📚 Exercise 22B: Mastering Differentiation Techniques | 练习22B：掌握微分技巧
Exercise 22B in the IB Mathematics curriculum typically consolidates core differentiation skills: from basic power rule to combined chain, product, and quotient rules. This article unpacks every sub-skill, provides worked reasoning, and highlights common pitfalls so that you can tackle any question in the exercise with confidence.
IB 数学课程中的练习 22B 通常用于巩固核心求导技能：从基本幂函数求导到复合的链式法则、积法则和商法则。本文拆解每一项子技能，给出推导思路，并指出常见误区，帮助你自信应对本次练习中的所有题目。
1. The Power Rule Foundation | 幂函数求导基础
The power rule states that if f(x) = xⁿ, then f'(x) = n xⁿ⁻¹. This holds for any real constant n, not just integers. Before tackling Exercise 22B, ensure you can apply it instantly to expressions like x⁵, x⁻², and √x rewritten as x½.
幂函数求导法则指出，若 f(x) = xⁿ，则 f'(x) = n xⁿ⁻¹。这适用于任意实数常数 n，而不仅仅是整数。在着手练习 22B 之前，务必能瞬间处理 x⁵、x⁻² 以及重写为 x½ 的 √x 等表达式。
- d/dx (x⁷) = 7x⁶
- d/dx (x⁻³) = -3x⁻⁴
- d/dx (√x) = d/dx (x½) = ½ x⁻½ = 1/(2√x)
A frequent mistake is forgetting to reduce the exponent properly when the function is written in radical form. Always rewrite roots as fractional exponents first.
常见错误是在根式形式下忘记正确降低指数。务必先将根式写成分数指数形式再求导。
2. Sum, Difference, and Constant Multiple Rules | 和、差与常数倍法则
Differentiation is linear: the derivative of a sum is the sum of the derivatives, and constants can be factored out. If h(x) = a·f(x) ± b·g(x), then h'(x) = a·f'(x) ± b·g'(x). In Exercise 22B, you often differentiate polynomials term by term.
求导是线性运算：和的导数为导数之和，常数可以提出。若 h(x) = a·f(x) ± b·g(x)，则 h'(x) = a·f'(x) ± b·g'(x)。在练习 22B 中，你经常需要逐项对多项式求导。
For example, differentiate f(x) = 4x³ − 7x² + 2x − 9. The derivative is f'(x) = 12x² − 14x + 2. Notice the constant term −9 disappears because the derivative of a constant is zero.
例如，求 f(x) = 4x³ − 7x² + 2x − 9 的导数。导数为 f'(x) = 12x² − 14x + 2。注意常数项 −9 消失，因为常数的导数为零。
3. Product Rule for Products of Functions | 函数乘积的积法则
When two differentiable functions u(x) and v(x) are multiplied, the product rule must be used: (uv)’ = u’v + uv’. Exercise 22B often includes products like (x²+1)(x³−2) or mixed polynomial-trigonometric forms.
当两个可导函数 u(x) 和 v(x) 相乘时，必须使用积法则：(uv)’ = u’v + uv’。练习 22B 中常出现 (x²+1)(x³−2) 或多项式与三角函数的混合乘积。
The choice of u and v can affect the amount of simplification needed. Usually, let u be the polynomial factor if it leads to simpler derivatives, but the rule works symmetrically. Always expand only after differentiating if the question allows.
u 和 v 的选择会影响后续化简的工作量。通常选多项式作为 u 会得到较简单的导数，但法则是对称的。若题目允许，最好在求导后再展开。
Example: differentiate h(x) = (3x²)(sin x). Let u = 3x², v = sin x. Then u’ = 6x, v’ = cos x. So h'(x) = 6x sin x + 3x² cos x.
例：求 h(x) = (3x²)(sin x) 的导数。令 u = 3x²，v = sin x，则 u’ = 6x，v’ = cos x。因此 h'(x) = 6x sin x + 3x² cos x。
4. Quotient Rule for Rational Expressions | 有理式的商法则
The quotient rule is essential when differentiating f(x) = u(x)/v(x). The formula is (u/v)’ = (u’v − uv’) / v². A common error is reversing the subtraction in the numerator, producing the wrong sign.
当求形如 f(x) = u(x)/v(x) 的导数时，商法则必不可少。公式为 (u/v)’ = (u’v − uv’) / v²。常见错误是颠倒分子中的减法顺序，导致符号出错。
In Exercise 22B, you might encounter rational functions such as (x²+1)/(x−2) or trigonometric quotients. Always identify u as the numerator and v as the denominator, differentiate each separately, then assemble carefully.
在练习 22B 中，你可能会遇到 (x²+1)/(x−2) 这类有理函数或三角分式。始终将分子定为 u，分母定为 v，分别求导，然后仔细代入公式。
For instance, differentiate f(x) = (2x)/(x²+1). u = 2x, v = x²+1. u’ = 2, v’ = 2x. f'(x) = [2(x²+1) − 2x(2x)] / (x²+1)² = (2x²+2 − 4x²)/(x²+1)² = (2 − 2x²)/(x²+1)².
