Differentiating Functions with Two or More Terms | 对含两项或更多项的函数求导

📚 Differentiating Functions with Two or More Terms | 对含两项或更多项的函数求导

In Edexcel A-Level Mathematics, differentiation is extended from single powers of x to functions that contain two or more terms. The key idea is that differentiation is a linear operation, so you can differentiate a sum or difference term by term. This article covers the essential rules and exam techniques for differentiating functions with several terms.

在爱德思 A-Level 数学中，求导从单个 x 的幂扩展到含有两项或更多项的函数。核心思想是求导是一种线性运算，因此你可以逐项对和或差求导。本文涵盖对含多个项的函数求导的基本规则与考试技巧。

---

1. The Sum Rule: Term-by-Term Differentiation | 和法则：逐项求导

If a function is written as a sum of two or more terms, you can differentiate each term separately and then add the results.

如果一个函数写成两项或更多项的和，你可以分别对每一项求导，然后把结果相加。

If y = u(x) + v(x), then dy/dx = du/dx + dv/dx

This is called the sum rule. It is the foundation for differentiating polynomials and many other algebraic functions.

这称为和法则。它是多项式以及许多其他代数函数求导的基础。

• Example: y = x⁵ + x³
• Differentiate term by term: dy/dx = 5x⁴ + 3x²

例如：y = x⁵ + x³，逐项求导得 dy/dx = 5x⁴ + 3x²。

---

2. The Difference Rule | 差法则：逐项相减求导

When a function is written as a difference of two or more terms, differentiate each term separately and keep the subtraction signs in place.

当一个函数写成两项或更多项的差时，分别对每一项求导，并保留减号。

If y = u(x) − v(x), then dy/dx = du/dx − dv/dx

y = x⁴ − 3x², then the derivative of x⁴ is 4x³ and the derivative of 3x² is 6x. Therefore:

例如，若 y = x⁴ − 3x²，则 x⁴ 的导数为 4x³，3x² 的导数为 6x。因此：

dy/dx = 4x³ − 6x

A common mistake is to change a minus sign into a plus sign. Always carry the original sign with each term.

一个常见错误是把减号变成加号。务必让每一项保留原来的符号。

---

3. The Constant Multiple Rule | 常数倍法则

If a term has a constant coefficient, the constant multiples the derivative of the variable part.

如果某一项有常数系数，该常数乘以变量部分的导数。

d/dx [a f(x)] = a f'(x)

+ 7. The term 5x³ differentiates to 5 × 3x² = 15x², and −2x differentiates to −2 × 1 = −2.

例如，考虑 y = 5x³ − 2x + 7。项 5x³ 求导得 5 × 3x² = 15x²，而 −2x 求导得 −2 × 1 = −2。

dy/dx = 15x² − 2

The constant multiplier does not need to be differentiated and should simply remain in front.

常数倍不需要单独求导，只需保留在前面。

---

4. Constant Terms Differentiate to Zero | 常数项求导为零

The derivative of any constant term is always zero. A constant does not change as x changes, so its rate of change is zero.

任何常数项的导数始终为零。常数不随 x 变化，因此其变化率为零。

d/dx [c] = 0

r instance, if y = 4x² + 3, then the derivative of 4x² is 8x and the derivative of 3 is 0.

例如，若 y = 4x² + 3，则 4x² 的导数为 8x，3 的导数为 0。

dy/dx = 8x

In exam work, do not leave a constant such as +3 in the derivative. It must disappear.

在考试中，不要在导数中保留 +3 这样的常数。它必须消失。

---

5. Rewriting Terms with Negative and Fractional Powers | 负指数和分数指数的改写

Many functions are not written as simple powers of x, so they must be rewritten first using the laws of indices.

许多函数并不是简单的 x 的幂，因此必须先利用指数法则改写。

• 1/x = x⁻¹
• 1/x² = x⁻²
• √x = x^½

例如：1/x = x⁻¹，1/x² = x⁻²，√x = x^½。

Consider y = √x + 1/x². Rewrite it as:

考虑 y = √x + 1/x²。将其改写为：

y = x^½ + x⁻²

Now differentiate term by term:

现在逐项求导：

dy/dx = ½ x⁻½ − 2x⁻³

You can leave the answer in this form, or rewrite it as 1/(2√x) − 2/x³ if the question asks for a simplified form.

你可以保留这种形式，如果题目要求化简，也可以写成 1/(2√x) − 2/x³。

---

广告 / Sponsored

6. Differentiating a Full Polynomial | 多项式的整体求导

To differentiate a polynomial with several terms, apply the power rule to each term: multiply by the exponent and reduce the exponent by 1.

要对含多个项的多项式求导，对每一项应用幂法则：乘以指数，再把指数减 1。

d/dx [a xⁿ] = a n xⁿ⁻¹

Worked example: differentiate y = 2x⁴ − 5x³ + 3x² − 7x + 9.

例题：求 y = 2x⁴ − 5x³ + 3x² − 7x + 9 的导数。

Term	Derivative
2x⁴	8x³
−5x³	−15x²
3x²	6x
−7x	−7
9	0

Therefore the derivative is:

因此导数为：

dy/dx = 8x³ − 15x² + 6x − 7

Always check that the power of each term has been reduced by exactly 1.

务必检查每一项的指数是否恰好减少了 1。

---

7. Finding the Gradient at a Point | 求曲线上一点的梯度

The first derivative dy/dx is the gradient function. To find the gradient of a curve at a specific point, substitute the x-coordinate into dy/dx.

一阶导数 dy/dx 是梯度函数。要求曲线上某一点的梯度，把该点的 x 坐标代入 dy/dx。

Example: find the gradient of y = 2x³ − 9x² + 12x + 1 at x = 2.

例题：求 y = 2x³ − 9x² + 12x + 1 在 x = 2 处的梯度。

First differentiate:

先求导：

dy/dx = 6x² − 18x + 12

Now substitute x = 2:

现在代入 x = 2：

dy/dx = 6(2)² − 18(2) + 12 = 24 − 36 + 12 = 0

A gradient of zero means the point is a stationary point. This can be a maximum, minimum, or point of inflection.

梯度为零意味着该点是一个驻点。它可能是极大值、极小值或拐点。

---

8. Equations of Tangents and Normals | 切线与法线方程

Once you know the gradient at a point, you can find the equation of the tangent and the normal.

一旦知道某一点的梯度，就可以求出切线和法线的方程。

Tangent: y − y₁ = m(x − x₁)

Normal: y − y₁ = (−1/m)(x − x₁)

Example: find the tangent and normal to y = x² − 3x + 4 at x = 1.

例题：求 y = x² − 3x + 4 在 x = 1 处的切线和法线方程。

so the point is (1, 2). The derivative is dy/dx = 2x − 3, so at x = 1 the gradient is −1.

在 x = 1 处，y = 1 − 3 + 4 = 2，因此点为 (1, 2)。导数为 dy/dx = 2x − 3，因此在 x = 1 处梯度为 −1。

Tangent:

切线：

y − 2 = −1(x − 1)

y = −x + 3

Normal gradient is 1:

法线梯度为 1：

y − 2 = 1(x − 1)

y = x + 1

The tangent and normal are perpendicular, so their gradients multiply to −1.

切线与法线互相垂直，因此它们的梯度乘积为 −1。

---

9. The Second Derivative | 二阶导数

Differentiating dy/dx again gives the second derivative,

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）