Derivatives of Inverse Functions | IB数学：反函数的导数求解

📚 Derivatives of Inverse Functions | IB数学：反函数的导数求解

In IB Mathematics (Analysis and Approaches and Applications and Interpretation), one important differentiation skill is finding the derivative of an inverse function. This topic links the concepts of inverse functions, implicit differentiation, and the chain rule, and it often appears in both Paper 1 and Paper 2 questions.

在IB数学（分析与方法、应用与解释）中，一个重要的求导技巧是求解反函数的导数。这一主题将反函数、隐函数求导和链式法则联系起来，常在Paper 1和Paper 2中出现。

1. The Inverse Function Rule | 反函数求导法则

If a function \( f \) is differentiable and has an inverse \( f^{-1} \), then the derivative of the inverse at a point \( x \) is given by the reciprocal of the derivative of \( f \) at the corresponding point \( f^{-1}(x) \). This is often written as:

如果函数 \( f \) 可导且存在反函数 \( f^{-1} \)，那么反函数在点 \( x \) 处的导数等于原函数 \( f \) 在对应点 \( f^{-1}(x) \) 处导数的倒数，即：

(f⁻¹)′(x) = 1 / f′(f⁻¹(x))

This formula is derived from the fundamental relationship \( f(f^{-1}(x)) = x \). Differentiating both sides with respect to \( x \) using the chain rule gives \( f'(f^{-1}(x)) \cdot (f^{-1})'(x) = 1 \), which leads directly to the rule.

该公式由恒等式 \( f(f^{-1}(x)) = x \) 推导而来。两边对 \( x \) 求导并应用链式法则，得到 \( f'(f^{-1}(x)) \cdot (f^{-1})'(x) = 1 \)，从而直接得到上述法则。

2. Understanding the Notation | 理解符号含义

It is crucial to distinguish between \( f^{-1}(x) \) meaning the inverse function and \( (f(x))^{-1} \) meaning the reciprocal. The derivative formula uses \( f^{-1} \) as the inverse function, not as \( 1/f \).

务必区分 \( f^{-1}(x) \) 表示反函数，而 \( (f(x))^{-1} \) 表示倒数。导数公式中的 \( f^{-1} \) 是反函数，而不是 \( 1/f \)。

For example, if \( f(x) = 2x + 3 \), then \( f^{-1}(x) = (x – 3)/2 \), not \( 1/(2x+3) \). The derivative of the inverse is \( 1/2 \), which matches the formula since \( f'(x) = 2 \) everywhere.

例如，若 \( f(x) = 2x + 3 \)，则其反函数为 \( f^{-1}(x) = (x – 3)/2 \)，而不是 \( 1/(2x+3) \)。反函数的导数为 \( 1/2 \)，与公式一致，因为 \( f'(x) = 2 \) 处处成立。

3. Derivation Using the Chain Rule | 利用链式法则推导

Start from the identity \( f(f^{-1}(x)) = x \). Let \( y = f^{-1}(x) \), so that \( f(y) = x \). Differentiate both sides with respect to \( x \):

从恒等式 \( f(f^{-1}(x)) = x \) 出发。令 \( y = f^{-1}(x) \)，则 \( f(y) = x \)。两边对 \( x \) 求导：

d/dx [f(y)] = d/dx [x] → f′(y) · dy/dx = 1

Thus \( dy/dx = 1 / f'(y) = 1 / f'(f^{-1}(x)) \). This proof is required for higher-level IB questions that ask students to “show that” the derivative of an inverse function takes this form.

因此 \( dy/dx = 1 / f'(y) = 1 / f'(f^{-1}(x)) \)。在IB高难度题目中，常要求学生“证明”反函数导数具有这种形式，这一推导过程非常重要。

4. A Simple Linear Example | 线性函数示例

Let \( f(x) = 3x + 5 \). Then \( f^{-1}(x) = (x – 5)/3 \), and \( (f^{-1})'(x) = 1/3 \). Using the formula:

设 \( f(x) = 3x + 5 \)，则 \( f^{-1}(x) = (x – 5)/3 \)，且 \( (f^{-1})'(x) = 1/3 \)。利用公式验证：

Since \( f'(x) = 3 \) for all \( x \), we have \( (f^{-1})'(x) = 1 / f'(f^{-1}(x)) = 1/3 \). This confirms the rule for a simple linear case.

