📚 Derivatives of Trigonometric Functions | 三角函数的导数
Understanding how to differentiate trigonometric functions is a cornerstone of IB Mathematics. These derivatives not only extend your calculus toolkit but also unlock a vast range of problems in physics, engineering, and periodic motion analysis. Mastering the rules for sin x, cos x, tan x, and their reciprocals, together with the chain rule, product rule, and quotient rule, empowers you to tackle any composite trigonometric function with confidence.
理解如何对三角函数求导是 IB 数学的基础。这些导数不仅扩展了你的微积分工具箱，还能解锁物理、工程和周期运动分析中的大量问题。掌握 sin x、cos x、tan x 及其倒数函数的求导法则，再配合链式法则、乘积法则和商法则，就能让你自信地处理任何复合三角函数。
1. Review of Key Limits and Derivative Definition | 关键极限与导数定义回顾
ions, we first recall two fundamental limits: lim (x→0) sin x / x = 1 and lim (x→0) (cos x – 1)/x = 0. These limits are essential when using the first principle definition f '(x) = lim (h→0) [f(x+h) – f(x)]/h.
为了推导三角函数的导数，我们首先回顾两个基本极限：lim (x→0) sin x / x = 1 和 lim (x→0) (cos x – 1)/x = 0。在使用第一原理定义 f '(x) = lim (h→0) [f(x+h) – f(x)]/h 时，这些极限至关重要。
The derivative definition will be applied directly to sin x and cos x. Once those are established, all other trigonometric derivatives can be obtained via the quotient rule and chain rule.
导数定义将直接应用于 sin x 和 cos x。一旦确立了这两个，所有其他三角函数的导数都可以通过商法则和链式法则得到。
2. Derivative of sin x from First Principles | 从第一原理推导 sin x 的导数
Let f(x) = sin x. Using the definition, f '(x) = lim (h→0) [sin(x+h) – dition formula sin(x+h) = sin x cos h + cos x sin h. This gives [sin x cos h + cos x sin h – sin x]/h = sin x (cos h – 1)/h + cos x (sin h / h).
设 f(x) = sin x。根据定义，f '(x) = lim (h→0) [sin(x+h) – sin x]/h。运用正弦加法公式 sin(x+h) = sin x cos h + cos x sin h，得到 [sin x cos h + cos x sin h – sin x]/h = sin x (cos h – 1)/h + cos x (sin h / h)。
Take the limit as h→0: (cos h – 1)/h → 0 and sin h / h → 1. So f '(x) = sin x · 0 + cos x · 1 = cos x. Thus, d/dx (sin x) = cos x.
取 h→0 时的极限：(cos h – 1)/h → 0，sin h / h → 1。因此 f '(x) = sin x · 0 + cos x · 1 = cos x。于是，d/dx (sin x) = cos x。
3. Derivative of cos x via Shift and First Principles | 通过平移和第一原理推导 cos x 的导数
We can find the derivative of cos x either by recognising that cos x = sin(π/2 – ng the chain rule, or directly from first principles. The direct limit lim (h→0) [cos(x+h) – cos x]/h employs the identity cos(x+h) = cos x cos h – sin x sin h.
我们可以通过认识到 cos x = sin(π/2 – x) 并使用链式法则来求 cos x 的导数，也可以直接从第一原理出发。直接极限 lim (h→0) [cos(x+h) – cos x]/h 使用了恒等式 cos(x+h) = cos x cos h – sin x sin h。
This leads to [cos x(cos h – 1) – sin x sin h]/h = cos x (cos h – 1)/h – sin x (sin h / h). Taking limits gives cos x · 0 – sin x · 1 = –sin x. Hence, d/dx (cos x) = –sin x.
由此得到 [cos x(cos h – 1) – sin x sin h]/h = cos x (cos h – 1)/h – sin x (sin h / h)。取极限得到 cos x · 0 – sin x · 1 = –sin x。因此，d/dx (cos x) = –sin x。
4. Derivative of tan x Using the Quotient Rule | 用商法则求 tan x 的导数
Since tan x = sin x / cos x, we apply the quotient rule: (u/v)' = (u' v – u v')/v². Here u = sin x, v = cos x, so u' = cos x, v' = –sin x.
由于 tan x = sin x / cos x，我们应用商法则：(u/v)' = (u' v – u v')/v²。这里 u = sin x，v = cos x，所以 u' = cos x，v' = –sin x。
Substituting, d/dx (tan x) = (cos x · cos x – sin x · (–sin x)) / cos² x = (cos² x + sin² x) / cos² x = 1 / cos² x = sec² x.
