Differentiating sin x and cos x | sin x 与 cos x 的求导

📚 Differentiating sin x and cos x | sin x 与 cos x 的求导

Understanding how to differentiate the sine and cosine functions is a cornerstone of A-Level calculus. These trigonometric derivatives underpin the analysis of oscillations, wave motion, and countless real-world phenomena. In this article you will explore the standard results, proofs from first principles, and a variety of techniques for differentiating more complex expressions involving sin x and cos x, including the use of the chain, product, and quotient rules.

掌握正弦和余弦函数的求导方法是 A-Level 微积分的基石。这些三角函数的导数支撑着对振动、波动以及无数现实世界现象的分析。本文将带你探索标准求导结果、从第一性原理出发的证明，以及多种对含 sin x 与 cos x 的复杂表达式进行求导的技巧，包括正确使用链式法则、乘积法则和商法则。

1. Key Derivatives and First Encounter | 关键导数与初识

ts you must memorise are: the derivative of sin x is cos x, and the derivative of cos x is –sin x. In Leibniz notation, d/dx (sin x) = cos x and d/dx (cos x) = – degrees. Radian measure ensures the gradient of the sine curve at the origin is exactly 1.

两个必须牢记的基本结果是：sin x 的导数是 cos x，cos x 的导数是 -sin x。用莱布尼茨记号表示，d/dx (sin x) = cos x，d/dx (cos x) = -sin x。要使这些结论成立，x 必须用弧度度量，而非角度。弧度制保证了正弦曲线在原点处的梯度恰好为 1。

2. Proof from First Principles | 从第一性原理证明

see where the derivatives come from, we can use the limit definition. For f(x) = sin x, f'(x) = lim_h→0 [sin(x+h) – sin x]/h. Using the identity sin(x+h) = sin x cos h + cos x sin h, this becomes lim_h→0 [sin x(cos h – 1) + cos x sin h]/h. Splitting the limit and using the small-angle approximations (cos h – 1)/h → 0 and sin h/h → 1 gives cos x.

要理解导数的来源，我们可以利用极限定义。对于 f(x) = sin x，f'(x) = lim_h→0 [sin(x+h) – sin x]/h。借助恒等式 sin(x+h) = sin x cos h + cos x sin h，原式变为 lim_h→0 [sin x(cos h – 1) + cos x sin h]/h。拆开极限并运用小角近似 (cos h – 1)/h → 0 和 sin h/h → 1，即得到 cos x。

A similar argument for cos x uses cos(x+h) = cos x cos h – sin x sin h. The derivative becomes lim_h→0 [cos x(cos h – 1) – sin x sin h]/h, which tends to – roofs highlight why radian measure is essential: the special limits sin h/h → 1 and (cos h – 1)/h → 0 are only true when angles are in radians.

对于 cos x，类似的方法使用 cos(x+h) = cos x cos h – sin x sin h。导数化为 lim_h→0 [cos x(cos h – 1) – sin x sin h]/h，当 h→0 时趋于 -sin x。这些证明凸显了弧度制的重要性：特殊极限 sin h/h → 1 和 (cos h – 1)/h → 0 只有在弧度下才成立。

3. Differentiating Simple Combinations | 简单组合的求导

ic derivatives, you can differentiate any linear combination. For instance, if y = 3 sin x – 5 cos x, then dy/dx = 3 cos x – 5(–sin x) = 3 cos x + 5 sin x. Constants multiply the derivative directly.

一旦掌握了基本导数，便可以求任何线性组合的导数。例如，若 y = 3 sin x – 5 cos x，则 dy/dx = 3 cos x – 5(-sin x) = 3 cos x + 5 sin x。常数直接乘以导数。

Also, be careful with arguments that are more than just x. If you see sin(2x) that requires the chain rule; do not simply write cos(2x). The chain rule is covered in the next section.

另外，需注意自变量不仅仅是 x 的情况。若见到 sin(2x)，那需要链式法则，不能直接写成 cos(2x)。链式法则将在下节介绍。

4. Chain Rule with sin and cos | 含 sin 与 cos 的链式法则

here u is a function of x, then dy/dx = cos(u) × du/dx. Similarly, for y = cos(u), dy/dx = – = sin(2x). Here u = 2x, so dy/dx = cos(2x) × 2 = 2 cos(2x). For y = cos(3x² + 1), dy/dx = –sin(3x² + 1) × 6x = –6x sin(3x² + 1).

当角度是 x 的函数时，链式法则必不可少。若 y = sin(u)，其中 u 是关于 x 的函数，则 dy/dx = cos(u) × du/dx。类似地，对于 y = cos(u)，dy/dx = -sin(u) × du/dx。一个经典例子是 y = sin(2x)。这里 u = 2x，所以 dy/dx = cos(2x) × 2 = 2 cos(2x)。对于 y = cos(3x² + 1)，dy/dx = -sin(3x² + 1) × 6x = -6x sin(3x² + 1)。

is sin(4x), and an even inner function is 4x. Then dy/dx = 3[sin(4x)]² × cos(4x) × 4 = 12 sin²(4x) cos(4x).

