📚 Differentiation from First Principles | 从第一原理求导
Differentiation from first principles is the foundation of calculus. It allows us to find the gradient of a curve at a point by considering the limiting value of the gradient of a chord. For Edexcel A-Level Mathematics, understanding this process is essential not only for proving standard derivatives but also for tackling exam questions that test your grasp of the underlying limit concept.
从第一原理求导是微积分的基础。它通过考虑弦的斜率的极限值来找出曲线上某一点的切线斜率。对Edexcel A-Level数学而言，理解这一过程不仅有助于证明标准导数公式，更是解答考查极限概念深层理解的考题所必备的能力。
1. What is Differentiation? | 什么是微分？
measures how a function changes as its input changes. In geometric terms, the derivative f' n the derivative is positive, the function is increasing; when negative, it is decreasing.
微分是一种衡量函数随自变量变化而变化的数学运算。从几何上看，导数 f'(x) 给出了曲线 y = f(x) 在任意点的切线斜率。导数为正时函数递增，为负时函数递减。
The instantaneous rate of change is obtained by taking the limit of the average rate of change over an interval as the interval shrinks to zero. This is precisely what 'first principles' achieves.
瞬时变化率是通过让一个区间内的平均变化率在区间趋近于零时取极限而得到的。这正是“第一原理”所实现的内容。
2. The Concept of a Limit | 极限的概念
A limit describes the value that a function approaches as the input approaches some value. When differentiating from first principles, we write limh→0 to indicate that h gets arbitrarily close to zero without actually being zero. This avoids division by zero while still capturing the behaviour near the point.
极限描述的是当自变量趋近于某个值时函数趋近的值。在第一原理求导中，我们写成 limh→0 来表示 h 无限接近于零但并不等于零。这既避免了除以零的问题，又刻画了函数在该点附近的行为。
ntal to differentiating trigonometric functions from first principles. Familiarity with such standard limits is a key skill.
例如，当 h → 0 时 (sin h)/h 的极限等于1，这个结果对于从第一原理求三角函数的导数至关重要。熟悉这些标准极限是一项关键技能。
3. The Difference Quotient | 差商
The difference quotient is the ratio (f(x+h) – f(x)) / h. It represents the gradient of the chord joining the points (x, f(x)) and (x+h, f(x+h)) on the curve. As h decreases, the chord gets shorter and its gradient approaches the tangent's gradient.
差商指的是比值 (f(x+h) – f(x)) / h。它表示连接曲线上两点 (x, f(x)) 和 (x+h, f(x+h)) 的弦的斜率。随着 h 减小，弦变短，其斜率趋近于切线的斜率。
Difference Quotient = (f(x + h) – f(x)) / h
差商 = (f(x + h) – f(x)) / h
tiable at x. For polynomials and many common functions, this limit always exists except at points of discontinuity or sharp corners.
如果该商在 h → 0 时的极限存在，我们就说函数在 x 处可导。对于多项式和许多常见函数，除非在不连续点或尖点处，该极限总是存在的。
4. General Formula for First Principles | 第一原理通式
The derivative of a function f(x) from first principles is defined as:
函数 f(x) 从第一原理求导的定义为：
f '(x) = limh→0 [ f(x + h) – f(x) ] / h
f '(x) = limh→0 [ f(x + h) – f(x) ] / h
then taking the limit h → 0. The simplification step usually involves cancelling h from the denominator.
这个公式是一切微分运算的万能钥匙。应用它时，先将具体的 f(x) 代入分子，进行代数化简，然后取 h → 0 的极限。化简步骤通常需要约去分母中的 h。
In Edexcel exams you may be asked to prove the derivative of xⁿ for small integer n, or for trigonometric functions sin x and cos x, using this limit definition. The working must be shown step by step.
在Edexcel考试中，你可能会被要求用该极限定义证明 xⁿ（n为小整数）或三角函数 sin x 和 cos x 的导数。过程必须逐步展示。
5. Differentiating f(x) = x² | 求导 f(x) = x²
f(x) = x². Then f(x+h) = (x+h)² = x² + 2xh + h². The difference quotient is (x² + 2xh + h² – x²) / h = (2xh + h²) / h = 2x + h. Taking the limit as h → 0 gives f '(x) = 2x.
