IB Math: The Sum Rule and Constant Multiple Rule | IB数学：和法则与常数倍法则

📚 IB Math: The Sum Rule and Constant Multiple Rule | IB数学：和法则与常数倍法则

Differentiation is one of the most powerful tools in IB Mathematics, appearing in both Analysis and Approaches (AA) and Applications and Interpretation (AI) courses. Before you can tackle product rule, quotient rule, or the chain rule, you must first master two elementary rules: the constant multiple rule and the sum rule. These rules allow you to differentiate functions term by term, which is the foundation of nearly every derivative you will ever compute.

微分是IB数学中最强大的工具之一，在分析与方法（AA）以及应用与解释（AI）课程中都会出现。在掌握积法则、商法则或链式法则之前，你必须先掌握两个基本规则：常数倍法则与和法则。这两个规则允许你逐项求导，这几乎是你将来计算所有导数的基础。

1. The Constant Multiple Rule | 常数倍法则

The constant multiple rule states that if a function is multiplied by a constant, its derivative is the constant multiplied by the derivative of the original function. In symbols, if c is a constant and f(x) is differentiable, then:

常数倍法则指出：如果一个函数乘以一个常数，那么它的导数等于这个常数乘以原函数的导数。用符号表示，若 c 为常数且 f(x) 可导，则：

d/dx [c·f(x)] = c · f'(x)

This rule reflects the fact that differentiation is a linear operation. It is intuitively reasonable: multiplying a function by 2 stretches the graph vertically by a factor of 2, and every tangent slope is also stretched by a factor of 2.

这条规则反映了微分是一种线性运算。从直观上很容易理解：将函数乘以 2，图像在竖直方向上被拉伸为原来的 2 倍，因此每一条切线的斜率也被拉伸为原来的 2 倍。

For example, if y = 5x², then dy/dx = 5 · (2x) = 10x. The constant 5 is simply carried along; we do not differentiate the 5. This is a common point of confusion for students who mistakenly try to apply the product rule to a constant times a function.

例如，若 y = 5x²，则 dy/dx = 5 · (2x) = 10x。常数 5 仅需原样保留，我们不会对 5 求导。这是学生中常见的困惑点，他们常常错误地把常数乘以函数的情况套用积法则。

Important: the constant multiple rule works for any constant, including negative constants and fractions. For example, d/dx [−3 sin x] = −3 cos x, and d/dx [(1/2)eˣ] = (1/2)eˣ.

重要提示：常数倍法则适用于任何常数，包括负数常数和分数。例如，d/dx [−3 sin x] = −3 cos x，以及 d/dx [(1/2)eˣ] = (1/2)eˣ。

2. The Sum Rule | 和法则

The sum rule states that the derivative of a sum of two or more functions is the sum of their derivatives. If f(x) and g(x) are both differentiable, then:

和法则指出：两个或多个函数之和的导数等于它们各自导数之和。若 f(x) 和 g(x) 都可导，则：

d/dx [f(x) + g(x)] = f'(x) + g'(x)

This rule is also a consequence of the limit definition of the derivative. When you compute the difference quotient for f(x) + g(x), the limit can be split into two separate difference quotients because the limit of a sum equals the sum of the limits (provided both limits exist).

这条规则同样是导数极限定义的结果。当你计算 f(x) + g(x) 的差商时，可以将极限拆分为两个独立的差商，因为和的极限等于极限的和（前提是两个极限都存在）。

For example, if y = x³ + 4x, then dy/dx = 3x² + 4. The derivative of x³ is 3x², the derivative of 4x is 4, and we simply add them together. No special cross-terms appear.

例如，若 y = x³ + 4x，则 dy/dx = 3x² + 4。x³ 的导数是 3x²，4x 的导数是 4，我们只需把它们相加。不会出现交叉项。

The sum rule also extends naturally to more than two functions. If y = f(x) + g(x) + h(x), then y’ = f'(x) + g'(x) + h'(x). You can differentiate a whole polynomial term by term without any additional complexity.

和法则还可以自然推广到两个以上的函数。若 y = f(x) + g(x) + h(x)，则 y’ = f'(x) + g'(x) + h'(x)。你可以逐项对整条多项式求导，而不会增加额外的复杂度。

3. The Difference Rule | 差法则

The difference rule is the sum rule applied to a subtraction. Because subtraction can be written as adding a negative, the derivative of a difference is the difference of the derivatives:

差法则可以看作和法则在减法情形下的应用。因为减法可以写成加上一个负数，所以差的导数等于导数的差：

d/dx [f(x) − g(x)] = f'(x) − g'(x)

In practice, the difference rule is often combined with the constant multiple rule. For example, to differentiate y = 3x⁴ − 2x², you find 12x³ for the first term and 4x for the second term, then subtract to get y’ = 12x³ − 4x.

