$$$\frac{7}{\sqrt{9 t^{4} + 4 t^{2} + 49}}$$$ 的導數

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 $$$c = 7$$$ 與 $$$f{\left(t \right)} = \frac{1}{\sqrt{9 t^{4} + 4 t^{2} + 49}}$$$：

$${\color{red}\left(\frac{d}{dt} \left(\frac{7}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right)\right)} = {\color{red}\left(7 \frac{d}{dt} \left(\frac{1}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right)\right)}$$

函數 $$$\frac{1}{\sqrt{9 t^{4} + 4 t^{2} + 49}}$$$ 是兩個函數 $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$ 與 $$$g{\left(t \right)} = 9 t^{4} + 4 t^{2} + 49$$$ 之複合 $$$f{\left(g{\left(t \right)} \right)}$$$。

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

$$7 {\color{red}\left(\frac{d}{dt} \left(\frac{1}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right)\right)} = 7 {\color{red}\left(\frac{d}{du} \left(\frac{1}{\sqrt{u}}\right) \frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right)\right)}$$

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

$$7 {\color{red}\left(\frac{d}{du} \left(\frac{1}{\sqrt{u}}\right)\right)} \frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right) = 7 {\color{red}\left(- \frac{1}{2 u^{\frac{3}{2}}}\right)} \frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right)$$

返回原變數：

$$- \frac{7 \frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right)}{2 {\color{red}\left(u\right)}^{\frac{3}{2}}} = - \frac{7 \frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right)}{2 {\color{red}\left(9 t^{4} + 4 t^{2} + 49\right)}^{\frac{3}{2}}}$$

和/差的導數等於導數的和/差：

$$- \frac{7 {\color{red}\left(\frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right)\right)}}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 {\color{red}\left(\frac{d}{dt} \left(9 t^{4}\right) + \frac{d}{dt} \left(4 t^{2}\right) + \frac{d}{dt} \left(49\right)\right)}}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$

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 $$$c = 4$$$ 與 $$$f{\left(t \right)} = t^{2}$$$：

$$- \frac{7 \left({\color{red}\left(\frac{d}{dt} \left(4 t^{2}\right)\right)} + \frac{d}{dt} \left(49\right) + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 \left({\color{red}\left(4 \frac{d}{dt} \left(t^{2}\right)\right)} + \frac{d}{dt} \left(49\right) + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$

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

$$- \frac{7 \left(4 {\color{red}\left(\frac{d}{dt} \left(t^{2}\right)\right)} + \frac{d}{dt} \left(49\right) + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 \left(4 {\color{red}\left(2 t\right)} + \frac{d}{dt} \left(49\right) + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$

常數的導數為$$$0$$$：

$$- \frac{7 \left(8 t + {\color{red}\left(\frac{d}{dt} \left(49\right)\right)} + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 \left(8 t + {\color{red}\left(0\right)} + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$

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 $$$c = 9$$$ 與 $$$f{\left(t \right)} = t^{4}$$$：

$$- \frac{7 \left(8 t + {\color{red}\left(\frac{d}{dt} \left(9 t^{4}\right)\right)}\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 \left(8 t + {\color{red}\left(9 \frac{d}{dt} \left(t^{4}\right)\right)}\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$

套用冪次法則 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$，取 $$$n = 4$$$：

$$- \frac{7 \left(8 t + 9 {\color{red}\left(\frac{d}{dt} \left(t^{4}\right)\right)}\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 \left(8 t + 9 {\color{red}\left(4 t^{3}\right)}\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$

化簡：

$$- \frac{7 \left(36 t^{3} + 8 t\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{14 t \left(9 t^{2} + 2\right)}{\left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$

因此，$$$\frac{d}{dt} \left(\frac{7}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right) = - \frac{14 t \left(9 t^{2} + 2\right)}{\left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$$。