链式法则可以多次运用。若 y = sin³(4x)，可写成 (sin(4x))³。外层是立方函数，内层是 sin(4x)，再内一层是 4x。于是 dy/dx = 3[sin(4x)]² × cos(4x) × 4 = 12 sin²(4x) cos(4x)。

5. Product Rule Involving sin x or cos x | 含 sin x 或 cos x 的乘积法则

differentiating a product like x² sin x, apply the product rule: if y = uv, then dy/dx = u'v + uv'. Let u = x², v = sin x, so u' = 2x, v' orm if possible, but at A-Level, expanded form is fine unless the question asks otherwise.

当求导如 x² sin x 的乘积时，使用乘积法则：若 y = uv，则 dy/dx = u'v + uv'。令 u = x²，v = sin x，则 u' = 2x，v' = cos x。于是 dy/dx = 2x sin x + x² cos x。尽量将导数写成因式分解形式，但在 A-Level 考试中，展开形式也可以，除非题目另有要求。

Another common case is e^x cos x. Set u = e^x, v = cos x. Then u' = e^x, v' = –sin x, giving dy/dx = e^x cos x – e^x sin x = e^x(cos x – sin x). The product rule often yields opportunities to simplify.

另一个常见情形是 e^x cos x。令 u = e^x，v = cos x，则 u' = e^x，v' = -sin x，得到 dy/dx = e^x cos x – e^x sin x = e^x(cos x – sin x)。乘积法则常常带来化简的机会。

6. Quotient Rule with Trigonometric Functions | 三角函数的商法则

or any fraction involving sin or cos, use the quotient rule. If y = u/v, then dy/dx = (vu' – uv')/v². Example: y = (sin x)/x. Let u = sin x, v = x, so u' = cos x, v' = 1. Then dy/dx = (x cos x – sin x × 1)/x² = (x cos x – sin x)/x².

对于 y = (sin x)/x 或任何包含 sin 与 cos 的分式，使用商法则。若 y = u/v，则 dy/dx = (vu' – uv')/v²。示例：y = (sin x)/x。令 u = sin x, v = x，则 u' = cos x, v' = 1，可得 dy/dx = (x cos x – sin x × 1)/x² = (x cos x – sin x)/x²。

t rule. Since tan x = sin x / cos x, let u = sin x, v = cos x. Then dy/dx = (cos x × cos x – sin x × (– is highlights how sin and cos derivatives come together to produce other standard results.

tan x 的导数就是通过商法则推导的。因为 tan x = sin x / cos x，令 u = sin x，v = cos x，则 dy/dx = (cos x × cos x – sin x × (-sin x))/cos² x = (cos² x + sin² x)/cos² x = 1/cos² x = sec² x。这体现出 sin 与 cos 的导数如何相互结合，产生其他标准结果。

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7. Second Derivatives and Further Differentiation | 二阶导数与进一步求导

imply means differentiating again. If y = sin(3x), dy/dx = 3 cos(3x). Then d²y/dx² = 3 × (–sin(3x)) × 3 = – ice that differentiating sin and cos repeatedly cycles through four forms: sin x → cos x → –sin x → –cos x → sin x.

学会一阶导数后，求二阶导数只需再次求导。若 y = sin(3x)，dy/dx = 3 cos(3x)，则 d²y/dx² = 3 × (-sin(3x)) × 3 = -9 sin(3x)。注意到反复对 sin 与 cos 求导会在四种形式间循环：sin x → cos x → -sin x → -cos x → sin x。

For more complex functions, such as y = e^2x cos x, you might need to use the product rule twice. It can produce expressions like e^2x(3 cos x – 4 sin x) after simplification. Practice helps you recognise patterns and avoid errors.

对于更复杂的函数，例如 y = e^2x cos x，可能需要两次使用乘积法则。经过化简后，可能会得到类似 e^2x(3 cos x – 4 sin x) 的式子。多加练习有助于识别规律并避免错误。

8. Using Trigonometric Identities Before Differentiating | 求导前巧用三角恒等式

sing trig identities before differentiating. For instance, y = sin² x can be written as (1 – cos(2x))/2 using the double-angle formula. Then dy/dx = –(1/2) × (– duct-rule result: 2 sin x cos x = sin(2x). Such simplifications reduce the risk of errors.

