$$$\frac{d^{2}}{dx^{2}} \left(\sqrt{x} + \frac{5}{\sqrt{x}}\right)$$$을(를) 구하세요

$$$\frac{d^{2}}{dx^{2}} \left(\sqrt{x} + \frac{5}{\sqrt{x}}\right)$$$을(를) 구하시오.

제1도함수 $$$\frac{d}{dx} \left(\sqrt{x} + \frac{5}{\sqrt{x}}\right)$$$를 구하세요

합/차의 도함수는 도함수들의 합/차이다:

$${\color{red}\left(\frac{d}{dx} \left(\sqrt{x} + \frac{5}{\sqrt{x}}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(\sqrt{x}\right) + \frac{d}{dx} \left(\frac{5}{\sqrt{x}}\right)\right)}$$

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

$${\color{red}\left(\frac{d}{dx} \left(\sqrt{x}\right)\right)} + \frac{d}{dx} \left(\frac{5}{\sqrt{x}}\right) = {\color{red}\left(\frac{1}{2 \sqrt{x}}\right)} + \frac{d}{dx} \left(\frac{5}{\sqrt{x}}\right)$$

상수배 법칙 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$을 $$$c = 5$$$와 $$$f{\left(x \right)} = \frac{1}{\sqrt{x}}$$$에 적용합니다:

$${\color{red}\left(\frac{d}{dx} \left(\frac{5}{\sqrt{x}}\right)\right)} + \frac{1}{2 \sqrt{x}} = {\color{red}\left(5 \frac{d}{dx} \left(\frac{1}{\sqrt{x}}\right)\right)} + \frac{1}{2 \sqrt{x}}$$

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

$$5 {\color{red}\left(\frac{d}{dx} \left(\frac{1}{\sqrt{x}}\right)\right)} + \frac{1}{2 \sqrt{x}} = 5 {\color{red}\left(- \frac{1}{2 x^{\frac{3}{2}}}\right)} + \frac{1}{2 \sqrt{x}}$$

간단히 하시오:

$$\frac{1}{2 \sqrt{x}} - \frac{5}{2 x^{\frac{3}{2}}} = \frac{x - 5}{2 x^{\frac{3}{2}}}$$

따라서, $$$\frac{d}{dx} \left(\sqrt{x} + \frac{5}{\sqrt{x}}\right) = \frac{x - 5}{2 x^{\frac{3}{2}}}$$$.

다음으로, $$$\frac{d^{2}}{dx^{2}} \left(\sqrt{x} + \frac{5}{\sqrt{x}}\right) = \frac{d}{dx} \left(\frac{x - 5}{2 x^{\frac{3}{2}}}\right)$$$

상수배 법칙 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$을 $$$c = \frac{1}{2}$$$와 $$$f{\left(x \right)} = \frac{x - 5}{x^{\frac{3}{2}}}$$$에 적용합니다:

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

$$$f{\left(x \right)} = x - 5$$$와 $$$g{\left(x \right)} = x^{\frac{3}{2}}$$$에 대해 몫의 미분법칙 $$$\frac{d}{dx} \left(\frac{f{\left(x \right)}}{g{\left(x \right)}}\right) = \frac{\frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} - f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)}{g^{2}{\left(x \right)}}$$$을 적용하십시오:

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

합/차의 도함수는 도함수들의 합/차이다:

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

상수의 도함수는 $$$0$$$입니다:

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

거듭제곱법칙 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$을 $$$n = \frac{3}{2}$$$에 적용합니다:

$$\frac{x^{\frac{3}{2}} \frac{d}{dx} \left(x\right) - \left(x - 5\right) {\color{red}\left(\frac{d}{dx} \left(x^{\frac{3}{2}}\right)\right)}}{2 x^{3}} = \frac{x^{\frac{3}{2}} \frac{d}{dx} \left(x\right) - \left(x - 5\right) {\color{red}\left(\frac{3 \sqrt{x}}{2}\right)}}{2 x^{3}}$$

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

$$\frac{x^{\frac{3}{2}} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} - \frac{3 \sqrt{x} \left(x - 5\right)}{2}}{2 x^{3}} = \frac{x^{\frac{3}{2}} {\color{red}\left(1\right)} - \frac{3 \sqrt{x} \left(x - 5\right)}{2}}{2 x^{3}}$$

간단히 하시오:

$$\frac{x^{\frac{3}{2}} - \frac{3 \sqrt{x} \left(x - 5\right)}{2}}{2 x^{3}} = \frac{15 - x}{4 x^{\frac{5}{2}}}$$

따라서, $$$\frac{d}{dx} \left(\frac{x - 5}{2 x^{\frac{3}{2}}}\right) = \frac{15 - x}{4 x^{\frac{5}{2}}}$$$.

따라서 $$$\frac{d^{2}}{dx^{2}} \left(\sqrt{x} + \frac{5}{\sqrt{x}}\right) = \frac{15 - x}{4 x^{\frac{5}{2}}}$$$.