$$$\sqrt{x} + \frac{5}{\sqrt{x}}$$$ 的導數

和/差的導數等於導數的和/差：

$${\color{red}\left(\frac{d}{dx} \left(\sqrt{x} + \frac{5}{\sqrt{x}}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(\sqrt{x}\right) + \frac{d}{dx} \left(\frac{5}{\sqrt{x}}\right)\right)}$$

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

$${\color{red}\left(\frac{d}{dx} \left(\sqrt{x}\right)\right)} + \frac{d}{dx} \left(\frac{5}{\sqrt{x}}\right) = {\color{red}\left(\frac{1}{2 \sqrt{x}}\right)} + \frac{d}{dx} \left(\frac{5}{\sqrt{x}}\right)$$

套用常數倍法則 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$，使用 $$$c = 5$$$ 與 $$$f{\left(x \right)} = \frac{1}{\sqrt{x}}$$$：

$${\color{red}\left(\frac{d}{dx} \left(\frac{5}{\sqrt{x}}\right)\right)} + \frac{1}{2 \sqrt{x}} = {\color{red}\left(5 \frac{d}{dx} \left(\frac{1}{\sqrt{x}}\right)\right)} + \frac{1}{2 \sqrt{x}}$$

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

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

化簡：

$$\frac{1}{2 \sqrt{x}} - \frac{5}{2 x^{\frac{3}{2}}} = \frac{x - 5}{2 x^{\frac{3}{2}}}$$

因此，$$$\frac{d}{dx} \left(\sqrt{x} + \frac{5}{\sqrt{x}}\right) = \frac{x - 5}{2 x^{\frac{3}{2}}}$$$。