例如，求 f(x) = (2x)/(x²+1) 的导数。u = 2x, v = x²+1。u’ = 2, v’ = 2x。f'(x) = [2(x²+1) − 2x(2x)] / (x²+1)² = (2x²+2 − 4x²)/(x²+1)² = (2 − 2x²)/(x²+1)²。
5. Chain Rule for Composite Functions | 复合函数的链式法则
The chain rule handles nested functions: if y = f(g(x)), then dy/dx = f'(g(x))·g'(x). This is arguably the most tested technique in Exercise 22B because it appears in almost every non-polynomial differentiation.
链式法则处理嵌套函数：若 y = f(g(x))，则 dy/dx = f'(g(x))·g'(x)。这或许是练习 22B 中考得最多的技巧，因为它几乎出现在每个非多项式求导中。
A systematic approach: identify the outer function and the inner function. Differentiate the outer function, keeping the inner untouched, then multiply by the derivative of the inner function. Write ‘let u = inner function’ if necessary.
系统方法是：识别外层函数和内层函数。对外层函数求导（保持内层不变），再乘以内层函数的导数。如有需要，可设 u = 内层函数。
Example: y = (3x²+5)⁴. Outer is u⁴, inner is 3x²+5. dy/dx = 4(3x²+5)³·(6x) = 24x(3x²+5)³.
例：y = (3x²+5)⁴。外层为 u⁴，内层为 3x²+5。dy/dx = 4(3x²+5)³·(6x) = 24x(3x²+5)³。
6. Combining Chain Rule with Product and Quotient | 链式法则与积、商法则的组合
Many questions in Exercise 22B require simultaneous application of the chain rule inside a product or quotient. For instance, f(x) = x²·√(2x+1) uses product rule where the second factor demands chain rule.
练习 22B 中有许多题目需要同时在积或商中使用链式法则。例如 f(x) = x²·√(2x+1) 要用积法则，而第二个因式又需要链式法则。
When facing a product like u(x)·v(g(x)), differentiate u normally, but for v'(g(x)) you must apply chain rule. Execute step by step: first write the structure, then fill in the derivatives carefully.
面对 u(x)·v(g(x)) 这类乘积时，对 u 正常求导，但对 v'(g(x)) 必须使用链式法则。逐步执行：先搭好结构，再小心填进导数。
Worked example: f(x) = x³·cos(2x). Let u = x³, v = cos(2x). u’ = 3x², v’ = −sin(2x)·2 = −2 sin(2x). Then f'(x) = 3x² cos(2x) + x³(−2 sin(2x)) = 3x² cos(2x) − 2x³ sin(2x).
完整示例：f(x) = x³·cos(2x)。令 u = x³, v = cos(2x)。u’ = 3x²，v’ = −sin(2x)·2 = −2 sin(2x)。则 f'(x) = 3x² cos(2x) + x³(−2 sin(2x)) = 3x² cos(2x) − 2x³ sin(2x)。
7. Differentiating Standard Trigonometric Functions | 标准三角函数的导数
Exercise 22B expects fluency with derivatives of sin x, cos x, tan x, and often sec x. d/dx(sin x) = cos x, d/dx(cos x) = −sin x, d/dx(tan x) = sec² x. Remember the chain rule versions: d/dx(sin(kx)) = k cos(kx).
练习 22B 要求你熟练掌握 sin x, cos x, tan x 以及常出现的 sec x 的导数。d/dx(sin x) = cos x，d/dx(cos x) = −sin x，d/dx(tan x) = sec² x。记住链式法则版本：d/dx(sin(kx)) = k cos(kx)。
When differentiating tan(3x), treat it as tan(u) with u=3x. The derivative is sec²(3x)·3 = 3 sec²(3x). Similarly, derivative of sin²x (often written as (sin x)²) requires chain rule: 2 sin x · cos x = sin(2x).
求 tan(3x) 的导数时，将其视为 tan(u)，其中 u=3x。导数为 sec²(3x)·3 = 3 sec²(3x)。类似地，sin²x（常写作 (sin x)²）的导数需要链式法则：2 sin x · cos x = sin(2x)。
8. Finding Equations of Tangents | 求切线方程
A major application in Exercise 22B is determining the equation of a tangent line at a given point. The derivative f'(a) gives the slope m; then use y − f(a) = m (x − a). Ensure your final equation is simplified to y = mx + c form.
练习 22B 的一个主要应用是求给定点处的切线方程。导数 f'(a) 给出斜率 m；然后使用 y − f(a) = m (x − a)。确保最终方程化简为 y = mx + c 形式。
Example: Find the tangent to f(x) = x³ − 2x at x=1. f(1) = −1, f'(x)=3x²−2, f'(1)=1. Tangent: y − (−1) = 1(x − 1) → y = x − 2.
例：求 f(x) = x³ − 2x 在 x=1 处的切线。f(1) = −1，f'(x)=3x²−2，f'(1)=1。切线：y − (−1) = 1(x − 1) → y = x − 2。
If the question asks for a tangent with a specific slope, set f'(x) equal to that slope and solve for x, then find the point and equation. This is a common twist in Exercise 22B.
