Finding the Derivatives of Tangent, Cotangent, Secant, and Cosecant Functions · 求正切、余切、正割和余割函数的导数
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| trigonometric/ˌtrɪɡənəʊˈmetrɪk/ | 三角 | sān jiǎo |
The other four trig functions
- We know $\sin'=\cos$ and $\cos'=-\sin$. The other four trig functions are all quotients of these.
- So the Quotient Rule derives their derivatives — no new magic, just careful algebra.
- $\tan x=\dfrac{\sin x}{\cos x}$, $\cot x=\dfrac{\cos x}{\sin x}$, $\sec x=\dfrac1{\cos x}$, $\csc x=\dfrac1{\sin x}$.
- Learn the four results, but know they come from the Quotient Rule.
另外四个三角函数
- 我们已知 $\sin'=\cos$、$\cos'=-\sin$。另外四个三角函数都是它们的商。
- 所以用商法则就能推出它们的导数——没有新魔法,只是细心的代数。
- $\tan x=\dfrac{\sin x}{\cos x}$,$\cot x=\dfrac{\cos x}{\sin x}$,$\sec x=\dfrac1{\cos x}$,$\csc x=\dfrac1{\sin x}$。
- 记住这四个结果,但要知道它们来自商法则。
These four derivatives are all obtained using the... · 这四个导数都是通过……获得的
Each of $\tan,\cot,\sec,\csc$ is a quotient of $\sin$/$\cos$. · 每个 $\tan,\cot,\sec,\csc$ 都是 $\sin$/$\cos$ 的商。
Deriving $\tan x$
- $\dfrac{d}{dx}[\tan x]=\dfrac{d}{dx}\!\left[\dfrac{\sin x}{\cos x}\right]$.
- Quotient Rule: $\dfrac{\cos x\cos x-\sin x(-\sin x)}{\cos^2 x}=\dfrac{\cos^2 x+\sin^2 x}{\cos^2 x}$.
- Since $\cos^2 x+\sin^2 x=1$: $=\dfrac{1}{\cos^2 x}=\sec^2 x$.
- So $\boxed{\dfrac{d}{dx}[\tan x]=\sec^2 x}$ and, similarly, $\dfrac{d}{dx}[\cot x]=-\csc^2 x$.
推导 $\tan x$
- $\dfrac{d}{dx}[\tan x]=\dfrac{d}{dx}\!\left[\dfrac{\sin x}{\cos x}\right]$。
- 商法则:$\dfrac{\cos x\cos x-\sin x(-\sin x)}{\cos^2 x}=\dfrac{\cos^2 x+\sin^2 x}{\cos^2 x}$。
- 因为 $\cos^2 x+\sin^2 x=1$:$=\dfrac{1}{\cos^2 x}=\sec^2 x$。
- 所以 $\boxed{\dfrac{d}{dx}[\tan x]=\sec^2 x}$,同理 $\dfrac{d}{dx}[\cot x]=-\csc^2 x$。
The sine wave whose ratio makes tan · 使 tan 成为比率的正弦波
$\tan x=\sin x/\cos x$; its derivative $\sec^2 x$ is always positive — tan is increasing wherever it is defined. · $\tan x=\sin x/\cos x$；其导数 $\sec^2 x$ 始终为正——tan 在其定义域内单调递增。
What is $\dfrac{d}{dx}[\tan x]$? · $\dfrac{d}{dx}[\tan x]$是什么？
From the Quotient Rule on $\sin/\cos$: $\sec^2 x$. · 从 $\sin/\cos$ 的商法则得出：$\sec^2 x$。
$\dfrac{d}{dx}[\cot x]=$ ____ $\csc^2 x$.
It is · 它是 $-\csc^2 x$. · 它是 $-\csc^2 x$。
The secant and cosecant pair
- $\dfrac{d}{dx}[\sec x]=\sec x\tan x$.
- $\dfrac{d}{dx}[\csc x]=-\csc x\cot x$.
- Pattern: the "co-" functions ($\cos,\cot,\csc$) all carry a minus sign in their derivative.
- Each of these four also follows from the Quotient Rule applied to a $\tfrac1{\cos}$ or $\tfrac1{\sin}$.
