一元链式法则The Single-Variable Chain Rule
输入经过两道处理后，局部变化率怎样传到最终输出？How does a local rate of change pass through two operations?
前置知识：Prerequisites: B07：平均变化率与导数。B07: average rates and derivatives.
复习前置：Review prerequisites: B07 · 从平均变化率到导数From Average Rate of Change to Derivatives
学完后你能：After this lesson you can: 拆分复合规则，在中间值处读取导数，并沿路径相乘。Split a composite rule, evaluate the outer derivative at the intermediate value, and multiply along the path.
动手观察Explore
改变输入 x，观察两道规则中的中间值、局部放大倍数和它们的乘积。Change input x and observe the intermediate value, the two local multipliers and their product.
从观察到解释From observation to explanation
拆成两道规则Split the rule
先算 u=2x+1，再算 y=u²。当 x=1 时，u=3，最终 y=9。先确认中间值，才能在正确位置读取外层变化率。First compute u=2x+1, then y=u². At x=1, u=3 and y=9. Find the intermediate value before reading the outer rate at the correct location.
沿路径相乘Multiply along the path
内层的变化率是 2；外层 y=u² 在 u=3 处的变化率是 2u=6。因此输入处的总变化率是 6×2=12。The inner rate is 2; the outer rate of y=u² at u=3 is 2u=6. The total rate at the input is therefore 6×2=12.
微小变化的传递Passing on a small change
输入略增 Δx 时，先有 Δu≈g′(x)Δx，再有 Δy≈f′(u)Δu；把两个近似接起来，就看见相乘的原因。For a small input change Δx, first Δu≈g′(x)Δx and then Δy≈f′(u)Δu. Combining the two approximations explains the multiplication.
公式与符号Formula and symbols
x 是输入，u 是中间值，y 是输出；g 与 f 是两道可导规则，g′(x) 是内层变化率，f′(u) 是外层在中间值 u 的变化率。x is the input, u the intermediate value and y the output; g and f are differentiable rules, g′(x) is the inner rate, and f′(u) is the outer rate evaluated at u.
留意这个区别Watch for this
外层导数必须在 u=g(x) 处求值，不能直接把 x 当成 u。某个局部导数为零时，这条路径的总导数也为零。Evaluate the outer derivative at u=g(x), not directly at x. If one local derivative is zero, the total derivative along this path is zero.
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