因为 \( f'(x) = 3 \) 对所有 \( x \) 成立，所以 \( (f^{-1})'(x) = 1 / f'(f^{-1}(x)) = 1/3 \)。这在线性简单情形下验证了法则。

5. Quadratic Function and Domain Restrictions | 二次函数与定义域限制

Consider \( f(x) = x^2 \) on the domain \( x \geq 0 \), so that its inverse is \( f^{-1}(x) = \sqrt{x} \). We know \( (\sqrt{x})’ = 1/(2\sqrt{x}) \). Let us verify this with the formula:

考虑定义域 \( x \geq 0 \) 上的函数 \( f(x) = x^2 \)，其反函数为 \( f^{-1}(x) = \sqrt{x} \)。我们知道 \( (\sqrt{x})’ = 1/(2\sqrt{x}) \)。下面用公式验证：

Here \( f'(x) = 2x \), and \( f^{-1}(x) = \sqrt{x} \). Therefore:

这里 \( f'(x) = 2x \)，且 \( f^{-1}(x) = \sqrt{x} \)，因此：

(f⁻¹)′(x) = 1 / f′(√x) = 1 / (2√x)

This matches the derivative of \( \sqrt{x} \). Note that the domain must be restricted so that the inverse exists.

这正好与 \( \sqrt{x} \) 的导数一致。注意必须限制定义域，反函数才存在。

6. Using the Formula at a Specific Point | 在某一点处使用公式

A common IB exam question gives a function \( f \), a point \( a \), and asks for \( (f^{-1})'(a) \). The key steps are:

IB考试中常见题型是给出函数 \( f \)，点 \( a \)，求 \( (f^{-1})'(a) \)。关键步骤为：

Find the value \( b \) such that \( f(b) = a \). This means \( b = f^{-1}(a) \).

求出满足 \( f(b) = a \) 的 \( b \) 值，即 \( b = f^{-1}(a) \)。
Evaluate \( f'(b) \).

计算 \( f'(b) \)。
Take the reciprocal: \( (f^{-1})'(a) = 1/f'(b) \).

取倒数：\( (f^{-1})'(a) = 1/f'(b) \)。

This method avoids explicitly finding the inverse function, which is especially useful when the inverse is difficult or impossible to express explicitly.

这种方法避免了显式求解反函数，尤其当反函数难以甚至无法显式表达时非常实用。

7. Worked Example: Cubic Function | 实例：三次函数

Let \( f(x) = x^3 + 2x + 1 \). It is given that \( f(1) = 4 \). Find \( (f^{-1})'(4) \).

设 \( f(x) = x^3 + 2x + 1 \)，已知 \( f(1) = 4 \)，求 \( (f^{-1})'(4) \)。

First, since \( f(1) = 4 \), we know \( f^{-1}(4) = 1 \). Next, \( f'(x) = 3x^2 + 2 \), so \( f'(1) = 3(1)^2 + 2 = 5 \). Hence:

首先，因为 \( f(1) = 4 \)，可知 \( f^{-1}(4) = 1 \)。其次，\( f'(x) = 3x^2 + 2 \)，所以 \( f'(1) = 3(1)^2 + 2 = 5 \)。因此：

(f⁻¹)′(4) = 1 / f′(1) = 1/5

This example shows how the formula can be applied without solving for the inverse explicitly.

这个例子说明，无需显式解出反函数即可应用公式。

8. Worked Example: Exponential and Logarithmic Functions | 实例：指数函数与对数函数

For \( f(x) = e^x \), its inverse is \( f^{-1}(x) = \ln x \). Using the formula:

对 \( f(x) = e^x \)，其反函数为 \( f^{-1}(x) = \ln x \)。利用公式：

Since \( f'(x) = e^x \), and \( f^{-1}(x) = \ln x \), we have:

因为 \( f'(x) = e^x \)，且 \( f^{-1}(x) = \ln x \)，所以：

(ln x)′ = 1 / e^(ln x) = 1/x

This provides a beautiful derivation of the derivative of the natural logarithm. IB students are expected to know both \( d/dx(e^x) = e^x \) and \( d/dx(\ln x) = 1/x \), and this connection is often tested.

这为自然对数的导数提供了一个简洁的推导。IB学生应同时掌握 \( d/dx(e^x) = e^x \) 和 \( d/dx(\ln x) = 1/x \)，两者之间的联系经常被考查。

9. Implicit Differentiation Connection | 与隐函数求导的联系

Finding \( dy/dx \) for an inverse function can also be done using implicit differentiation. If \( y = f^{-1}(x) \), then \( f(y) = x \). Differentiating implicitly:

求反函数的导数 \( dy/dx \) 也可以用隐函数求导完成。若 \( y = f^{-1}(x) \)，则 \( f(y) = x \)。隐式求导：

d/dx [f(y)] = d/dx [x] → f′(y) · dy/dx = 1

Therefore \( dy/dx = 1/f'(y) \). This is exactly the same result as before. In IB exams, implicit differentiation is a standard tool, and recognizing this connection can help solve problems more flexibly.

