1. Constant multiples of functions (函数的常数倍)

It’s easy to deal with a constant multiple of a function: you just multiply by the constant after you differentiate.
对函数的常数倍进行求导，只需在函数求导后，用常数乘以该函数的导数。

We know the derivative of $x^2$ is $2x$; so the derivative of $7x^2$ is 7 times $2x$, or $14x$.

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

$$\begin{array}{r}\frac{\text{d}}{\text{d}x}(7x^2)=7\times \frac{\text{d}}{\text{d}x}(x^2)=7\times 2x=14x\end{array}$$

The derivative of $-x^2$ is $-2x$, since you can think of the minus out front as multiplication by -1.

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

Similarly, to find the derivative of $13x^4$, multiply 13 by 4, giving a coefficient of 52, and then knock the power down by one to get $52x^3$.

$$\begin{array}{r}\frac{\text{d}}{\text{d}x}(13x^4)=13\times \frac{\text{d}}{\text{d}x}(x^4)=13\times 4x^3=52x^3\end{array}$$

2. Sums and differences of functions (函数和与函数差)

It’s even easier to differentiate sums and differences of functions: just differentiate each piece and then add or subtract.
对函数和与函数差求导，只需对每一部分求导，然后再相加或相减。

$$\begin{array}{rl}\frac{\text{d}}{\text{d}x}(3x^5-2x^2+\frac{7}{\sqrt{x}}+2)& =\frac{\text{d}}{\text{d}x}(3x^5-2x^2+\frac{7}{x^{1/2}}+2)\\ & =\frac{\text{d}}{\text{d}x}(3x^5-2x^2+7x^{-1/2}+2)\\ & =\frac{\text{d}}{\text{d}x}(3x^5)+\frac{\text{d}}{\text{d}x}(-2x^2)+\frac{\text{d}}{\text{d}x}(7x^{-1/2})+\frac{\text{d}}{\text{d}x}(2)\\ & =3\times \frac{\text{d}}{\text{d}x}(x^5)-2\times \frac{\text{d}}{\text{d}x}(x^2)+7\times \frac{\text{d}}{\text{d}x}(x^{-1/2})+\frac{\text{d}}{\text{d}x}(2)\\ & =3\times 5x^4-2\times 2x+7\times (-1/2)x^{-1/2-1}+0\\ & =15x^4-4x-\frac{7}{2}{x}^{-3/2}\\ & =15x^4-4x-\frac{7}{2}\frac{1}{x^{3/2}}\\ & =15x^4-4x-\frac{7}{2}\frac{1}{x^{2/2}{x}^{1/2}}\\ & =15x^4-4x-\frac{7}{2}\frac{1}{x\sqrt{x}}\end{array}$$

$$\begin{array}{r}{x}^{5/2}=x^{4/2}{x}^{1/2}=x^2{x}^{1/2}=x^2\sqrt{x}\\ x^{7/2}=x^{6/2}{x}^{1/2}=x^3{x}^{1/2}=x^3\sqrt{x}\end{array}$$

3. Products of functions via the product rule (通过乘积法则求积函数的导数)

• 乘积法则 (版本 1)

Product rule (version 1): if $h(x)=f(x)g(x)$, then $h^{\prime }(x)=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$.

That is, you take the derivative of $f$ and multiply it by $g$ (not the derivative of $g$). Then you also have to take the derivative of $g$ and multiply it by $f$. Finally, add the two things together.

$h(x)=f(x)g(x)=(x^5+2x-1)(3x^8-2x^7-x^4-3x)$
$f(x)=x^5+2x-1,f^{\prime }(x)=5x^4+2$
$g(x)=3x^8-2x^7-x^4-3x,g^{\prime }(x)=24x^7-14x^6-4x^3-3$

$$\begin{array}{rl}{h}^{\prime }(x)& =f^{\prime }(x)g(x)+f(x)g^{\prime }(x)\\ & =(5x^4+2)(3x^8-2x^7-x^4-3x)+(x^5+2x-1)(24x^7-14x^6-4x^3-3)\end{array}$$

• 乘积法则 (版本 2)

We can take the above form of the product rule and make some replacements: first, $u$ replaces $f(x)$, so that $\text{d}u/\text{d}x$ replaces $f^{\prime }(x)$; we also do the same thing with $v$ and $g(x)$.

Product rule (version 2): if $y=uv$, then $\frac{\text{d}y}{\text{d}x}=\frac{\text{d}u}{\text{d}x}v+u\frac{\text{d}v}{\text{d}x}$.

$y=uv=(x^3+2x)(3x+\sqrt{x}+1)$
$u=x^3+2x,\frac{\text{d}u}{\text{d}x}=3x^2+2$
$v=3x+\sqrt{x}+1,\frac{\text{d}v}{\text{d}x}=3+\frac{1}{2}{x}^{-1/2}=3+\frac{1}{2x^{1/2}}=3+\frac{1}{2\sqrt{x}}$

$$\begin{array}{rl}\frac{\text{d}y}{\text{d}x}& =\frac{\text{d}u}{\text{d}x}v+u\frac{\text{d}v}{\text{d}x}\\ & =(3x^2+2)(3x+\sqrt{x}+1)+(x^3+2x)(3+\frac{1}{2\sqrt{x}})\end{array}$$

• 乘积法则 (三个变量)

Product rule (three variables): if $y=uvw$, then $\frac{\text{d}y}{\text{d}x}=\frac{\text{d}u}{\text{d}x}vw+u\frac{\text{d}v}{\text{d}x}w+uv\frac{\text{d}w}{\text{d}x}$.

