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$$$\sqrt{x}$$$의 이차 도함수
사용자 입력
$$$\frac{d^{2}}{dx^{2}} \left(\sqrt{x}\right)$$$을(를) 구하시오.
풀이
제1도함수 $$$\frac{d}{dx} \left(\sqrt{x}\right)$$$를 구하세요
거듭제곱법칙 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$을 $$$n = \frac{1}{2}$$$에 적용합니다:
$${\color{red}\left(\frac{d}{dx} \left(\sqrt{x}\right)\right)} = {\color{red}\left(\frac{1}{2 \sqrt{x}}\right)}$$따라서, $$$\frac{d}{dx} \left(\sqrt{x}\right) = \frac{1}{2 \sqrt{x}}$$$.
다음으로, $$$\frac{d^{2}}{dx^{2}} \left(\sqrt{x}\right) = \frac{d}{dx} \left(\frac{1}{2 \sqrt{x}}\right)$$$
상수배 법칙 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$을 $$$c = \frac{1}{2}$$$와 $$$f{\left(x \right)} = \frac{1}{\sqrt{x}}$$$에 적용합니다:
$${\color{red}\left(\frac{d}{dx} \left(\frac{1}{2 \sqrt{x}}\right)\right)} = {\color{red}\left(\frac{\frac{d}{dx} \left(\frac{1}{\sqrt{x}}\right)}{2}\right)}$$거듭제곱법칙 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$을 $$$n = - \frac{1}{2}$$$에 적용합니다:
$$\frac{{\color{red}\left(\frac{d}{dx} \left(\frac{1}{\sqrt{x}}\right)\right)}}{2} = \frac{{\color{red}\left(- \frac{1}{2 x^{\frac{3}{2}}}\right)}}{2}$$따라서, $$$\frac{d}{dx} \left(\frac{1}{2 \sqrt{x}}\right) = - \frac{1}{4 x^{\frac{3}{2}}}$$$.
따라서 $$$\frac{d^{2}}{dx^{2}} \left(\sqrt{x}\right) = - \frac{1}{4 x^{\frac{3}{2}}}$$$.
정답
$$$\frac{d^{2}}{dx^{2}} \left(\sqrt{x}\right) = - \frac{1}{4 x^{\frac{3}{2}}}$$$A
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