正割与余割一对
- $\dfrac{d}{dx}[\sec x]=\sec x\tan x$。
- $\dfrac{d}{dx}[\csc x]=-\csc x\cot x$。
- 规律:"余-"函数($\cos,\cot,\csc$)的导数都带一个负号。
- 这四个也都可由对 $\tfrac1{\cos}$ 或 $\tfrac1{\sin}$ 应用商法则得到。
What is $\dfrac{d}{dx}[\sec x]$? · $\dfrac{d}{dx}[\sec x]$是什么？
$\frac{d}{dx}[\sec x]=\sec x\tan x$.
The derivatives of $\cot x$ and $\csc x$ both carry a minus sign. · $\cot x$ 和 $\csc x$ 的导数都带有负号。
$-\csc^2 x$ and $-\csc x\cot x$ — the "co-" functions carry a minus. · $-\csc^2 x$ 和 $-\csc x\cot x$——“co-”函数带有负号。
Using them with other rules
- These trigonometric derivatives combine with the product and quotient rules like any others.
- $\dfrac{d}{dx}[x\tan x]=\tan x+x\sec^2 x$ (Product Rule).
- $\dfrac{d}{dx}[3\sec x-\cot x]=3\sec x\tan x+\csc^2 x$.
- Keep the four boxed results handy; they turn up throughout the rest of the course.
与其他规则并用
- 这些三角导数像其他导数一样,与乘积法则、商法则组合。
- $\dfrac{d}{dx}[x\tan x]=\tan x+x\sec^2 x$(乘积法则)。
- $\dfrac{d}{dx}[3\sec x-\cot x]=3\sec x\tan x+\csc^2 x$。
- 把这四个框起来的结果放在手边;它们贯穿本课程其余部分。
Select all · 所有 correct derivatives. · 选择所有正确的导数。
The last is wrong: $\frac{d}{dx}[\sec x]=\sec x\tan x$, not $\sec^2 x$. · 最后一项是错误的：$\frac{d}{dx}[\sec x]=\sec x\tan x$，而非 $\sec^2 x$。
Watch the minus signs on the "co-" functions: $\frac{d}{dx}[\cot x]=-\csc^2 x$ and $\frac{d}{dx}[\csc x]=-\csc x\cot x$ both carry a minus, matching $\cos'=-\sin$. Mixing up $\sec^2 x$ (from $\tan$) with $\csc^2 x$ (from $\cot$) is another frequent slip — pair each derivative with its source.
注意"余-"函数上的负号:$\frac{d}{dx}[\cot x]=-\csc^2 x$ 与 $\frac{d}{dx}[\csc x]=-\csc x\cot x$ 都带负号,与 $\cos'=-\sin$ 相配。把 $\sec^2 x$(来自 $\tan$)与 $\csc^2 x$(来自 $\cot$)搞混是另一个常见失误——把每个导数与它的来源配对。
Differentiate $y=x^2\sec x$.
- Product Rule with $f=x^2$, $g=\sec x$: $f'=2x$, $g'=\sec x\tan x$.
- $y'=f'g+fg'=2x\sec x+x^2\sec x\tan x$.
- Factor: $y'=x\sec x\,(2+x\tan x)$.
对 $y=x^2\sec x$ 求导。
- 乘积法则,$f=x^2$、$g=\sec x$:$f'=2x$、$g'=\sec x\tan x$。
- $y'=f'g+fg'=2x\sec x+x^2\sec x\tan x$。
- 提取因式:$y'=x\sec x\,(2+x\tan x)$。
The four extra trigonometric derivatives all follow from the Quotient Rule: $\frac{d}{dx}[\tan x]=\sec^2 x$, $\frac{d}{dx}[\cot x]=-\csc^2 x$, $\frac{d}{dx}[\sec x]=\sec x\tan x$, $\frac{d}{dx}[\csc x]=-\csc x\cot x$. The "co-" functions carry a minus. Combine them freely with the product and quotient rules.
另外四个三角导数都来自商法则:$\frac{d}{dx}[\tan x]=\sec^2 x$、$\frac{d}{dx}[\cot x]=-\csc^2 x$、$\frac{d}{dx}[\sec x]=\sec x\tan x$、$\frac{d}{dx}[\csc x]=-\csc x\cot x$。"余-"函数带负号。自由地与乘积法则、商法则组合。