设 f(x) = x²。那么 f(x+h) = (x+h)² = x² + 2xh + h²。差商为 (x² + 2xh + h² – x²) / h = (2xh + h²) / h = 2x + h。当 h → 0 时取极限，得 f '(x) = 2x。
This result confirms the power rule for this case. The algebra is straightforward, and the key step is cancelling a factor of h after expanding the brackets. Always show the expansion clearly.
这一结果验证了该情形下的幂法则。代数运算十分直接，关键一步是在展开括号后约去公因子 h。务必清晰展示展开过程。
6. Differentiating f(x) = x³ | 求导 f(x) = x³
) = (x+h)³ = x³ + 3x²h + 3xh² + h³. The difference quotient becomes (x³ + 3x²h + 3xh² + h³ – x²h + 3xh² + h³) / h = 3x² + 3xh + h². As h → 0, the terms containing h vanish, yielding f '(x) = 3x².
对于 f(x) = x³，我们计算 f(x+h) = (x+h)³ = x³ + 3x²h + 3xh² + h³。差商变为 (x³ + 3x²h + 3xh² + h³ – x³) / h = (3x²h + 3xh² + h³) / h = 3x² + 3xh + h²。当 h → 0 时，含 h 的项消失，得到 f '(x) = 3x²。
xⁿ⁻¹. Proving the general case requires the binomial theorem, but the method is identical.
注意出现的模式：x² 的导数是 2x，x³ 的导数是 3x²。一般地，对于 f(x) = xⁿ，导数为 nxⁿ⁻¹。证明一般情形需要用到二项式定理，但方法完全相同。
7. Differentiating f(x) = kx (Linear Functions) | 求导 f(x) = kx (线性函数)
where k is a constant. Then f(x+h) = k(x+h) = kx + kh. The difference quotient is (kx + kh – kx) / h = kh / h = k. Hence the limit as h → 0 is simply k, so f '(x) = k.
考虑 f(x) = kx，其中 k 为常数。那么 f(x+h) = k(x+h) = kx + kh。差商为 (kx + kh – kx) / h = kh / h = k。因此 h → 0 时的极限就是 k，因此 f '(x) = k。
This shows that the gradient of a straight line is constant. The first principles approach confirms what we already know: the derivative of a linear function is its slope.
这表明一条直线的斜率是常数。第一原理的方法证实了我们已经知道的事实：线性函数的导数就是其斜率。
8. Differentiating f(x) = c (Constant Function) | 求导常数函数
If f(x) = c, where c is a constant, then f(x+h) = c as well. The difference quotient is (c – h = 0 / h = 0. As h → 0, the limit remains 0. Thus the derivative of any constant is zero.
若 f(x) = c，c 为常数，那么 f(x+h) 也等于 c。差商为 (c – c) / h = 0 / h = 0。当 h → 0 时，极限仍为 0。因此任何常数的导数都是零。
Geometrically, a constant function is a horizontal line with zero slope. The first principles argument is a simple but rigorous justification of that fact.
从几何上看，常数函数的图像是一条斜率为零的水平线。第一原理的论证为这一事实提供了简洁而严谨的证明。
9. Trigonometric Functions: f(x) = sin x | 三角函数：f(x) = sin x 的导数
cos B + cos A sin B. Set f(x) = sin x. Then f(x+h) = sin(x+h) = sin x cos h + cos x sin h.
从第一原理证明 sin x 的导数需要用到和角公式 sin(A+B) = sin A cos B + cos A sin B。令 f(x) = sin x，则 f(x+h) = sin(x+h) = sin x cos h + cos x sin h。
The difference quotient is [sin x cos h + cos x sin h – sin x] / h. Rearranges to sin x (cos h – 1)/h + cos x (sin h)/h. Taking limit h → 0, we use lim (sin h)/h = 1 and lim (cos h – 1)/h = 0, yielding f '(x) = cos x.
