Differentiating Exponential and Logarithmic Functions | 指数函数与对数函数的微分

📚 Differentiating Exponential and Logarithmic Functions | 指数函数与对数函数的微分

Differentiating exponential and logarithmic functions is a core topic in A-Level Pure Mathematics, forming the foundation for solving complex calculus problems. In Edexcel's specification, understanding the derivatives of e^x and ln x, along with the chain, product, and quotient rules, is essential for success in differentiation-related questions.

指数函数与对数函数的微分是A-Level纯数学的核心内容，为解决复杂微积分问题奠定基础。在Edexcel考试中，掌握 e^x 和 ln x 的导数，以及链式法则、乘积法则和商法则，对顺利解答微分相关题目至关重要。

1. The Derivative of e^x | e^x 的导数

The natural exponential function f(x)=e^x has the unique property that its derivative is the function itself: f '(x)=e^x. This means the rate of change of e^x at any point is equal to the value of e^x at that point.

自然指数函数 f(x)=e^x 具有独特的性质，即其导数就是它本身：f '(x)=e^x。这意味着 e^x 在任意点的变化率等于该点的函数值。

d/dx (e^x) = e^x

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2. Differentiating e^kx | e^kx 的微分

When the exponent is a linear function kx, the derivative is found using the chain rule: d/dx (e^kx) = k e^kx. The coefficient k multiplies the original exponential function.

当指数为线性函数 kx 时，利用链式法则求导：d/dx (e^kx) = k e^kx。系数 k 乘上原指数函数。

d/dx (e^kx) = k e^kx

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3. The Derivative of ln x | ln x 的导数

ntiates to 1/x. This result is valid for x>0, and it is the inverse relationship with e^x that produces this simple derivative.

自然对数函数 f(x)=ln x 的导数为 1/x。该结果适用于 x>0，正是由于它与 e^x 互为反函数，才得到如此简洁的导数。

d/dx (ln x) = 1/x

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4. Differentiating ln(kx) | ln(kx) 的微分

ant, its derivative is 0, so d/dx (ln(kx)) = 1/x. Surprisingly, the constant k disappears.

利用对数性质，ln(kx)=ln k + ln x。由于 ln k 为常数，其导数为0，因此 d/dx (ln(kx)) = 1/x。出乎意料地，常数 k 消失了。

d/dx (ln(kx)) = 1/x

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5. Chain Rule with Exponential Functions | 链式法则与指数函数

For more complex exponentials like y=e^u, where u is a function of x, the chain rule gives dy/dx = e^u × du/dx. For example, differentiate e^{sin x}: let u=sin x, then dy/dx = e^{sin x} × cos x.

对于形如 y=e^u（其中 u 为 x 的函数）的更复杂指数函数，链式法则给出 dy/dx = e^u × du/dx。例如，对 e^{sin x} 求导：令 u=sin x，则 dy/dx = e^{sin x} × cos x。

dy/dx = e^u du/dx

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6. Chain Rule with ln(u) | 链式法则与 ln(u)

So to differentiate ln(3x²+1), set u=3x²+1, giving dy/dx = (1/(3x²+1)) × 6x = 6x/(3x²+1).

类似地，对于 y=ln(u)，dy/dx = (1/u) × du/dx。因此，对 ln(3x²+1) 求导，设 u=3x²+1，得 dy/dx = (1/(3x²+1)) × 6x = 6x/(3x²+1)。

d/dx [ln(u)] = (1/u) du/dx

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7. Product Rule Involving Exponentials and Logs | 含指数与对数的乘积法则

When differentiating products like xe^x, apply the product rule: if y=uv, then dy/dx = u'v + uv'. For y=x e^x, u=x, v=e^x, so dy/dx = 1·e^x + x·e^x = e^x(1+x).

