$$$\frac{1}{\sqrt{5 t^{2} + 1}}$$$の導関数

関数$$$\frac{1}{\sqrt{5 t^{2} + 1}}$$$は、2つの関数$$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$と$$$g{\left(t \right)} = 5 t^{2} + 1$$$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:

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

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:

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

元の変数に戻す:

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

和/差の導関数は、導関数の和/差である:

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

定数の導数は$$$0$$$です:

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

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 $$$c = 5$$$ と $$$f{\left(t \right)} = t^{2}$$$ に対して適用します:

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

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:

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

したがって、$$$\frac{d}{dt} \left(\frac{1}{\sqrt{5 t^{2} + 1}}\right) = - \frac{5 t}{\left(5 t^{2} + 1\right)^{\frac{3}{2}}}$$$。