Cari terbitan pertama bagi $$ f(x)=\frac{2}{x+4} $$ dengan menggunakan prinsip pertama. Find the first derivative of $$ f(x)=\frac{2}{x+4} $$ by using the first principle.

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要使用**第一性原理**求函数 $$ f(x) = \frac{2}{x+4} $$ 的导数，我们按照导数的定义进行计算：

$$
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
$$

**步骤如下：**

1. **计算 $$ f(x+h) $$：**
   $$
   f(x+h) = \frac{2}{(x + h) + 4} = \frac{2}{x + 4 + h}
   $$

2. **计算 $$ f(x+h) - f(x) $$：**
   $$
   f(x+h) - f(x) = \frac{2}{x + 4 + h} - \frac{2}{x + 4}
   $$

3. **通分并简化：**
   $$
   f(x+h) - f(x) = 2 \left( \frac{1}{x + 4 + h} - \frac{1}{x + 4} \right) = 2 \left( \frac{(x + 4) - (x + 4 + h)}{(x + 4)(x + 4 + h)} \right) = \frac{-2h}{(x + 4)(x + 4 + h)}
   $$

4. **将差商除以 $$ h $$：**
   $$
   \frac{f(x+h) - f(x)}{h} = \frac{-2h}{h (x + 4)(x + 4 + h)} = \frac{-2}{(x + 4)(x + 4 + h)}
   $$

5. **取极限 $$ h \to 0 $$：**
   $$
   f'(x) = \lim_{h \to 0} \frac{-2}{(x + 4)(x + 4 + h)} = \frac{-2}{(x + 4)^2}
   $$

**最终结果：**
$$
f'(x) = -\frac{2}{(x + 4)^2}
$$
$$ f(x) = \frac{2}{x+4} $$ 的导数是 $$ f'(x) = -\frac{2}{(x+4)^2} $$。

Cari terbitan pertama bagi \( f(x)=\frac{2}{x+4} \) dengan menggunakan prinsip pertama. Find the first derivative of \( f(x)=\frac{2}{x+4} \) by using the first principle.

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