$$$\frac{1}{\sqrt{5 t^{2} + 1}}$$$ 的導數

函數 $$$\frac{1}{\sqrt{5 t^{2} + 1}}$$$ 是兩個函數 $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$ 與 $$$g{\left(t \right)} = 5 t^{2} + 1$$$ 之複合 $$$f{\left(g{\left(t \right)} \right)}$$$。

應用鏈式法則 $$$\frac{d}{dt} \left(f{\left(g{\left(t \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dt} \left(g{\left(t \right)}\right)$$$：

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

套用冪次法則 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$，取 $$$n = - \frac{1}{2}$$$：

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

返回原變數：

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

和/差的導數等於導數的和/差：

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

常數的導數為$$$0$$$：

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

套用常數倍法則 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$，使用 $$$c = 5$$$ 與 $$$f{\left(t \right)} = t^{2}$$$：

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

套用冪次法則 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$，取 $$$n = 2$$$：

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

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