在实际操作中，差法则常与常数倍法则结合。例如，对 y = 3x⁴ − 2x² 求导，第一项得到 12x³，第二项得到 4x，相减得到 y’ = 12x³ − 4x。

It is helpful to remember that every function you meet in IB Mathematics that is a sum or difference of power functions, exponential, trigonometric, or logarithmic functions can be differentiated by applying these two rules along with the known derivatives of basic functions.

请记住：在IB数学中遇到的所有幂函数、指数函数、三角函数或对数函数的和与差，都可以通过运用这两条规则以及基本函数的已知导数来求导。

4. Combining with the Power Rule | 与幂法则结合

For standard IB questions, you will frequently combine the sum rule, the constant multiple rule, and the power rule. The power rule states that if y = xⁿ, then dy/dx = n·xⁿ⁻¹. The rules together give:

对于标准的IB考题，你常常需要结合和法则、常数倍法则和幂法则。幂法则指出：若 y = xⁿ，则 dy/dx = n·xⁿ⁻¹。这些规则结合起来给出：

d/dx [a·xⁿ + b·xᵐ] = a·n·xⁿ⁻¹ + b·m·xᵐ⁻¹

Notice that each term is differentiated independently: multiply by the exponent, reduce the exponent by one, and keep the constant coefficient intact before the multiplication.

注意每一项都独立求导：乘以指数，指数减一，并保持常数系数在乘法前不变。

For example, differentiate y = 2x⁵ − 7x³ + 4x − 9. Term by term:

例如，求 y = 2x⁵ − 7x³ + 4x − 9 的导数。逐项来看：

Derivative of 2x⁵ is 2 × 5x⁴ = 10x⁴.
Derivative of −7x³ is −7 × 3x² = −21x².
Derivative of 4x is 4.
Derivative of −9 is 0.

2x⁵ 的导数是 2 × 5x⁴ = 10x⁴。
−7x³ 的导数是 −7 × 3x² = −21x²。
4x 的导数是 4。
−9 的导数是 0。

Therefore y’ = 10x⁴ − 21x² + 4. The constant term disappears because the derivative of a constant is always zero.

因此 y’ = 10x⁴ − 21x² + 4。常数项消失，因为常数的导数总是零。

5. Working with Negative and Fractional Exponents | 负指数与分数指数

In IB Mathematics, especially in the Analysis and Approaches SL and HL courses, you must be comfortable differentiating expressions with negative and fractional exponents. The power rule works for all real exponents, not just positive integers.

在IB数学中，尤其是分析与方法SL和HL课程，你必须熟悉对负指数和分数指数的表达式求导。幂法则适用于所有实数指数，而不仅仅是正整数。

For instance, to differentiate y = 1/x², first rewrite it as y = x⁻². Then apply the power rule together with the constant multiple rule:

例如，要对 y = 1/x² 求导，先将其改写为 y = x⁻²。然后结合幂法则和常数倍法则：

dy/dx = −2x⁻³ = −2/x³

Similarly, if y = 3/√x, rewrite as y = 3x^(−1/2). Then dy/dx = 3 × (−1/2) x^(−3/2) = −3/(2x^{3/2}).

类似地，若 y = 3/√x，改写为 y = 3x^(−1/2)。则 dy/dx = 3 × (−1/2) x^(−3/2) = −3/(2x^{3/2})。

Always simplify roots into fractional powers before differentiating. This makes the sum rule and constant multiple rule far easier to apply accurately and reduces the chance of arithmetic mistakes.

在求导之前，一定要先把根式化为分数指数幂。这样能使和法则与常数倍法则应用得更准确，并减少计算失误的机会。

6. Trigonometric, Exponential, and Logarithmic Terms | 三角函数、指数函数与对数项

The sum rule and constant multiple rule are not limited to polynomials. They apply to any differentiable functions, including trigonometric, exponential, and logarithmic functions.

