$$$\frac{1}{\sqrt{5 t^{2} + 1}}$$$의 도함수

함수 $$$\frac{1}{\sqrt{5 t^{2} + 1}}$$$는 두 함수 $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$와 $$$g{\left(t \right)} = 5 t^{2} + 1$$$의 합성함수 $$$f{\left(g{\left(t \right)} \right)}$$$이다.

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

$${\color{red}\left(\frac{d}{dt} \left(\frac{1}{\sqrt{5 t^{2} + 1}}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\frac{1}{\sqrt{u}}\right) \frac{d}{dt} \left(5 t^{2} + 1\right)\right)}$$

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

$${\color{red}\left(\frac{d}{du} \left(\frac{1}{\sqrt{u}}\right)\right)} \frac{d}{dt} \left(5 t^{2} + 1\right) = {\color{red}\left(- \frac{1}{2 u^{\frac{3}{2}}}\right)} \frac{d}{dt} \left(5 t^{2} + 1\right)$$

역치환:

$$- \frac{\frac{d}{dt} \left(5 t^{2} + 1\right)}{2 {\color{red}\left(u\right)}^{\frac{3}{2}}} = - \frac{\frac{d}{dt} \left(5 t^{2} + 1\right)}{2 {\color{red}\left(5 t^{2} + 1\right)}^{\frac{3}{2}}}$$

합/차의 도함수는 도함수들의 합/차이다:

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

상수의 도함수는 $$$0$$$입니다:

$$- \frac{{\color{red}\left(\frac{d}{dt} \left(1\right)\right)} + \frac{d}{dt} \left(5 t^{2}\right)}{2 \left(5 t^{2} + 1\right)^{\frac{3}{2}}} = - \frac{{\color{red}\left(0\right)} + \frac{d}{dt} \left(5 t^{2}\right)}{2 \left(5 t^{2} + 1\right)^{\frac{3}{2}}}$$

상수배 법칙 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$을 $$$c = 5$$$와 $$$f{\left(t \right)} = t^{2}$$$에 적용합니다:

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

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

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

따라서, $$$\frac{d}{dt} \left(\frac{1}{\sqrt{5 t^{2} + 1}}\right) = - \frac{5 t}{\left(5 t^{2} + 1\right)^{\frac{3}{2}}}$$$.