📚 Derivatives of Logarithmic Functions | 对数函数的导数
Logarithmic functions appear extensively in calculus, from simple growth models to advanced integration techniques. Understanding how to differentiate logarithms is essential for any IB Mathematics student. This revision guide covers the derivative of ln x, logarithms with arbitrary bases, the chain rule, logarithmic differentiation, and common pitfalls, with plenty of worked examples.
对数函数在微积分中应用广泛，从简单的增长模型到复杂的积分技巧均有涉及。掌握对数函数的求导方法是每位 IB 数学学生必备的技能。本复习指南将涵盖 ln x 的导数、任意底数的对数、链式法则、对数求导法以及常见误区，并配有大量例题解析。
1. Introduction to Logarithmic Differentiation | 对数函数求导简介
Differentiating logarithmic functions builds on your knowledge of exponentials, limits, and the chain rule. The natural logarithm ln x is particularly important because its derivative takes the simple form 1/x, and it serves as a bridge to logarithms of other bases via the change-of-base formula.
求对数函数的导数需要你综合运用指数函数、极限和链式法则的知识。其中自然对数 ln x 尤为重要，因为它的导数形式简洁，仅为 1/x，并且通过换底公式可将其推广到任意底数的对数。
Before diving into formulas, recall the definition: y = logₐ x ⇔ aʸ = x, where a > 0, a ≠ 1, and x > 0. The special case a = e (Euler's number) gives the natural logarithm ln x.
在深入公式之前，回顾一下定义：y = logₐ x ⇔ aʸ = x，其中 a > 0 且 a ≠ 1，同时 x > 0。当底数 a = e（欧拉数）时，便得到自然对数 ln x。
2. The Derivative of ln x | 自然对数 ln x 的导数
The derivative of the natural logarithm function is one of the most elegant results in calculus.
自然对数函数的导数是微积分中最简洁优美的结果之一。
d/dx (ln x) = 1/x, for x > 0
nition or more elegantly by implicit differentiation: starting from y = ln x, rewrite as ey = x. Differentiate both sides with respect to x, giving ey (dy/dx) = 1, so dy/dx = 1/(ey) = 1/x.
该公式表明曲线 y = ln x 在任意点 x 处的斜率恰好等于 x 的倒数。这一结果可利用极限定义推导，更巧妙的方法是隐函数求导：由 y = ln x 变形为 ey = x，两边同时对 x 求导得 ey (dy/dx) = 1，因此 dy/dx = 1/(ey) = 1/x。
leteness, we often extend the domain to all nonzero x by using the absolute value: d/dx ln|x| x ≠ 0. This is particularly useful when the argument of the logarithm may become negative.
为完整起见，我们常借助绝对值将定义域扩展到所有非零实数：当 x ≠ 0 时，d/dx ln|x| = 1/x。当对数的真数可能取负值时，这一形式尤为重要。
3. Derivative of logₐ x | 以 a 为底的对数的导数
For a logarithm with an arbitrary base a (a > 0, a ≠ 1), the derivative is easily obtained using the change-of-base formula.
对于任意底数 a（a > 0, a ≠ 1）的对数，其导数可通过换底公式轻松得到。
d/dx (logₐ x) = 1 / (x ln a)
rentiation gives (1/ln a) * d/dx (ln x) = (1/ln a) * (1/x) = 1/(x ln a). The factor 1/ln a 'scales' the derivative of ln x.
证明：logₐ x = (ln x) / (ln a)。由于 1/ln a 为常数，求导得 (1/ln a) * d/dx (ln x) = (1/ln a) * (1/x) = 1/(x ln a)。可见因子 1/ln a 对 ln x 的导数进行了缩放。
= 1, so the formula reduces to 1/x as expected. For a = 10, d/dx (log₁₀ x) = 1/(x ln 10).
注意当 a = e 时，ln e = 1，公式就退化为 1/x，符合预期。当 a = 10 时，d/dx (log₁₀ x) = 1/(x ln 10)。
4. Chain Rule with Natural Logarithms | 自然对数与链式法则
When the argument of a logarithm is itself a function of x, the chain rule must be applied.
当对数的真数本身是 x 的函数时，必须使用链式法则求导。
d/dx [ln(u(x))] = u'(x) / u(x), provided u(x) > 0
This rule follows directly from the derivative of ln x: the outer derivative is 1/u, multiplied by the derivative of the inner function u. The result is a quotient: the derivative of u over u itself.
这一法则直接源于 ln x 的导数：外层导数为 1/u，再乘以内层函数 u 的导数。最终结果是一个分式：u 的导数除以 u 本身。
ample, if y = ln(3x + 1), set u = 3x + 1. Then dy/dx = (1/(3x+1)) * d/dx(3x+1) = 3/(3x+1).
例如，设 y = ln(3x + 1)，令 u = 3x + 1，则 dy/dx = (1/(3x+1)) * d/dx(3x+1) = 3/(3x+1)。
s to logarithms of any base when combined with the constant factor: d/dx [logₐ (u(x))] = u'(x) / (u(x) ln a).
