📚 Derivatives of inverse functions | 反函数的导数
In IB Mathematics, one of the elegant applications of the chain rule appears when we differentiate inverse functions. Instead of finding the inverse explicitly and then differentiating, a powerful formula connects the derivative of a function and the derivative of its inverse. This topic builds on your understanding of composite functions and implicit differentiation, and it leads directly to the derivatives of inverse trigonometric functions, which are essential in calculus.
在IB数学中，当我们对反函数进行求导时，链式法则展现出一个极为优美的应用。我们无需先显式求出反函数再求导，一个强大的公式将函数导数与其反函数导数紧密联系起来。这一主题建立在你对复合函数和隐函数求导的理解之上，并直接引出反三角函数的导数，这在整个微积分中都是至关重要的。
1. What is an inverse function? | 什么是反函数？
nd be differentiable, the original function must be one-to-one on an interval, and its derivative must not be zero at the corresponding point. Remember the domain and range swap roles: the domain of f becomes the range of f⁻¹, and vice versa.
给定一个函数 f，它的反函数 f⁻¹ 会逆转 f 的作用。如果 f(a) = b，那么 f⁻¹(b) = a。为了让反函数存在并且可导，原函数在某个区间内必须是一一对应的，并且在相应点处其导数不能为零。记住定义域和值域互换了角色：f 的定义域变成了 f⁻¹ 的值域，反之亦然。
2. The key relationship between function and inverse | 函数与其反函数之间的关键关系
here g = f⁻¹, we can obtain the derivative formula without knowing the explicit form of g.
反函数的定义性质是：对于各自定义域内的所有 x，都有 f(f⁻¹(x)) = x 以及 f⁻¹(f(x)) = x。这个恒等式是求反函数导数的出发点。将 f(g(x)) = x 两边同时对 x 求导（其中 g = f⁻¹），我们就可以在不知道 g 的具体表达式的情况下得到导数公式。
3. Deriving the inverse function derivative formula | 反函数求导公式的推导
¹(x). Then f(g(x)) = x. Differentiate both sides with respect to x using the chain rule: f'(g(x)) * g'(x) = 1. Provided f'(g(x)) ≠ 0, we can solve for g'(x):
g'(x) = 1 / f'(g(x)) or (f⁻¹)'(x) = 1 / f'(f⁻¹(x))
This simple formula states that the derivative of the inverse function at a point is the reciprocal of the derivative of the original function evaluated at the corresponding point.
设 g(x) = f⁻¹(x)，那么 f(g(x)) = x。使用链式法则对 x 求导：f'(g(x)) * g'(x) = 1。只要 f'(g(x)) ≠ 0，我们就可以解出 g'(x)：
g'(x) = 1 / f'(g(x)) 或 (f⁻¹)'(x) = 1 / f'(f⁻¹(x))
这个简洁的公式表明，反函数在某点的导数等于原函数在对应点处的导数的倒数。
4. Interpretation using the chain rule | 用链式法则进行诠释
t y = f⁻¹(x), which means f(y) = x. Differentiate f(y) = x implicitly with respect to x: f'(y) * dy/dx = 1, so dy/dx = 1 / f'(y). Since y = f⁻¹(x), we again obtain (f⁻¹)'(x) = 1 / f'x)). This method is particularly useful when dealing with inverse trigonometric functions.
还有一种理解方式，非常富有启发性：令 y = f⁻¹(x)，这就意味着 f(y) = x。对 f(y) = x 进行隐函数求导，关于 x：f'(y) * dy/dx = 1，因此 dy/dx = 1 / f'(y)。由于 y = f⁻¹(x)，我们再次得到 (f⁻¹)'(x) = 1 / f'(f⁻¹(x))。在处理反三角函数时，这种方法特别实用。
5. Derivative of arcsin x | 反正弦函数的导数
this in terms of x, use the identity cos² y = 1 − sin² y. Since sin y = x, we get cos y = √1 − x². For the principal values in [−π/2, π/2], cos y is non‑negative, so
d/dx (arcsin x) = 1 / √1 − x², for |x| < 1.
令 y = arcsin x，则 x = sin y。关于 x 隐式求导：1 = cos y * dy/dx。因此 dy/dx = 1 / cos y。为了用 x 来表示，利用恒等式 cos² y = 1 − sin² y。由于 sin y = x，得到 cos y = √1 − x²。在主值范围 [−π/2, π/2] 内，cos y 非负，所以
d/dx (arcsin x) = 1 / √1 − x²，其中 |x| < 1。
6. Derivative of arccos x | 反余弦函数的导数
Differentiating implicitly gives 1 = −sin y * dy/dx, so dy/dx = −1 / sin y. Using sin y = √1 − cos² y = √1 − x², and noting that for the principal values in [0, π], sin y is non‑negative, we get
d/dx (arccos x) = −1 / √1 − x², for |x| < 1.
. Indeed, arccos x = π/2 − arcsin x, so differentiating this identity confirms the result.
对于 y = arccos x，有 x = cos y。隐式求导：1 = −sin y * dy/dx，所以 dy/dx = −1 / sin y。利用 sin y = √1 − cos² y = √1 − x²，并注意到主值范围 [0, π] 上 sin y 非负，得到
d/dx (arccos x) = −1 / √1 − x²，其中 |x| < 1。
注意与 arcsin x 相比多了一个负号。事实上，arccos x = π/2 − arcsin x，因此对这个恒等式求导便可验证该结果。
7. Derivative of arctan x | 反正切函数的导数
s² y in terms of x, recall the identity sec² y = 1 + tan² y, so sec² y = 1 + x². Therefore
d/dx (arctan x) = 1 / (1 + x²).
