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Mathematics · 术语表

什么是 derivative?

Definition 14.1 University Mathematics — Year 1 · 第 14 章 — Differentiation

f ⁣:I→Rf \colon I \to \R is differentiable at x0∈Ix_0 \in I when the difference quotient f(x)−f(x0)x−x0\frac{f(x) - f(x_0)}{x - x_0} has a (finite) limit as x→x0x \to x_0; the limit is written f′(x0)f'(x_0). Equivalently:

f(x0+h)=f(x0)+f′(x0) h+h ε(h),ε(h)→h→00,f(x_0 + h) = f(x_0) + f'(x_0)\,h + h\,\varepsilon(h), \qquad \varepsilon(h) \xrightarrow[h \to 0]{} 0 ,

the graph then admitting the tangent line y=f(x0)+f′(x0)(x−x0)y = f(x_0) + f'(x_0)(x - x_0). Differentiability at x0x_0 implies continuity at x0x_0 (read the display). ff is differentiable on II when it is at every point; ff is of class C1C^1 when moreover f′f' is continuous, and of class CkC^k when ff can be differentiated kk times with f(k)f^{(k)} continuous.

示例

Example 14.2

The converse of “differentiable ⇒\Rightarrow continuous” fails: ∣ ⋅ ∣\abs{\,\cdot\,} at 00. More surprisingly, differentiable does not imply C1C^1: the function f(x)=x2sin⁡1xf(x) = x^2 \sin\frac 1x (f(0)=0f(0) = 0) is differentiable everywhere, with f′(0)=0f'(0) = 0, but f′(x)=2xsin⁡1x−cos⁡1xf'(x) = 2x \sin\frac1x - \cos\frac 1x has no limit at 00 (Exercise 14.2).

Example 14.3 (Differentiable at exactly one point)

Let f(x)=x2f(x) = x^2 for x∈Qx \in \Q and f(x)=0f(x) = 0 for x∉Qx \notin \Q. At 00: ∣f(h)−0h∣≤∣h∣→0\bigl|\frac{f(h) - 0}{h}\bigr| \leq \abs h \to 0, so ff is differentiable at 00 with f′(0)=0f'(0) = 0. At any x0≠0x_0 \neq 0, ff is not even continuous: rational and irrational sequences converging to x0x_0 send ff to x02≠0x_0^2 \neq 0 and to 00 respectively (density, Theorem 10.14). So differentiability is a genuinely pointwise notion: it can hold at one point of R\R and nowhere else. The moral for practice: statements like the monotonicity criterion or Rolle require the derivative on an interval — possessing f′(x0)f'(x_0) at isolated points, however many, supports no global conclusion whatsoever.

Example 14.6 (Inverse derivatives, twice)

The theorem recomputes the classical derivatives with no limit work. For ln⁡=exp⁡−1\ln = \exp^{-1}: at y=exy = \eu^x,

(ln⁡)′(y)=1exp⁡′(x)=1ex=1y,(\ln)'(y) = \frac{1}{\exp'(x)} = \frac{1}{\eu^{x}} = \frac1y ,

valid for every y>0y > 0 since exp⁡′=exp⁡\exp' = \exp never vanishes. For arctan⁡=tan⁡−1\arctan = \tan^{-1}: at y=tan⁡xy = \tan x,

(arctan⁡)′(y)=11+tan⁡2x=11+y2,(\arctan)'(y) = \frac{1}{1 + \tan^2 x} = \frac{1}{1 + y^2} ,

using tan⁡′=1+tan⁡2>0\tan' = 1 + \tan^2 > 0. The closing insight: the formula converts knowledge about a function into knowledge about its inverse at the price of one substitution — and the substitution (x=ln⁡yx = \ln y, x=arctan⁡yx = \arctan y) is exactly the statement that the two variables live on opposite sides of the bijection.

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