$$$\frac{1}{\sqrt{x^{2} - 3 x + 9}}$$$的导数

函数$$$\frac{1}{\sqrt{x^{2} - 3 x + 9}}$$$是两个函数$$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$和$$$g{\left(x \right)} = x^{2} - 3 x + 9$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。

应用链式法则 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：

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

应用幂次法则 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$，其中 $$$n = - \frac{1}{2}$$$:

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

返回到原变量：

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

和/差的导数等于导数的和/差：

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

常数的导数是$$$0$$$:

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

应用幂次法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，其中 $$$n = 2$$$:

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

对 $$$c = 3$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：

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

应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dx} \left(x\right) = 1$$$：

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

化简：

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

因此，$$$\frac{d}{dx} \left(\frac{1}{\sqrt{x^{2} - 3 x + 9}}\right) = \frac{\frac{3}{2} - x}{\left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$$。