有时先利用三角恒等式化简再求导会更简单。例如，y = sin² x 可用倍角公式写成 (1 – cos(2x))/2。那么 dy/dx = –(1/2) × (-sin(2x)) × 2 = sin(2x)。这与乘积法则的结果 2 sin x cos x = sin(2x) 一致。这样的简化能降低出错风险。

2) × 2 cos(2x) = cos(2x). Recognising these shortcuts is highly valued in exam conditions.

类似地，y = sin x cos x 可写成 (1/2) sin(2x)，求导得 (1/2) × 2 cos(2x) = cos(2x)。在考试中善用这些简便方法是备受推崇的。

9. Combining Rules: Harder Examples | 综合运用：进阶例题

+ 1). This involves the product rule and the chain rule. Let u = x³, v = sin(5x+1). Then u' = 3x², v' in(5x+1) + 5x³ cos(5x+1). Always present your final answer clearly and factorise if asked.

一道典型的 A-Level 考题可能要求求 y = x³ sin(5x + 1) 的导数。这同时涉及乘积法则和链式法则。令 u = x³, v = sin(5x+1)，则 u' = 3x²，v' = cos(5x+1) × 5。于是 dy/dx = 3x² sin(5x+1) + x³ × 5 cos(5x+1) = 3x² sin(5x+1) + 5x³ cos(5x+1)。最终答案要书写清晰，若题目要求可进行因式分解。

is y = (cos x)/(1 + sin x). Using the quotient rule with u = cos x, v = 1 + sin x gives u' = –sin x, v' = cos x. Then dy/dx = [ (1+sin x)(–sin x) – cos x(cos x) ] / (1+sin x)² = [ –sin x – sin² x – cos² x ] / (1+sin x)² = [ –sin x – (sin² x + cos² x) ] / (1+sin x)² = (–sin x – 1)/(1+sin x)² = –(1+sin x)/(1+sin x)² = –1/(1+sin x). Such simplifications show a deep understanding.

另一个高难度组合是 y = (cos x)/(1 + sin x)。使用商法则，u = cos x，v = 1 + sin x，得 u' = -sin x，v' = cos x。于是 dy/dx = [(1+sin x)(-sin x) – cos x(cos x)] / (1+sin x)² = [ -sin x – sin² x – cos² x ] / (1+sin x)² = [ -sin x – (sin² x + cos² x) ] / (1+sin x)² = (-sin x – 1)/(1+sin x)² = –(1+sin x)/(1+sin x)² = –1/(1+sin x)。这类化简体现了对知识的深刻理解。

10. Applications: Tangents, Normals, and Rates of Change | 应用：切线、法线与变化率

tiation of sin and cos is central to coordinate geometry problems. For a curve y = 2 sin x – x = π/6 is found by evaluating dy/dx = 2 cos x + 2 sin(2x). Substitute x = π/6: cos(π/6) = √3/2, sin(π/3) = √3/2, so gradient = 2(√3/2) + 2(√3/2) = 2√3. The equation of the tangent then follows using y – y₁ = m(x – x₁).

sin 和 cos 的求导在坐标几何问题中处于核心地位。对于曲线 y = 2 sin x – cos(2x)，在 x = π/6 处的梯度可通过求 dy/dx = 2 cos x + 2 sin(2x) 并代入得到。代入 x = π/6：cos(π/6) = √3/2，sin(π/3) = √3/2，故梯度为 2(√3/2) + 2(√3/2) = 2√3。接着利用 y – y₁ = m(x – x₁) 就能写出切线方程。

acement is s = 5 sin(2t), the velocity v = ds/dt = 10 cos(2t) and acceleration a = dv/dt = – (2t). Recognising these as simple harmonic motion helps in physics and mechanics contexts.

相关变化率问题常涉及 sin 和 cos。若某质点位移为 s = 5 sin(2t)，则速度 v = ds/dt = 10 cos(2t)，加速度 a = dv/dt = -20 sin(2t)。认识到这是简谐运动，有助于物理和力学背景下的题目。

11. Summary and Common Pitfalls | 总结与常见错误

The core derivatives are:

Common errors include forgetting the negative sign when differentiating cos x, missing the chain rule multiplier, and placing terms incorrectly in the quotient rule (remember it is 'bottom times derivative of top minus top times derivative of bottom, all over bottom squared').

常见错误包括：cos x 求导时漏掉负号，忘记乘以链式法则中的 du/dx，以及在商法则中分子顺序写反（请记住“分母乘以分子导数 减 分子乘以分母导数，整体除以分母的平方”）。

Always check whether the argument is simply x or a more complex function. Practise with a variety of questions until the process becomes automatic. Your confidence in differentiating trigonometric functions will pay dividends when tackling integration, differential equations, and higher-level applications.

始终检查自变量是否仅为 x，还是更复杂的函数。通过多样化的练习，直到求导过程变得自然而然。你对三角函数求导的熟练和自信，将为后续学习积分、微分方程以及更高层次的应用带来巨大回报。

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