从平均变化率到导数From Average Rate of Change to Derivatives
怎样从一段区间的平均变化，描述某一点附近的变化速度？How can the average change over an interval describe change near one point?
前置知识：Prerequisites: B01：函数与坐标。B01: functions and coordinates.
复习前置：Review prerequisites: B01 · 函数、坐标与参数Functions, Coordinates & Parameters
学完后你能：After this lesson you can: 计算割线斜率，解释导数的局部意义，并用切线估算附近值。Calculate secant slopes, explain the local derivative, and estimate nearby values with a tangent.
动手观察Explore
在 f(x)=x² 上选择起点和间隔，比较割线与切线。逐渐减小间隔，并切换起点的左右侧。Choose a starting point and interval on f(x)=x². Compare the secant and tangent as the interval shrinks on either side.
从观察到解释From observation to explanation
先看两点Start with two points
两点间的平均变化率是输出差除以输入差，也是连接两点的割线斜率。对于 f(x)=x²，x=1 与 x=2 之间，斜率为 (4−1)/(2−1)=3。Average rate of change is output difference divided by input difference, the slope of the secant through two points. For f(x)=x² between x=1 and x=2, the slope is (4−1)/(2−1)=3.
缩小间隔Shrink the interval
把第二点移近 x=1，割线斜率从 3 逐渐靠近 2。二次函数在该点的切线斜率是 2：输入在附近增加 0.1，输出约增加 2×0.1=0.2。Move the second point toward x=1 and the secant slope approaches 2. The tangent slope there is 2: a nearby input increase of 0.1 raises the output by about 2×0.1=0.2.
局部直线A local line
当 h 很小时，f(x+h)≈f(x)+f′(x)h。这个估算只在起点附近可靠；导数为正表示局部上升，为负表示局部下降。For small h, f(x+h)≈f(x)+f′(x)h. This estimate is local: a positive derivative indicates local increase, and a negative derivative local decrease.
公式与符号Formula and symbols
x 是起点，h 是非零的输入间隔；分数是平均变化率。若左右两侧都趋近同一个有限值，这个值就是导数 f′(x)。x is the starting input and h a nonzero interval; the fraction is the average rate of change. If both sides approach the same finite value, that value is the derivative f′(x).
留意这个区别Watch for this
导数要求左右变化率趋向同一有限值。|x| 在 0 左侧斜率为 −1、右侧为 1，因此在那里不可导；极小间隔还可能遭遇计算机舍入误差。A derivative requires both sides to approach the same finite value. |x| has left slope −1 and right slope 1 at 0, so it is not differentiable there; extremely tiny intervals can also suffer numerical rounding.
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