根据导数的定义计算导函数

1. Finding derivatives using the definition (使用定义求导)

1.1. We want to differentiate $f(x)=1/x$ with respect to $x$

对 $f(x)=1/x$ 关于 $x$ 求导

The definition of the derivative (导数的定义) is
$$\begin{array}{r}{f}^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\end{array}$$

If you just replace $h$ by 0 in the fraction, you end up with the indeterminate form $\frac{0}{0}$.
如果只是用 0 替换 $h$，结果就会得到一个 $\frac{0}{0}$ 的不定式。

So in our case we have
$$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{\frac{1}{x+h}-\frac{1}{x}}{h}\\ & =\underset{h\to 0}{\lim }\frac{\frac{x-(x+h)}{(x+h)x}}{h}\\ & =\underset{h\to 0}{\lim }\frac{-h}{h(x+h)x}\\ & =\underset{h\to 0}{\lim }\frac{-1}{(x+h)x}\\ & =\frac{-1}{(x+0)x}\\ & =-\frac{1}{x^2}\end{array}$$

$$\begin{array}{r}\frac{\text{d}}{\text{d}x}(\frac{1}{x})=-\frac{1}{x^2}\end{array}$$

1.2. We want to differentiate $f(x)=\sqrt{x}$ with respect to $x$

对 $f(x)=\sqrt{x}$ 关于 $x$ 求导

Let’s multiply top and bottom by the conjugate of the numerator (分子和分母同时乘以分子的共轭表达式) to get
$$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{\sqrt{x+h}-\sqrt{x}}{h}\\ & =\underset{h\to 0}{\lim }\frac{\sqrt{x+h}-\sqrt{x}}{h}\times \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\ & =\underset{h\to 0}{\lim }\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}\\ & =\underset{h\to 0}{\lim }\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\\ & =\underset{h\to 0}{\lim }\frac{1}{\sqrt{x+h}+\sqrt{x}}\\ & =\frac{1}{\sqrt{x+0}+\sqrt{x}}\\ & =\frac{1}{2\sqrt{x}}\end{array}$$

$$\begin{array}{r}\frac{\text{d}}{\text{d}x}(\sqrt{x})=\frac{1}{2\sqrt{x}}\end{array}$$

1.3. We want to differentiate $f(x)=\sqrt{x}+x^2$ with respect to $x$

对 $f(x)=\sqrt{x}+x^2$ 关于 $x$ 求导

$$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{(\sqrt{x+h}+(x+h{)}^2)-(\sqrt{x}+x^2)}{h}\\ & =\underset{h\to 0}{\lim }\frac{\sqrt{x+h}-\sqrt{x}+(x+h{)}^2-x^2}{h}\\ & =\underset{h\to 0}{\lim }\frac{\sqrt{x+h}-\sqrt{x}+2xh+h^2}{h}\\ & =\underset{h\to 0}{\lim }(\frac{\sqrt{x+h}-\sqrt{x}}{h}+\frac{2xh+h^2}{h})\\ & =\underset{h\to 0}{\lim }(\frac{\sqrt{x+h}-\sqrt{x}}{h}+2x+h)\\ & =\underset{h\to 0}{\lim }\frac{\sqrt{x+h}-\sqrt{x}}{h}+\underset{h\to 0}{\lim }(2x+h)\\ & =\frac{1}{2\sqrt{x}}+\underset{h\to 0}{\lim }(2x+h)\\ & =\frac{1}{2\sqrt{x}}+(2x+0)\\ & =\frac{1}{2\sqrt{x}}+2x\end{array}$$

1.4. We want to differentiate $f(x)=x^n$ with respect to $x$, where $n$ is some positive integer

对 $f(x)=x^n$ 关于 $x$ 求导，其中 $n$ 是某个正整数

$$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{(x+h{)}^n-x^n}{h}\end{array}$$

$$\begin{array}{r}(x+h{)}^n=(x+h)(x+h)\dots (x+h)\end{array}$$

If you take the term $x$ from each factor, there are $n$ of them, so you get one term $x^n$ in the product.
如果从每一个因子中提取项 $x$，将会有 $n$ 个 $x$，因而会在乘积中得到 $x^n$ 这一项。

$$\begin{array}{r}(x+h{)}^n=(x+h)(x+h)\dots (x+h)=x^n+含有因子h的项\end{array}$$

If you take the term $h$ from the first factor and $x$ from the others, then you have one $h$ and $(n-1)$ copies of $x$, so you get $h{x}^{n-1}$ when you multiply them all together.
如果从第一个因子中提取 $h$，然后从其他因子中提取 $x$，那样就会有一个 $h$ 和 $(n-1)$ 个 $x$，因此当将它们都乘起来的时候会得到 $h{x}^{n-1}$。还有其他的方法来选择一个 $h$ 和其余的 $x$ (可以从第二个因子里提取 $h$，然后从其他因子中提取 $x$；或者从第三个因子里提取 $h$，然后从其他因子中提取 $x$，如此等等)。