和法则与常数倍法则不仅适用于多项式。它们适用于任何可导函数，包括三角函数、指数函数和对数函数。

Here are the essential derivative formulas you should memorize for IB:

以下是IB中你需要牢记的基本导数公式：

d/dx [sin x] = cos x
d/dx [cos x] = −sin x
d/dx [eˣ] = eˣ
d/dx [ln x] = 1/x

d/dx [sin x] = cos x
d/dx [cos x] = −sin x
d/dx [eˣ] = eˣ
d/dx [ln x] = 1/x

For example, differentiate y = 4 sin x + 3eˣ. The derivative is y’ = 4 cos x + 3eˣ. The constant 4 multiplies cos x, and the constant 3 multiplies eˣ; the sum rule lets us write the two derivatives side by side.

例如，求 y = 4 sin x + 3eˣ 的导数。结果为 y’ = 4 cos x + 3eˣ。常数 4 乘以 cos x，常数 3 乘以 eˣ；和法则允许我们把两个导数并列写出。

7. Worked Examples: Polynomial and Radical Functions | 例题：多项式与根式函数

Let us work through a complete example that mirrors a typical IB question. Differentiate the function f(x) = 2x³ − 5x² + 7x − 11.

让我们完整地完成一个与典型IB题目相似的例子。求函数 f(x) = 2x³ − 5x² + 7x − 11 的导数。

Apply the constant multiple rule to each term:

对每一项应用常数倍法则：

f'(x) = 2(3x²) − 5(2x) + 7(1) − 0 = 6x² − 10x + 7

This answer is fully simplified. Notice that we did not differentiate the 2, the −5, or the 7 themselves; they are constants attached to each function term.

这个答案已经完整化简。注意我们没有对 2、−5 或 7 本身求导；它们是附加在各项上的常数。

Now consider g(x) = 3x² + 4√x. Rewrite √x as x^(1/2), so g(x) = 3x² + 4x^(1/2). Then:

现在考虑 g(x) = 3x² + 4√x。将 √x 改写为 x^(1/2)，于是 g(x) = 3x² + 4x^(1/2)。那么：

g'(x) = 6x + 4 × (1/2) x^(−1/2) = 6x + 2/√x

In the second term, the constant multiple rule requires us to multiply 4 by the derivative of x^(1/2), which is (1/2)x^(−1/2), giving 2x^(−1/2).

在第二项中，常数倍法则要求我们将 4 乘以 x^(1/2) 的导数 (1/2)x^(−1/2)，得到 2x^(−1/2)。

8. Worked Examples: Rational and Exponential Functions | 例题：有理函数与指数函数

When you encounter a rational function such as y = (x² + 1)/x, you should simplify it algebraically before differentiating. Divide each term in the numerator by x:

当你遇到有理函数如 y = (x² + 1)/x 时，应先进行代数化简再求导。将分子中的每一项除以 x：

y = x + 1/x = x + x⁻¹

Then differentiate using the sum rule and power rule:

然后利用和法则与幂法则求导：

dy/dx = 1 − x⁻² = 1 − 1/x²

If you had tried to use the quotient rule without simplifying, you would obtain the same answer, but simplification reduces the risk of algebra errors and is often expected in IB mark schemes.

如果你不化简而直接使用商法则，也能得到相同的答案，但化简可以降低代数错误的风险，并且往往是IB评分方案所期望的做法。

For an exponential function, consider h(x) = 7eˣ − 2x⁴. The derivative is h'(x) = 7eˣ − 8x³. Since the derivative of eˣ is eˣ itself, the constant 7 simply carries along.

对于一个指数函数，考虑 h(x) = 7eˣ − 2x⁴。其导数为 h'(x) = 7eˣ − 8x³。由于 eˣ 的导数仍是 eˣ 本身，常数 7 只是原样保留。

9. Common Mistakes to Avoid | 常见错误

IB examiners consistently report that students lose marks on differentiation questions because of small but avoidable errors. The most frequent mistake related to these rules is incorrectly differentiating the constant itself.

IB考官常报告说，学生在求导题目中因为一些微小但可避免的错误而丢分。与这两条规则最相关的常见错误是对常数本身进行求导。

For example, when differentiating y = 6x³, some students write dy/dx = 6 × 3x² × 0? This is wrong. The derivative of 6x³ is simply 6 × 3x² = 18x². The derivative of the constant 6 inside a product is not taken when using the constant multiple rule.

例如，当对 y = 6x³ 求导时，有些学生会写出 dy/dx = 6 × 3x² × 0？这是错误的。6x³ 的导数就是 6 × 3x² = 18x²。在使用常数倍法则时，不需要对乘积中的常数6求导。

Another common error is forgetting to differentiate the constant term in a polynomial. If y = x² + 5, then dy/dx = 2x, not 2x + 5. Since the derivative of 5 is zero, it disappears.