差商为 [sin x cos h + cos x sin h – sin x] / h。整理得 sin x (cos h – 1)/h + cos x (sin h)/h。取 h → 0 时的极限，利用 lim (sin h)/h = 1 和 lim (cos h – 1)/h = 0，得到 f '(x) = cos x。
f(x) = sin x => f '(x) = cos x
f(x) = sin x => f '(x) = cos x
10. The Derivative of cos x from First Principles | 从第一原理求 cos x 的导数
Let f(x) = cos x. Using cos(x+h) = cos x cos h – sin x sin h, the difference quotient is [cos x cos h – sin x sin h – cos x] / h = cos x (cos h – 1)/h – in x (sin h)/h. Applying the same standard limits, the cos x term vanishes and we obtain f '(x) = –sin x.
设 f(x) = cos x。利用 cos(x+h) = cos x cos h – sin x sin h，差商为 [cos x cos h – sin x sin h – cos x] / h = cos x (cos h – 1)/h – sin x (sin h)/h。应用相同的标准极限，含 cos x 的项消失，得到 f '(x) = –sin x。
Note the negative sign. This is a common source of error. The derivative of cos x is –sin x, not sin x. Remembering this distinction is crucial for subsequent topics such as integration and kinematic equations.
注意负号。这是一个常见的错误来源。cos x 的导数是 –sin x，而不是 sin x。记住这一区别对后续的积分和运动学方程等内容至关重要。
11. Common Mistakes and Exam Tips | 常见错误与应试技巧
One frequent mistake is forgetting to expand brackets fully or mishandling the algebra in the difference quotient. Always write f(x+h) carefully and subtract f(x) before dividing by h. Check that after simplification the denominator cancels.
一个常见错误是忘记完全展开括号，或在处理差商的代数运算时出错。务必仔细写出 f(x+h)，先减去 f(x) 再除以 h。确保化简后分母被约去。
Another pitfall is misapplying limits, especially (cos h – hink it tends to 1. Practise the derivation of both sin x and cos x until the steps flow naturally. Edexcel examiners expect clear logical flow and correct limit statements.
另一个陷阱是错误使用极限，尤其是 (cos h – 1)/h → 0。有些学生错误地认为它趋近于1。反复练习 sin x 和 cos x 的推导，直到步骤流畅自然。Edexcel考官希望看到清晰的逻辑和正确的极限陈述。
Even if you make an arithmetic slip later, you can secure method marks. Also, never write 'h = 0' when taking the limit; always use the limit notation correctly.
在考试中，分数通常分配给初始的表达建立、展开、化简以及最后取极限的过程。即使最后出现计算失误，仍可获得方法分。此外，取极限时千万不要写“h = 0”，务必正确使用极限符号。
12. Summary and Further Practice | 总结与练习
The first principles method defines the derivative as the limit of a difference quotient. The key results you must be able to prove are:
第一原理方法将导数定义为差商的极限。你必须能够证明的关键结论如下：
- f(x) = x² gives f '(x) = 2x | f(x) = x² 得 f '(x) = 2x
- f(x) = x³ gives f '(x) = 3x² | f(x) = x³ 得 f '(x) = 3x²
- f(x) = sin x gives f '(x) = cos x | f(x) = sin x 得 f '(x) = cos x
- f(x) = cos x gives f '(x) = –sin x | f(x) = cos x 得 f '(x) = –sin x
Use the following table to consolidate the standard derivatives obtained from first principles:
用下表巩固从第一原理得出的标准导数：
| Function f(x) | Derivative f '(x) | 函数 f(x) | 导数 f '(x) |
|---|---|---|---|
| c (constant) | 0 | c (常数) | 0 |
| kx | k | kx | k |
| x² | 2x | x² | 2x |
| x³ | 3x² | x³ | 3x² |
| sin x | cos x | sin x | cos x |
| cos x | –sin x | cos x | –sin x |
Mastering differentiation from first principles gives you deep insight into why the rules work, not just how to apply them. Practise with a range of polynomial and trigonometric functions, and always include the limit statement in your final answer.
掌握第一原理求导法能让你深入理解这些法则为什么成立，而不仅仅是知道怎么用。请用各种多项式和三角函数进行练习，并在最终答案中始终包含极限表达式。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图