结合常数因子，该法则可推广到任意底数的对数：d/dx [logₐ (u(x))] = u'(x) / (u(x) ln a)。
5. Derivative of ln|u(x)| | ln|u(x)| 的导数
The absolute value version yields an even more compact formula that works regardless of the sign of u(x), as long as u(x) ≠ 0.
带绝对值的对数形式给出了一个更为紧凑的公式，不论 u(x) 的正负，只要 u(x) ≠ 0 便适用。
d/dx (ln|u(x)|) = u'(x) / u(x)
This is particularly handy when integrating or when dealing with rational functions wrapped in a logarithm. The absolute value ensures the logarithm's argument remains positive, and the derivative neatly eliminates the sign issue.
这在积分运算或处理包裹对数符号的有理函数时尤为便利。绝对值保证了对数的真数为正，而求导后符号问题恰好被巧妙消除。
For instance, if y = ln|3x - 5| rry about where 3x - 5 is positive or negative, except for the domain restriction x ≠ 5/3.
例如，若 y = ln|3x - 5|，则 dy/dx = 3 / (3x - 5)。除了定义域要求 x ≠ 5/3，无需操心 3x - 5 的正负。
6. Logarithmic Differentiation Technique | 对数求导法
ithmic differentiation is a powerful method used when a function is of the form y = f(x)ᵍ⁽ˣ dure involves taking the natural logarithm of both sides, then differentiating implicitly.
对数求导法是一个强有力的技巧，适用于 y = f(x)ᵍ⁽ˣ⁾ 这类函数或复杂的函数乘除组合。其步骤是先对等式两边取自然对数，再进行隐函数求导。
5) Solve for dy/dx by multiplying both sides by y, and replace y with the original expression.
步骤如下：(1) 写出 y = f(x)。(2) 两边取自然对数：ln y = ln f(x)。(3) 需要时用对数性质将右侧化简成 ln 项的和与差。(4) 两边对 x 进行隐函数求导：(1/y) dy/dx = 右侧的导数。(5) 两边乘以 y 解出 dy/dx，并将 y 替换回原函数表达式。
This technique transforms exponentiation into multiplication and products into sums, making differentiation much easier.
该方法将指数运算转化为乘法运算，将乘除转化为加减，从而使求导变得极为简便。
7. Worked Examples: Basic and Chain Rule | 例题解析：基本导数与链式法则
Example 1: Differentiate y = ln(x² + 4).
例 1： 求 y = ln(x² + 4) 的导数。
Solution: Let u = x² + 4, then u' = 2x. Using the chain rule, dy/dx = u'/u = (2x)/(x² + 4).
解：令 u = x² + 4，则 u' = 2x。由链式法则，dy/dx = u'/u = (2x)/(x² + 4)。
Example 2: Find d/dx [log₅ (sin x)], for sin x > 0.
例 2： 求 d/dx [log₅ (sin x)]，其中 sin x > 0。
Solution: The derivative of log₅ (u) is u'/(u ln 5). Here u = sin x, u' = cos x. So derivative = (cos x)/(sin x · ln 5) = cot x / ln 5.
解：log₅ (u) 的导数为 u'/(u ln 5)。这里 u = sin x, u' = cos x。因此导数为 (cos x)/(sin x · ln 5) = cot x / ln 5。
Example 3: y = ln(√x²+1). Use log properties to simplify first.
例 3： y = ln(√x²+1)。先用对数性质进行化简。
Solution: y = ln((x²+1)1/2 /(x²+1). This is much quicker than using the chain rule directly on the original function.
解：y = ln((x²+1)1/2) = (1/2) ln(x²+1)。求导得 dy/dx = (1/2) * (2x)/(x²+1) = x/(x²+1)。这比直接对原函数使用链式法则快得多。
8. Worked Examples: Logarithmic Differentiation | 例题解析：对数求导法
Example 4: Differentiate y = xˣ for x > 0.
例 4： 求 y = xˣ（x > 0）的导数。
ntiate implicitly: (1/y) dy/dx = ln x + x*(1/x) = ln x + 1. Hence dy/dx = y (ln x + 1) = xˣ (ln x + 1).
解：两边取对数：ln y = x ln x。隐函数求导：(1/y) dy/dx = ln x + x*(1/x) = ln x + 1。因此 dy/dx = y (ln x + 1) = xˣ (ln x + 1)。
Example 5: Find dy/dx if y = (x²+1)³ eˣ / √2x-5.
例 5： 若 y = (x²+1)³ eˣ / √2x-5，求 dy/dx。
y/dx = y [ 6x/(x²+1) + 1 - 1/(2x-5) ]. Then substitute back the original expression for y.
解：利用对数求导法处理这个复杂的乘除组合。ln y = 3 ln(x²+1) + x - (1/2) ln(2x-5)。求导得：(1/y) dy/dx = 3*(2x)/(x²+1) + 1 - (1/2)*(2)/(2x-5)。因此 dy/dx = y [ 6x/(x²+1) + 1 - 1/(2x-5) ]，再将 y 的原始表达式代回即可。
9. Common Mistakes to Avoid | 常见错误与提示
Mistake 1: Forgetting the chain rule. The derivative of ln(2x) is not 1/(2x); it
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