \right)} = x^{5} + 2$$$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:

$$\frac{12 x^{2}}{x^{3} + 4} + 2 {\color{red}\left(\frac{d}{dx} \left(\ln\left(x^{5} + 2\right)\right)\right)} = \frac{12 x^{2}}{x^{3} + 4} + 2 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x^{5} + 2\right)\right)}$$

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

$$\frac{12 x^{2}}{x^{3} + 4} + 2 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x^{5} + 2\right) = \frac{12 x^{2}}{x^{3} + 4} + 2 {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x^{5} + 2\right)$$

역치환:

$$\frac{12 x^{2}}{x^{3} + 4} + \frac{2 \frac{d}{dx} \left(x^{5} + 2\right)}{{\color{red}\left(u\right)}} = \frac{12 x^{2}}{x^{3} + 4} + \frac{2 \frac{d}{dx} \left(x^{5} + 2\right)}{{\color{red}\left(x^{5} + 2\right)}}$$

합/차의 도함수는 도함수들의 합/차이다:

$$\frac{12 x^{2}}{x^{3} + 4} + \frac{2 {\color{red}\left(\frac{d}{dx} \left(x^{5} + 2\right)\right)}}{x^{5} + 2} = \frac{12 x^{2}}{x^{3} + 4} + \frac{2 {\color{red}\left(\frac{d}{dx} \left(x^{5}\right) + \frac{d}{dx} \left(2\right)\right)}}{x^{5} + 2}$$

상수의 도함수는 $$$0$$$입니다:

$$\frac{12 x^{2}}{x^{3} + 4} + \frac{2 \left({\color{red}\left(\frac{d}{dx} \left(2\right)\right)} + \frac{d}{dx} \left(x^{5}\right)\right)}{x^{5} + 2} = \frac{12 x^{2}}{x^{3} + 4} + \frac{2 \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{5}\right)\right)}{x^{5} + 2}$$

거듭제곱법칙 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$을 $$$n = 5$$$에 적용합니다:

$$\frac{12 x^{2}}{x^{3} + 4} + \frac{2 {\color{red}\left(\frac{d}{dx} \left(x^{5}\right)\right)}}{x^{5} + 2} = \frac{12 x^{2}}{x^{3} + 4} + \frac{2 {\color{red}\left(5 x^{4}\right)}}{x^{5} + 2}$$

따라서, $$$\frac{d}{dx} \left(4 \ln\left(x^{3} + 4\right) + 2 \ln\left(x^{5} + 2\right)\right) = \frac{10 x^{4}}{x^{5} + 2} + \frac{12 x^{2}}{x^{3} + 4}$$$.

따라서 $$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}} = \frac{10 x^{4}}{x^{5} + 2} + \frac{12 x^{2}}{x^{3} + 4}$$$.

따라서 $$$\frac{d}{dx} \left(H{\left(x \right)}\right) = \left(\frac{10 x^{4}}{x^{5} + 2} + \frac{12 x^{2}}{x^{3} + 4}\right) H{\left(x \right)} = 2 x^{2} \left(x^{3} + 4\right)^{3} \left(x^{5} + 2\right) \left(11 x^{5} + 20 x^{2} + 12\right).$$$