对 xe^x 这类乘积求导时，应用乘积法则：若 y=uv，则 dy/dx = u'v + uv'。对于 y=x e^x，令 u=x, v=e^x，得 dy/dx = 1·e^x + x·e^x = e^x(1+x)。

dy/dx = u'v + uv'

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8. Quotient Rule with Exponentials | 含指数的商法则

For quotients like e^x/(x+1), use the quotient rule: if y=u/v, then dy/dx = (u'v - uv')/v². Let u=e^x, v=x+1, then dy/dx = (e^x(x+1) - e^x·1)/(x+1)² = e^xx/(x+1)².

对于 e^x/(x+1) 这类分式，使用商法则：若 y=u/v，则 dy/dx = (u'v - uv')/v²。令 u=e^x, v=x+1，得 dy/dx = (e^x(x+1) - e^x·1)/(x+1)² = e^xx/(x+1)²。

dy/dx = (u'v - uv')/v²

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9. Second Derivatives of Exponential Functions | 指数函数的二阶导数

Because the derivative of e^x is e^x, all higher-order derivatives of e^x are also e^x. For e^kx, the second derivative is k² e^kx. This makes exponential functions particularly straightforward to differentiate repeatedly.

由于 e^x 的导数就是 e^x，所以 e^x 的所有高阶导数仍然是 e^x。对于 e^kx，二阶导数为 k² e^kx。这使得指数函数在多次求导时格外简洁。

d²y/dx² = e^x ; d²/dx² (e^kx) = k² e^kx

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10. Solving Simple Differential Equations | 解简单微分方程

The relationship dy/dx = y implies y = Ae^x, where A is a constant. Similarly, dy/dx = ky gives y = Ae^kx. These appear when modelling exponential growth or decay, such as population growth or radioactive decay.

由 dy/dx = y 可得 y = Ae^x，其中 A 为常数。类似地，dy/dx = ky 给出 y = Ae^kx。此类方程出现在指数增长或衰减模型中，如人口增长或放射性衰变。

If dy/dx = ky, then y = Ae^kx

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11. Tangents and Normals | 切线与法线

To find the equation of a tangent to y=e^x the gradient: dy/dx = e⁰ = 1. The point is (0,1), so the tangent equation is y-1 = 1(x-0) ⇒ y = x+1. The normal gradient is -1, giving y-1 = -1(x-0) ⇒ y = -x+1.

求 y=e^x 在 x=0 处的切线方程，先计算斜率：dy/dx = e⁰ = 1。点为 (0,1)，故切线方程为 y-1 = 1(x-0) ⇒ y = x+1。法线斜率为 -1，得法线方程 y-1 = -1(x-0) ⇒ y = -x+1。

Tangent: y = x+1 ; Normal: y = -x+1

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12. Common Pitfalls and Exam Tips | 常见错误与应试技巧

Students often forget to use the chain rule for e^kx and incorrectly write the derivative as e^kx rather than k e^kx. Also, the derivative of ln(kx) is often mistaken for (1/(kx))·k, which simplifies to 1/x, so the answer is correct but the reasoning may be flawed if the chain rule is not applied properly. Always simplify expressions and factorise where possible to secure method marks.

学生常忘记对 e^kx 使用链式法则，误把导数写作 e^kx 而不是 k e^kx。另外，ln(kx) 的导数常被误认为 (1/(kx))·k，化简后得 1/x，答案虽正确，但若链式法则应用不当，推理过程可能出错。考试中应尽量化简表达式并因式分解，以确保获得方法分数。

Another frequent mistake is forgetting that ln x is only defined for x>0, so the derivative 1/x is only valid on that domain. When using the product or quotient rules, carefully identify u and v, and double-check the signs in the numerator of the quotient rule. Finally, practise differentiating combinations such as x²e^3x or ln(cos x) to build confidence.

另一个常见错误是忘记 ln x 仅在 x>0 时有定义，因此导数 1/x 也只在该区间有效。在使用乘积或商法则时，要仔细识别 u 和 v，并反复检查商法则分子中的符号。最后，通过练习诸如 x²e^3x 或 ln(cos x) 的组合函数求导来增强信心。 Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

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