代入可得 d/dx (tan x) = (cos x · cos x – sin x · (–sin x)) / cos² x = (cos² x + sin² x) / cos² x = 1 / cos² x = sec² x。
Thus, the derivative of tan x is sec² x. This is often memorised and used extensively in integration as well.
因此，tan x 的导数是 sec² x。这个结果经常被记忆并在积分中广泛使用。
5. Derivatives of sec x, csc x, and cot x | sec x、csc x 和 cot x 的导数
, and cot x = cos x/sin x, we can differentiate each with the quotient rule or chain rule.
利用定义 sec x = 1/cos x、csc x = 1/sin x 和 cot x = cos x/sin x，我们可以用商法则或链式法则对每一个进行求导。
For sec x: d/dx (1/cos x) = (0·cos x – 1·(–sin x)) / cos² x = sin x / cos² x = (1/cos x)(sin x/cos x) = sec x tan x.
对于 sec x：d/dx (1/cos x) = (0·cos x – 1·(–sin x)) / cos² x = sin x / cos² x = (1/cos x)(sin x/cos x) = sec x tan x。
For csc x: d/dx (1/sin x) = (0·sin x – 1·cos x) / sin² x = –cos x / sin² x = –csc x cot x.
对于 csc x：d/dx (1/sin x) = (0·sin x – 1·cos x) / sin² x = –cos x / sin² x = –csc x cot x。
For cot x: d/dx (cos x/sin x) = (–sin x·sin x – cos x·cos x) / sin² x = –(sin² x + cos² x)/sin² x = –1/sin² x = –csc² x.
对于 cot x：d/dx (cos x/sin x) = (–sin x·sin x – cos x·cos x) / sin² x = –(sin² x + cos² x)/sin² x = –1/sin² x = –csc² x。
The results are summarised in the table below.
结果总结在下表中。
| Function f(x) | Derivative f '(x) |
| sin x | cos x |
| cos x | –sin x |
| tan x | sec² x |
| sec x | sec x tan x |
| csc x | –csc x cot x |
| cot x | –csc² x |
6. Chain Rule with Trigonometric Functions | 三角复合函数的链式法则
hen dy/dx = cos(u) · du/dx. Similar patterns hold for all other trigonometric derivatives.
当三角函数的变量不是简单的 x，而是 x 的一个函数时，我们使用链式法则：如果 y = sin(u) 且 u = g(x)，那么 dy/dx = cos(u) · du/dx。类似的模式适用于所有其他三角函数的导数。
For example, differentiate y = cos(3x²). Let u = 3x², then du/dx = 6x. So dy/dx = –sin(u) · 6x = –6x sin(3x²). Another example: y = tan(√x). dy/dx = sec²(√x) · (1/(2√x)).
例如，对 y = cos(3x²) 求导。设 u = 3x²，则 du/dx = 6x。所以 dy/dx = –sin(u) · 6x = –6x sin(3x²)。另一个例子：y = tan(√x)，dy/dx = sec²(√x) · (1/(2√x))。
rite as [sin(2x+1)]³. Then dy/dx = 3[sin(2x+1)]² · cos(2x+1) · 2 = 6 sin²(2x+1) cos(2x+1).
链式法则可以扩展到嵌套函数，比如 y = sin³(2x+1)。可以写成 [sin(2x+1)]³。那么 dy/dx = 3[sin(2x+1)]² · cos(2x+1) · 2 = 6 sin²(2x+1) cos(2x+1)。
7. Product Rule and Quotient Rule Applications | 乘积法则与商法则的应用
are frequently multiplied by polynomials or other functions. The product rule states: (uv)' = u' v + u v'. For instance, differentiate f(x) = x² sin x. Here u = x², v = sin x. Then f '(x) = 2x sin x + x² cos x.
三角函数经常与多项式或其他函数相乘。乘积法则表述为 (uv)' = u' v + u v'。例如，对 f(x) = x² sin x 求导。这里 u = x²，v = sin x，则 f '(x) = 2x sin x + x² cos x。
ivided by another or by a polynomial. For y = sin x / x, identify u = sin x, v = x. Then y' = (cos x · x – sin x · 1) / x² = (x cos x – sin x)/x².
当三角函数除以另一个函数或多项式时，使用商法则。对于 y = sin x / x，设 u = sin x，v = x，则 y' = (cos x · x – sin x · 1) / x² = (x cos x – sin x)/x²。
Combining rules can be tested. For y = eˣ cos x, product rule gives y' = eˣ cos x + eˣ (–sin x) = eˣ (cos x – sin x). With trigonometric identities, you can sometimes simplify the result further.