若题目要求一条具有特定斜率的切线，令 f'(x) 等于该斜率，解出 x，再找出对应点和方程。这是练习 22B 中常见的变体。
9. Normal Lines and Perpendicularity | 法线与垂直关系
The normal line to a curve at a point is perpendicular to the tangent. Its slope mₙ is given by mₙ = −1 / f'(a), provided f'(a) ≠ 0. Many students forget the negative reciprocal and simply keep the same slope.
曲线在某点处的法线垂直于切线。其斜率 mₙ 由 mₙ = −1 / f'(a) 给出，前提是 f'(a) ≠ 0。许多学生忘记取负倒数，而直接沿用相同斜率。
Once you have the normal’s slope and the point, use the same point-slope formula. Exam questions sometimes ask for the normal first, so read the prompt carefully.
得到法线的斜率和点后，沿用相同的点斜式公式。考试题有时会先求法线，因此务必仔细审题。
Example: For f(x) = √x at x=4, f(4)=2, f'(x)= 1/(2√x), f'(4)=1/4. Tangent slope = 1/4, normal slope = −4. Normal equation: y − 2 = −4(x − 4) → y = −4x + 18.
例：对于 f(x) = √x 在 x=4 处，f(4)=2，f'(x)= 1/(2√x)，f'(4)=1/4。切线斜率为 1/4，法线斜率为 −4。法线方程：y − 2 = −4(x − 4) → y = −4x + 18。
10. Second Derivatives and Concavity | 二阶导数与凹凸性
The second derivative f”(x) is obtained by differentiating the first derivative again. In Exercise 22B, you may be asked to compute f”(x) directly or to use it to discuss concavity and points of inflection.
二阶导数 f”(x) 由一阶导数再求导得到。在练习 22B 中，可能会要求直接计算 f”(x) 或利用它讨论函数的凹凸性和拐点。
A function is concave up where f”(x) > 0 and concave down where f”(x) < 0. Potential inflection points occur where f''(x)=0 or fails to exist, and concavity changes.
当 f”(x) > 0 时函数上凹，f”(x) < 0 时下凹。潜在拐点出现在 f''(x)=0 或不存在且凹凸性发生改变的位置。
For example, f(x) = x³ − 3x. f'(x) = 3x² − 3, f”(x) = 6x. Setting f”(x) = 0 gives x=0. For x<0, f''(x)<0 (concave down); for x>0, f”(x)>0 (concave up). Hence x=0 is an inflection point.
例如，f(x) = x³ − 3x。f'(x) = 3x² − 3，f”(x) = 6x。令 f”(x)=0 得 x=0。当 x<0 时 f''(x)<0（下凹）；当 x>0 时 f”(x)>0（上凹）。因此 x=0 为拐点。
11. Rates of Change in Context | 情境中的变化率
Exercise 22B sometimes frames derivatives as rates of change: velocity, acceleration, or growth rates. If s(t) is displacement, then s'(t) is velocity and s”(t) is acceleration. Always include units when the question provides a real-world context.
练习 22B 有时将导数置于变化率的情境中：速度、加速度或增长率。若 s(t) 表示位移，则 s'(t) 为速度，s”(t) 为加速度。当题目给出实际情境时，要始终包含单位。
When asked for the rate at a specific instant, evaluate the derivative at that t-value. For related rates, identify the relationship between variables and differentiate implicitly with respect to time.
当被要求求某瞬时的速率时，需计算该 t 值处的导数值。对于相关变化率，先确定变量间的关系，然后对时间进行隐式求导。
Example: The position of a particle is s(t) = t³ − 6t² + 9t metres. Velocity v(t) = 3t² − 12t + 9 m/s. At t=2, velocity is 3(4)−24+9 = −3 m/s (moving backwards).
例：质点位置 s(t) = t³ − 6t² + 9t 米。速度 v(t) = 3t² − 12t + 9 米/秒。在 t=2 时，速度为 3(4)−24+9 = −3 米/秒（反向运动）。
12. Common Pitfalls and How to Avoid Them | 常见陷阱及规避方法
In Exercise 22B, the most frequent errors include: forgetting to apply the chain rule when the argument is not just x; misplacing signs in the quotient rule numerator; and incorrectly differentiating constants or radical forms. Writing every step reduces mistakes.
在练习 22B 中，最常见的错误包括：当自变量不仅是 x 时忘记使用链式法则；商法则分子中符号错位；以及错误地对常数或根式求导。逐步书写可以降低错误率。
Another trap is simplifying too early — differentiate first, then simplify. Also watch out for hidden products: x·eˣ is a product, not a simple power. Practice recognizing the structure before differentiating.
另一个陷阱是过早化简——先求导，再化简。还要注意隐藏的乘积：x·eˣ 是一个乘积，而非简单的幂函数。求导前要先练习辨识结构。
Finally, always double-check whether the question requires the derivative function or the derivative at a point. Misreading this distinction costs marks unnecessarily.
最后，务必仔细核对题目要求的是导函数还是某点处的导数值。误读这层区别会白白丢分。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply