📚 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
For example, if 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)
For example, consider y = 5x³ − 2x + 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
For 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。
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。
At x = 1, y = 1 − 3 + 4 = 2, 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,
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