因此 \( dy/dx = 1/f'(y) \)，与之前的结论完全相同。在IB考试中，隐函数求导是常用工具，认识到这种联系有助于更灵活地解题。

10. Common IB Exam Question Types | IB常见题型

Typical questions involving derivatives of inverse functions include:

涉及反函数导数的典型题型包括：

Given a function and a point on its inverse, find the slope of the tangent to the inverse at that point.

给定函数及其反函数上的一点，求反函数在该点处切线的斜率。
Given \( f \) and \( f'(x) \), evaluate \( (f^{-1})'(a) \) using the formula.

给定 \( f \) 和 \( f'(x) \)，利用公式计算 \( (f^{-1})'(a) \)。
Proof questions asking students to derive \( (f^{-1})'(x) = 1/f'(f^{-1}(x)) \).

要求学生推导 \( (f^{-1})'(x) = 1/f'(f^{-1}(x)) \) 的证明题。
Application to trigonometric inverse functions such as \( \arcsin x \) and \( \arctan x \).

应用于反三角函数，如 \( \arcsin x \) 和 \( \arctan x \)。

These questions often appear in Paper 1 without a calculator, so algebraic accuracy is essential.

这些问题常出现在不使用计算器的Paper 1中，因此代数准确性至关重要。

11. Derivatives of Inverse Trigonometric Functions | 反三角函数的导数

The inverse function rule can be used to derive well-known formulas for inverse trigonometric functions. For example, let \( y = \arcsin x \), so \( \sin y = x \). Differentiate implicitly:

反函数法则可用于推导反三角函数的常见导数公式。例如，令 \( y = \arcsin x \)，则 \( \sin y = x \)。隐式求导：

cos y · dy/dx = 1 → dy/dx = 1 / cos y

Using \( \cos^2 y + \sin^2 y = 1 \) and \( \sin y = x \), we have \( \cos y = \sqrt{1 – x^2} \) (for the principal range). Thus:

利用 \( \cos^2 y + \sin^2 y = 1 \) 以及 \( \sin y = x \)，得到 \( \cos y = \sqrt{1 – x^2} \)（在主值范围内）。因此：

d/dx(arcsin x) = 1 / √(1 – x²)

Similarly, for \( y = \arctan x \), \( \tan y = x \), and using \( d/dy(\tan y) = \sec^2 y \), one obtains:

类似地，对 \( y = \arctan x \)，\( \tan y = x \)，利用 \( d/dy(\tan y) = \sec^2 y \)，可得：

d/dx(arctan x) = 1 / (1 + x²)

These formulas are listed in the IB formula booklet for Analysis and Approaches, but understanding their derivation helps with retention and application.

这些公式列在IB分析与方法的公式手册中，但理解其推导有助于记忆和应用。

12. Summary and Exam Tips | 总结与考试技巧

To master derivatives of inverse functions, remember the core formula:

要掌握反函数求导，请牢记核心公式：

(f⁻¹)′(x) = 1 / f′(f⁻¹(x))

Always check that the inverse exists: the function must be one-to-one on the given domain.

始终检查反函数是否存在：函数在给定定义域上必须是一一对应。
When asked for \( (f^{-1})'(a) \), find \( b = f^{-1}(a) \) by solving \( f(b) = a \).

当要求 \( (f^{-1})'(a) \) 时，通过解 \( f(b) = a \) 来求 \( b = f^{-1}(a) \)。
Do not confuse \( f^{-1} \) with \( (f)^{-1} \); the notation is context-dependent.

不要将 \( f^{-1} \) 与 \( 1/f \) 混淆；符号含义取决于上下文。
Practice both the formula and implicit differentiation approaches, as IB marks may reward clear method.

同时练习公式法和隐函数求导法，因为IB评分可能看重方法的清晰性。

With consistent practice, inverse function derivative problems become straightforward and often provide a quick win in exams.

通过持续练习，反函数求导问题将变得十分直接，通常能在考试中成为快速得分的题目。

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Derivatives of Inverse Functions | IB数学：反函数的导数求解