Quick confession? I never fully learned the trig derivatives. Sure, I memorized $\sin' = \cos$ and $\cos' = -\sin $ like everyone else, but the derivative of tangent? Cosecant? Forget it, magic spells.
After years of searching, there's a middle ground between tedious derivation and rote memorization.Aha moment: all trig functions change using the same process: (sign)(scale)(swapped function).
Here's the Table of Trig Derivatives we'll learn to fill out:
As background, learn tovisualize the trig functions, and how they're related by the Pythagorean Theorem and similarity:
Part 1: Learn the table
First, let's learn to make the table, one column at a time:
- Function:The function to derive (sin, cos, tan, cot, sec, csc)
- Sign:The "primary" functions are positive, and the "co" (complementary) functions are negative
- Scale:The hypotenuse (red) used by each function
- Swap:Theotherfunction in each Pythagorean triangle (sin ⇄ cos, tan ⇄ sec, cot ⇄ csc)
- Derivative:Multiply to find the derivative
Tada! This procedure somehow finds derivatives for trig fucntions. Learning tips:
- Think "triple S": sign, scale, swap
- You've likely memorized $\sin' = \cos$ and $\cos' = -\sin$. Fill in those rows to kickstart the process.
通常情况下，我更喜欢洞察而不是记忆。但实际上，你们问的是三角导数因为你们要考试，我现在想帮你们。
Like a multiplication table, after filling in the entries, we notice patterns. Could $\sin' = \cos$ and $\csc' = -\csc \cot$ have something in common?
You bet.
Part 2: Visualize the derivatives
What's the derivative of sine?
Theformal approachis to plug $\sin(x)$ into the definition of derivative, do the algebra, and see that $\cos(x)$ pops out. Accurate, but unsatisfying. If $\sec(x)$ was the derivative, would you notice something was amiss? Probably not.
Here's what's happening geometrically:
The derivative of sine means "How much does our height change when I change my angle?"
I see it like this: we have a starting angle, $x$. We increase it a smidge ($dx$), which we can placealongour unit circle (since radians aredistance traveled沿着周长)。
We then draw a mini-triangle based on $dx$, similar to the large one, which shows the height and width change as we move along the perimeter.
大三角有比例$\text{红}:\text{蓝}:\text{绿}= 1:\cos: \sin$。迷你三角形也有类似的比例，最长的边是$dx$而不是1。Therefore, the mini lengths are:
$\text{mini red} : \text{mini blue} : \text{mini green} = dx : \cos dx : \sin dx$
Since mini blue is the change in sine, and mini green the change in cosine, we have:
$\sin'(x) = \text{height change} = \text{mini blue} = \cos(x) dx$
$\cos'(x) = \text{width change} = (-1) \cdot \text{mini green} = - \sin(x) dx$
Notice the negative sign with $\cos'$, since mini-green points to the left (negative).
Quick Aside: How the Columns Work
“迷你三角”策略适用于所有三角函数。There are 3 factors:
Q1: What's the sign?
The trig co-functions are the original function applied to thecomplementary angle.
目测一下，我们看到参数$x$和$(90 - x)$被散布。TheChain Ruleis whispering (screaming?) that the derivatives should be opposite, right?
Let's try it:
$\cos'(x) = [\sin(90 - x)]' = [\sin'(90 - x)][(90-x)'] = \cos(90-x)(-1)$ $= \sin(x)(-1) = -\sin(x) $
Yep, we got a negative sign.
What happened? We converted $\cos$ into its $\sin$ form (turning the angle into the complement), took the derivative, got the $-1$ term, and converted back. All the cofunctions have a similar pattern, giving us a negative sign in the table.
(Note: the negative sign means the cofunction changesoppositethe original function, not that the derivative isless than zero. Cosine increases when sine is negative.)
Q2: What's the scale?
Sine and cosine live on the unit circle (radius 1). The other functions use a radius of secant (tan/sec) or cosecant (cot/csc).
Q3: What's the swapped function?
