$$$\frac{1}{\sqrt{x^{2} - 3 x + 9}}$$$の導関数

関数$$$\frac{1}{\sqrt{x^{2} - 3 x + 9}}$$$は、2つの関数$$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$と$$$g{\left(x \right)} = x^{2} - 3 x + 9$$$の合成$$$f{\left(g{\left(x \right)} \right)}$$$である。

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

$${\color{red}\left(\frac{d}{dx} \left(\frac{1}{\sqrt{x^{2} - 3 x + 9}}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\frac{1}{\sqrt{u}}\right) \frac{d}{dx} \left(x^{2} - 3 x + 9\right)\right)}$$

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:

$${\color{red}\left(\frac{d}{du} \left(\frac{1}{\sqrt{u}}\right)\right)} \frac{d}{dx} \left(x^{2} - 3 x + 9\right) = {\color{red}\left(- \frac{1}{2 u^{\frac{3}{2}}}\right)} \frac{d}{dx} \left(x^{2} - 3 x + 9\right)$$

元の変数に戻す:

$$- \frac{\frac{d}{dx} \left(x^{2} - 3 x + 9\right)}{2 {\color{red}\left(u\right)}^{\frac{3}{2}}} = - \frac{\frac{d}{dx} \left(x^{2} - 3 x + 9\right)}{2 {\color{red}\left(x^{2} - 3 x + 9\right)}^{\frac{3}{2}}}$$

和/差の導関数は、導関数の和/差である:

$$- \frac{{\color{red}\left(\frac{d}{dx} \left(x^{2} - 3 x + 9\right)\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = - \frac{{\color{red}\left(\frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(3 x\right) + \frac{d}{dx} \left(9\right)\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

定数の導数は$$$0$$$です:

$$- \frac{{\color{red}\left(\frac{d}{dx} \left(9\right)\right)} - \frac{d}{dx} \left(3 x\right) + \frac{d}{dx} \left(x^{2}\right)}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = - \frac{{\color{red}\left(0\right)} - \frac{d}{dx} \left(3 x\right) + \frac{d}{dx} \left(x^{2}\right)}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

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

$$- \frac{{\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} - \frac{d}{dx} \left(3 x\right)}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = - \frac{{\color{red}\left(2 x\right)} - \frac{d}{dx} \left(3 x\right)}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

定数倍の法則 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ を $$$c = 3$$$ と $$$f{\left(x \right)} = x$$$ に対して適用します:

$$- \frac{2 x - {\color{red}\left(\frac{d}{dx} \left(3 x\right)\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = - \frac{2 x - {\color{red}\left(3 \frac{d}{dx} \left(x\right)\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

$$$n = 1$$$ を用いて冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を適用すると、すなわち $$$\frac{d}{dx} \left(x\right) = 1$$$:

$$- \frac{2 x - 3 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = - \frac{2 x - 3 {\color{red}\left(1\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

簡単化せよ:

$$- \frac{2 x - 3}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = \frac{\frac{3}{2} - x}{\left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

したがって、$$$\frac{d}{dx} \left(\frac{1}{\sqrt{x^{2} - 3 x + 9}}\right) = \frac{\frac{3}{2} - x}{\left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$$。