Just add up $uvw$ three times, but put a $\text{d}/\text{d}x$ in front of a different variable in each term. The same trick works for four or more variables, every variable gets differentiated once!
把 $uvw$ 加三次，但对于每一项，要将 $\text{d}/\text{d}x$ 放在不同的变量之前。同样的方法适用于四个或更多个变量，每一个变量都要进行一次微分运算。

$y=uvw=(x^2+1)(x^2+3x)(x^5+2x^4+7)$
$u=x^2+1,\frac{\text{d}u}{\text{d}x}=2x$
$v=x^2+3x,\frac{\text{d}v}{\text{d}x}=2x+3$
$w=x^5+2x^4+7,\frac{\text{d}w}{\text{d}x}=5x^4+8x^3$

$$\begin{array}{rl}\frac{\text{d}y}{\text{d}x}& =\frac{\text{d}u}{\text{d}x}vw+u\frac{\text{d}v}{\text{d}x}w+uv\frac{\text{d}w}{\text{d}x}\\ & =(2x)(x^2+3x)(x^5+2x^4+7)+(x^2+1)(2x+3)(x^5+2x^4+7)+(x^2+1)(x^2+3x)(5x^4+8x^3)\end{array}$$

4. Quotients of functions via the quotient rule (通过商法则求商函数的导数)

• 除法法则 (版本 1)

Quotient rule (version 1): if $h(x)=\frac{f(x)}{g(x)}$, then $h^{\prime }(x)=\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{(g(x){)}^2}$.

Notice that the numerator of the right-hand fraction is the same as the numerator in the product rule, except with a minus instead of a plus.
除了正号变成了负号外，等号右边分式的分子与乘积法则中的分子是一样的。

$h(x)=f(x)/g(x)=\frac{2x^3-3x+1}{x^5-8x^3+2}$
$f(x)=2x^3-3x+1,f^{\prime }(x)=6x^2-3$
$g(x)=x^5-8x^3+2,g^{\prime }(x)=5x^4-24x^2$

$$\begin{array}{rl}{h}^{\prime }(x)& =\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{(g(x){)}^2}\\ & =\frac{(6x^2-3)(x^5-8x^3+2)-(2x^3-3x+1)(5x^4-24x^2)}{(x^5-8x^3+2{)}^2}\end{array}$$

• 除法法则 (版本 2)

Quotient rule (version 2): if $y=\frac{u}{v}$, then $\frac{\text{d}y}{\text{d}x}=\frac{\frac{\text{d}u}{\text{d}x}v-u\frac{\text{d}v}{\text{d}x}}{v^2}$.

$y=u/v=\frac{3x^2+1}{2x^8-7}$
$u=3x^2+1,\frac{\text{d}u}{\text{d}x}=6x$
$v=2x^8-7,\frac{\text{d}v}{\text{d}x}=16x^7$

$$\begin{array}{rl}\frac{\text{d}y}{\text{d}x}& =\frac{\frac{\text{d}u}{\text{d}x}v-u\frac{\text{d}v}{\text{d}x}}{v^2}\\ & =\frac{(6x)(2x^8-7)-(3x^2+1)(16x^7)}{(2x^8-7{)}^2}\end{array}$$

5. Composition of functions via the chain rule (通过链式求导法则求复合函数的导数)

• 链式求导法则 (版本 1)

Chain rule (version 1): if $h(x)$ = $f(g(x))$, then $h^{\prime }(x)=f^{\prime }(g(x))g^{\prime }(x)$.

$f(u)=u^{99},f^{\prime }(u)=99u^{98}$
$g(x)=x^2+1,g^{\prime }(x)=2x$
$h(x)=f(g(x))=f(x^2+1)=(x^2+1{)}^{99}$

$$\begin{array}{rl}{h}^{\prime }(x)& =f^{\prime }(g(x))g^{\prime }(x)\\ & =99(g(x){)}^{98}(2x)\\ & =99(x^2+1{)}^{98}(2x)\\ & =198x(x^2+1{)}^{98}\end{array}$$

• 链式求导法则 (版本 2)

Chain rule (version 2): if $y$ is a function of $u$, and $u$ is a function of $x$, then $\frac{\text{d}y}{\text{d}x}=\frac{\text{d}y}{\text{d}u}\frac{\text{d}u}{\text{d}x}$.

$y=u^{99},\frac{\text{d}y}{\text{d}u}=99u^{98}$
$u=x^2+1,\frac{\text{d}u}{\text{d}x}=2x$
$h(x)=f(g(x))=f(x^2+1)=(x^2+1{)}^{99}$

$$\begin{array}{rl}\frac{\text{d}y}{\text{d}x}& =\frac{\text{d}y}{\text{d}u}\frac{\text{d}u}{\text{d}x}\\ & =99u^{98}\times 2x\\ & =198x{u}^{98}\\ & =198x(x^2+1{)}^{98}\end{array}$$

Now you just need to tidy it up by replacing $u$ by $x^2+1$ to see that we have $\text{d}y/\text{d}x=198x(x^2+1{)}^{98}$.
使用 $x^2+1$ 替换 $u$，得到 $\text{d}y/\text{d}x=198x(x^2+1{)}^{98}$。

6. A nasty example (难以处理的例子)

7. Justification of the product rule and the chain rule (乘积法则和链式求导法则的理由)

justification /ˌdʒʌstɪfɪˈkeɪʃn/：n. 正当理由

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