我们通过缩小原来的三角形来做一个迷你三角形，然后旋转，使$dx$匹配长度为$1$的边。如果旋转之后，原来的颜色(函数)指向相同的方向，那就奇怪了。
The change must be based on the other function in the triangle (sine's change is based on cosine, cosine on sine, tangent on secant, etc.)
Also, it would be strange for a function to grow based on its own current value, right? (Hold that thought.)
Derivatives of tangent and secant
Ok, let's draw the mini triangles for tangent and secant:
- First, we draw the $dx$ mini-triangle on the unit circle (like sin/cos).
- 接下来，我们滑动/缩放小三角形以适应“割线”半径:$dx$在割线圆上变成$\sec(x) dx$。
- 最后，旋转小三角形，使已知的$1$边(蓝色)匹配我们的$\sec(x)dx$的变化。
Ok. So how big are the sides of the mini triangle?
$\text{mini blue}: \text{mini green} : \text{mini red} = 1 : \tan(x) : \sec(x)$
We know $\text{mini blue} = \sec(x) dx$, so we just scale up the other sides by that amount:
- $d\sec = \text{mini green} = \tan(x) [\text{mini blue}] = \tan(x) \sec(x)dx = \sec(x) \tan(x) dx$
- $d\tan= \text{mini red} = \sec(x) [\text{mini blue}] = \sec(x) \sec(x)dx = \sec^2(x) dx$
Nice! I like how this matches the sine/cosine process. We're just measuring sides in the mini-triangle.
Derivatives of Cosecant and Cotangent
For completeness, here's csc/cot:
注意在小三角中的$d\cot$和$d\csc$是如何在大三角中逆着它们的正边移动的。Using the same sign, scale, swap process, we get:
$\cot' = (-)(\csc)(\csc) = -\csc^2$
$\csc' = (-)(\csc)(\cot) = -\csc \cot$
Colorizing the sides really helps me link the mini-triangle back to the original.
Now, we didn't have to draw this all out: we already know $\tan'$ and $\sec'$. Using the chain rule and complementary functions, we can do:
$\cot(x)' = [\tan(90-x)]' = \tan'(90-x)(90 - x)' = \sec^2(90-x)(-1) = -\csc^2(x)$
$\csc(x)' = [\sec(90-x)]' = \sec'(90-x)(90 - x)' = \sec(90-x)\tan(90-x)(-1)$ $= -\csc(x)\cot(x)$
Summary: What do the derivatives mean?
Blindly memorizing trig derivatives doesn't teach you much.
The deeper intuition:Trig derivatives are based on 3 effects: the sign, the radius (scale), and the other function.
所以美元\ tan ' = \交会^ 2美元,认为它是$ \ tan ' = (+) (\ sec) (\ sec)美元,即美元(\文本}{标志)(\文本{规模})(\文本{交换函数})美元。见鬼，你甚至可以看到$\cos' = (-)(1)(\sin)$。
If you can complete the derivative table and draw the mini-triangles, you'll have amuchbetter understanding of trig than I ever did.
快乐数学。
Appendix: Combined Diagram
It's a bit busy, but here's all the mini-triangles together:
再一次，直觉:这些迷你红/绿/蓝三角形(都是相似的!)显示了变化。
Appendix: Exponential Behavior
Remember how we didn't think a derivative should be based on the same function? Well,
$\tan' = \sec^2 = 1 + \tan^2$
which means tangent growsfasterthanexponential: it grows based on its own square (vs. "just" the current value).
我们看到$\tan(x)$在比赛中击败了$e^x$，指数也不逊色!(I'm looking forward to "this is growing tangentially" to be the new catchphrase.)
Appendix: Other mini-triangle layouts
There's other ways we could arrange the mini-triangle. I think it's easiest when the change along the perimeter is mapped to the side of length 1.
But, when finding $\tan'$, you could use the red/blue/green ratios from the 1/cos/sin triangle and find:
$\tan' = \sec \frac{1}{\cos} = \sec^2$
The scale of $\sec$ is the same, but the unknown side is of length $\frac{1}{\cos} dx$. Different approach, same result.