In fact, there are $n$ ways you could pick one $h$ and the rest $x$, so you actually have $n$ copies of $h{x}^{n-1}$. Together, this makes $nh{x}^{n-1}$.
事实上，有 $n$ 种方法来选取一个 $h$ 和其余的 $x$，因此实际上有 $n$ 个 $h{x}^{n-1}$ 加在一起，会得到 $nh{x}^{n-1}$。

Every other term in the expansion has at least two copies of $h$, so every other term has a factor of $h^2$.
在展开式中，每隔一项至少有两个 $h$，因此每隔一项就含有一个带 $h^2$ 的因子。

$$\begin{array}{r}(x+h{)}^n=(x+h)(x+h)\dots (x+h)=x^n+nh{x}^{n-1}+含有因子h^2的项\end{array}$$

Term is just a polynomial in $x$ and $h$.
项是含有 $x$ 和 $h$ 的多项式。

$$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{(x+h{)}^n-x^n}{h}\\ & =\underset{h\to 0}{\lim }\frac{x^n+nh{x}^{n-1}+h^2\times (\text{term})-x^n}{h}\\ & =\underset{h\to 0}{\lim }\frac{nh{x}^{n-1}+h^2\times (\text{term})}{h}\\ & =\underset{h\to 0}{\lim }(n{x}^{n-1}+h\times (\text{term}))\\ & =n{x}^{n-1}+0\times (\text{term})\\ & =n{x}^{n-1}\end{array}$$

The $x^n$ terms cancel, and then we can cancel out a factor of $h$.

$$\begin{array}{r}\frac{\text{d}}{\text{d}x}(x^n)=n{x}^{n-1}\text{when}n\text{is a positive integer}\end{array}$$

1.5. We want to differentiate $f(x)=x^a$ with respect to $x$, when $a$ is any real number at all

对 $f(x)=x^a$ 关于 $x$ 求导，其中 $a$ 是任意实数

In words, you are simply taking the power, putting a copy of it out front as the coefficient, and then knocking the power down by 1.
提取次数，将它放在最前面作系数，然后再将次数减少 1。

$$\begin{array}{r}\frac{\text{d}}{\text{d}x}(x^a)=a{x}^{a-1}\text{when}a\text{ is any real number at all}\end{array}$$

When $a=0$, then $x^a$ is the constant function 1. The derivative is then $0x^{-1}$, which is just 0.
当 $a=0$ 时，$x^a$ 是常数函数 1，其导数是 $0x^{-1}$，结果是 0。

$$\begin{array}{r}如果C是常数，那么\frac{\text{d}}{\text{d}x}(C)=0。\end{array}$$

If $a=1$, then $x^a$ is just $x$. According to the formula, the derivative
is $1x^0$, which is the constant function 1.
当 $a=1$ 时，$x^a$ 是 $x$，其导数是 $1x^0$，也就是常数函数 1。

$$\begin{array}{r}\frac{\text{d}}{\text{d}x}(x)=1\end{array}$$

When $a=2$, then we see that the derivative of $x^2$ with respect to $x$ is $2x^1$, which is just $2x$.

When $a=-1$, we can use our formula to see that the derivative of $x^{-1}$ is $-1\times x^{-2}$. In fact, this just says that the derivative of $1/x$ is $-1/x^2$.

$$\begin{array}{rl}\frac{\text{d}}{\text{d}x}(\sqrt{x})& =\frac{\text{d}}{\text{d}x}(x^{1/2})\\ & =\frac{1}{2}{x}^{1/2-1}\\ & =\frac{1}{2}{x}^{-1/2}\\ & =\frac{1}{2}\times \frac{1}{x^{1/2}}\\ & =\frac{1}{2}\times \frac{1}{\sqrt{x}}\\ & =\frac{1}{2\sqrt{x}}\end{array}$$

$$\begin{array}{rl}\frac{\text{d}}{\text{d}x}(\sqrt[3]{x})& =\frac{\text{d}}{\text{d}x}(x^{1/3})\\ & =\frac{1}{3}{x}^{1/3-1}\\ & =\frac{1}{3}{x}^{-2/3}\\ & =\frac{1}{3}\times \frac{1}{x^{2/3}}\\ & =\frac{1}{3}\times \frac{1}{\sqrt[3]{x^2}}\\ & =\frac{1}{3\sqrt[3]{x^2}}\end{array}$$

References

[1] Yongqiang Cheng, https://yongqiang.blog.csdn.net/
[2] 普林斯顿微积分读本 (修订版), https://m.ituring.com.cn/book/1623