This result holds for all real x. Because the denominator is always positive, the derivative respects the global monotonicity of arctan x.
令 y = arctan x，则 x = tan y。隐式求导：1 = sec² y * dy/dx。因此 dy/dx = 1 / sec² y = cos² y。为了用 x 表示 cos² y，利用恒等式 sec² y = 1 + tan² y，所以 sec² y = 1 + x²。从而得到
d/dx (arctan x) = 1 / (1 + x²)。
该结果对所有实数 x 都成立。因为分母始终为正，导数体现了 arctan x 在整个定义域上的单调性。
8. Applying the formula at a specific point | 在特定点处应用公式
Suppose f(x) = x³ + 2x and you are asked to find (f⁻¹)' plicitly finding the inverse. First, note that f(1) = 1³ + 2*1 = 3, so f⁻¹(3) = 1. Next, f'(x) = 3x² + 2, so f'(1) = 5. Using (f⁻¹)'(x) = 1 / f'(f⁻¹(x)), we get (f⁻¹)'(3) = 1 / f'(1) = 1/5.
假设 f(x) = x³ + 2x，要求 (f⁻¹)'(3) 而无需显式求出反函数。首先，注意到 f(1) = 1³ + 2*1 = 3，因此 f⁻¹(3) = 1。接着，f'(x) = 3x² + 2，所以 f'(1) = 5。套用公式 (f⁻¹)'(x) = 1 / f'(f⁻¹(x))，得到 (f⁻¹)'(3) = 1 / f'(1) = 1/5。
9. Derivative of an inverse function expressed explicitly | 显式反函数的导数
If f is simple enough to invert explicitly, you can verify the formula. Take f(x) = ex, whose inverse is f⁻¹(x) = ln x. Our formula gives (ln x)' = 1 / eln x tive. The formula works seamlessly for any invertible differentiable function, but it is especially valuable when the inverse is hard to write down.
如果 f 足够简单，可以显式求出反函数，你也能验证这个公式。以 f(x) = ex 为例，其反函数为 f⁻¹(x) = ln x。公式给出 (ln x)' = 1 / eln x = 1/x，这与我们已知的导数完全吻合。对于任何可逆的可微函数，这个公式都完美适用，而当反函数很难显式表达时，它就显得尤为珍贵。
10. A worked example combining several ideas | 综合例题
Let g(x) = arcsin(2x). Find g'(x). Use the chain rule: g'(x) = 1 / √(1 − (2x)²) * d/dx(2x) = 2 / √1 − 4x², for |2x| < 1, i.e., |x| < 1/2. As another example, differentiate h(x) = arctan(√x). Then h'(x) = 1 / (1 + (√x)²) * d/dx (x1/2) = 1 / (1 + x) * (1/(2√x)) = 1 / (2√x (1 + x)), for x > 0.
设 g(x) = arcsin(2x)，求 g'(x)。使用链式法则：g'(x) = 1 / √(1 − (2x)²) * d/dx(2x) = 2 / √1 − 4x²，要求 |2x| < 1，即 |x| < 1/2。再举一个例子，对 h(x) = arctan(√x) 求导：h'(x) = 1 / (1 + (√x)²) * d/dx (x1/2) = 1 / (1 + x) * (1/(2√x)) = 1 / (2√x (1 + x))，其中 x > 0。
11. Common pitfalls and how to avoid them | 常见错误与如何避免
Many students forget to check that the derivative of f at the point does not vanish; otherwise the inverse function is not differentiable there. Also, when working with inverse trigonometric functions, the domain restrictions are crucial. For arcsin and arccos the derivative formula only holds for |x| < 1, and for arctan it holds for all real x, but composition with another function may impose further restrictions. Always state the domain of the derivative clearly.
许多学生忘记检查 f 在该点的导数不能为零，否则反函数在该点不可导。另外，在处理反三角函数时，定义域的限制至关重要。对于 arcsin 和 arccos，导数公式仅在 |x| < 1 时成立；对于 arctan，虽然对所有实数 x 都成立，但与其他函数复合时可能会施加额外的限制。一定要清楚地标明导数的定义域。
12. Summary and IB exam tips | 总结与IB考试技巧
The inverse function derivative formula (f⁻¹)'(x) = 1 / f' igonometric functions, you should memorise the three core derivatives: d/dx(arcsin x) = 1/√1−x², d/dx(arccos x) = −1/√1−x² exams, questions often ask for the derivative at a particular point rather than a general expression, so practising the point-wise application saves time and reduces algebraic strain.
反函数求导公式 (f⁻¹)'(x) = 1 / f'(f⁻¹(x)) 是链式法则的直接推论。对于标准的反三角函数，你需要记住三个核心导数：d/dx(arcsin x) = 1/√1−x²、d/dx(arccos x) = −1/√1−x² 和 d/dx(arctan x) = 1/(1+x²)。在IB考试中，题目经常要求求特定点处的导数，而非一般表达式，因此熟练掌握在具体点上的应用能节省时间并减少代数运算的负担。
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