综合运用多个法则是常见的考查方式。对于 y = eˣ cos x，乘积法则给出 y' = eˣ cos x + eˣ (–sin x) = eˣ (cos x – sin x)。借助三角恒等式，有时可以进一步化简结果。
8. Second Derivatives and Beyond | 二阶及更高阶导数
igonometric function is obtained by differentiating the first derivative. For y = sin x, y' = cos x, so y'' = –sin x. This demonstrates that sin x satisfies the differential equation y'' + y = 0.
三角函数的二阶导数由对一阶导数再次求导得到。对于 y = sin x，y' = cos x，所以 y'' = –sin x。这表明 sin x 满足微分方程 y'' + y = 0。
Similarly, y = cos x gives y' = –sin x and y'' = –cos x, also leading to y'' + y = 0. For y = tan x, y' = sec² x. The second derivative y'' = 2 sec x · sec x tan x = 2 sec² x tan x, which can also be written as 2 tan x / cos² x.
类似地，y = cos x 给出 y' = –sin x，y'' = –cos x，同样满足 y'' + y = 0。对于 y = tan x，y' = sec² x，二阶导数 y'' = 2 sec x · sec x tan x = 2 sec² x tan x，也可以写作 2 tan x / cos² x。
this pattern is useful in problems involving Maclaurin series and differential equations.
对于 sin 和 cos，高阶导数每四步循环一次：d⁴/dx⁴ sin x = sin x。理解这个规律在涉及麦克劳林级数和微分方程的问题中很有用。
9. Common Mistakes and How to Avoid Them | 常见错误与避免方法
A frequent error is forgetting the negative sign in the derivative of cos x: d/dx(cos x) = – ite 1/ cos² x or tan² x instead of sec x tan x. Remember, sec x tan x is the correct form.
一个常见错误是忘记 cos x 导数中的负号：d/dx(cos x) = –sin x，而非 sin x。务必反复检查符号。另一个错误是错求 sec x 的导数；有些学生写成 1/cos² x 或 tan² x，而正确形式是 sec x tan x。请记住，sec x tan x 才是正确形式。
in(5x), the derivative is 5 cos(5x), not cos(5x). Always write the inner derivative explicitly.
使用链式法则时，忘记乘以内层函数的导数是一个典型失误。对于 y = sin(5x)，导数是 5 cos(5x)，而不是 cos(5x)。务必明确写出内层导数。
ctly applying the product rule order can also cause sign errors. Stick to the formula (fg)' = f ' g + f g ' terms involve cos or sin derivatives. Practice with structured steps helps build accuracy.
错误地应用乘积法则的顺序也可能导致符号错误。应严格遵循 (fg)' = f ' g + f g '，并在项中包含 cos 或 sin 的导数时留意符号。通过分步骤的练习有助于提高准确性。
10. Worked Examples and Applications | 应用例题与解析
Example 1: Find the equation of the tangent to y = tan x at x = π/4. First, y(π/4) = 1. Derivative: y' = sec² x. At x = π/4, sec²(π/4) = (√2)² = 2. Tangent line: y – 1 = 2(x – π/4).
例 1：求 y = tan x 在 x = π/4 处的切线方程。首先，y(π/4) = 1。导数：y' = sec² x。在 x = π/4 处，sec²(π/4) = (√2)² = 2。切线为：y – 1 = 2(x – π/4)。
Example 2: Differentiate f(x) = ln(sin x). Use chain rule: f 'x) = (1/sin x) · cos x = cot x. This elegant result appears often in integration problems.
例 2：对 f(x) = ln(sin x) 求导。运用链式法则：f '(x) = (1/sin x) · cos x = cot x。这个简洁的结果经常出现在积分问题中。
Example 3: Find the stationary points of f(x) = x cos x on [0, 2π]. f '(x) = cos x – be solved numerically, e.g., using a GDC. The critical values occur where cos x = x sin x.
例 3：求 f(x) = x cos x 在 [0, 2π] 上的驻点。f '(x) = cos x – x sin x = 0。这个超越方程可使用 GDC 数值求解。临界点满足 cos x = x sin x。
Such examples illustrate how trigonometric derivatives are combined with algebraic techniques to explore function behaviour, including increasing/decreasing intervals and concavity.
这些例子说明如何将三角导数与代数技巧相结合，以探究函数性态，包括增减区间和凹凸性。
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