另一个常见错误是忘记对多项式中的常数项求导。若 y = x² + 5，则 dy/dx = 2x，而不是 2x + 5。因为 5 的导数是零，所以它消失了。

A third error is misapplying the sum rule to products or quotients. The sum rule cannot be used on f(x)·g(x) or f(x)/g(x). For products, you need the product rule; for quotients, you need the quotient rule. Only sums and differences allow term-by-term differentiation.

第三个错误是将和法则错误地应用于积或商。和法则不能用于 f(x)·g(x) 或 f(x)/g(x)。对于积，你需要积法则；对于商，你需要商法则。只有和与差才允许逐项求导。

10. Interpreting the Derivative in Context | 结合情境理解导数

In IB Applications and Interpretation (AI), differentiation is often used to model rates of change in real-world contexts. The sum rule and constant multiple rule let you differentiate composite models quickly.

在IB应用与解释（AI）课程中，导数常被用来建立现实情境中的变化率模型。和法则与常数倍法则让你能够快速求导复合模型。

For example, suppose a company’s profit in thousands of dollars is modeled by P(t) = 50t − 3t², where t is time in years. The marginal profit is P'(t) = 50 − 6t. This is obtained by differentiating each term separately.

例如，假设一家公司的利润（以千美元计）由 P(t) = 50t − 3t² 建模，其中 t 是以年为单位的时间。边际利润为 P'(t) = 50 − 6t。这正是通过逐项求导得到的。

If the company also receives a fixed annual grant of 20 thousand dollars, the profit model becomes P(t) = 50t − 3t² + 20. The derivative is still P'(t) = 50 − 6t because the derivative of the constant 20 is zero. This shows that fixed costs or fixed incomes do not affect the rate of change.

如果公司还获得每年2万美元的固定补助，利润模型变为 P(t) = 50t − 3t² + 20。导数仍然是 P'(t) = 50 − 6t，因为常数 20 的导数是零。这显示固定成本或固定收入不会影响变化率。

11. Exam-Style Practice Questions | 模拟考试练习题

Try these IB-style questions before checking the answers below. Each question requires the sum rule or the constant multiple rule, often combined with the power rule.

在查看下面的答案之前，请先尝试这些IB风格的题目。每题都需要用到和法则或常数倍法则，并常常结合幂法则。

Question 1 Differentiate f(x) = 4x³ − 7x + 2.

问题1 求 f(x) = 4x³ − 7x + 2 的导数。

Question 2 Differentiate g(x) = 5 sin x + 3 cos x.

问题2 求 g(x) = 5 sin x + 3 cos x 的导数。

Question 3 If y = (2x² + 3x)/x, find dy/dx and simplify.

问题3 若 y = (2x² + 3x)/x，求 dy/dx 并化简。

Solutions:

解答：

Question 1: f'(x) = 12x² − 7.
Question 2: g'(x) = 5 cos x − 3 sin x.
Question 3: First simplify y = 2x + 3, so dy/dx = 2.

问题1：f'(x) = 12x² − 7。
问题2：g'(x) = 5 cos x − 3 sin x。
问题3：先化简 y = 2x + 3，所以 dy/dx = 2。

12. Summary and Final Tips | 总结与要点

The constant multiple rule and the sum rule are the building blocks of differentiation in IB Mathematics. Together they enable you to differentiate any function that is a sum or difference of simpler known functions, term by term.

常数倍法则和和法则是IB数学中微分运算的基石。它们共同使你能够对任何可写成若干已知函数之和或差的函数进行逐项求导。

Remember three core facts: first, constants are carried along, not differentiated; second, the derivative of a sum is the sum of derivatives; third, always simplify expressions such as fractions or radicals before differentiating when possible.

请记住三个核心要点：第一，常数只保留，不求导；第二，和的导数等于导数的和；第三，在求导前应尽可能先化简表达式，例如分数或根式。

Practice these rules until they feel automatic. In the IB exam, quick and accurate differentiation frees up time for the more complex techniques such as the chain rule, product rule, and integration by parts. Mastering these simple rules is not just about getting easy marks; it is about building the fluency you need for every later topic in calculus.

请反复练习这些规则，直到它们成为本能。在IB考试中，快速而准确的求导能为更复杂的技巧如链式法则、积法则和分部积分法腾出时间。掌握这些简单规则不仅是为了轻松得分，更是为了建立起你在之后所有微积分主题中所需要的流畅运算能力。

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IB Math: The Sum Rule and Constant Multiple Rule | IB数学：和法则与常数倍法则