$$$\frac{7}{\sqrt{9 t^{4} + 4 t^{2} + 49}}$$$の導関数

定数倍の法則 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$ を $$$c = 7$$$ と $$$f{\left(t \right)} = \frac{1}{\sqrt{9 t^{4} + 4 t^{2} + 49}}$$$ に対して適用します:

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

関数$$$\frac{1}{\sqrt{9 t^{4} + 4 t^{2} + 49}}$$$は、2つの関数$$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$と$$$g{\left(t \right)} = 9 t^{4} + 4 t^{2} + 49$$$の合成$$$f{\left(g{\left(t \right)} \right)}$$$である。

連鎖律 $$$\frac{d}{dt} \left(f{\left(g{\left(t \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dt} \left(g{\left(t \right)}\right)$$$ を適用する:

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

冪法則 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ を $$$n = - \frac{1}{2}$$$ に対して適用する:

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

元の変数に戻す:

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

和/差の導関数は、導関数の和/差である:

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

定数倍の法則 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$ を $$$c = 4$$$ と $$$f{\left(t \right)} = t^{2}$$$ に対して適用します:

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

冪法則 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ を $$$n = 2$$$ に対して適用する:

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

定数の導数は$$$0$$$です:

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

定数倍の法則 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$ を $$$c = 9$$$ と $$$f{\left(t \right)} = t^{4}$$$ に対して適用します:

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

冪法則 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ を $$$n = 4$$$ に対して適用する:

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

簡単化せよ:

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

したがって、$$$\frac{d}{dt} \left(\frac{7}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right) = - \frac{14 t \left(9 t^{2} + 2\right)}{\left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$$。