Chain rules for nested functions and restricted inputs
Введено| English | Значение |
|---|---|
| composite derivative/ˈkɒmpəzɪt dɪˈrɪvətɪv/ | производная сложной функции |
How do we keep every link in its rate of change?
- A sensor converts a changing input through several formulas.
- This lesson studies composite derivative 复合函数导数: The outer derivative evaluated at the inner output, multiplied by the derivative of that inner input.
The method
- For y=f(u) and u=g(x), dy/dx=f′(g(x))g′(x).
- Evaluate the outer derivative at the actual inner expression.
- For several nested functions, multiply every inner-rate factor.
- Domain restrictions come from each stage; the differentiated expression cannot restore a forbidden original input.
For z=e^(x²), why is z′(0)=0 even though the exponential derivative is never zero?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
z′=e^(x²)·2x. The outer factor e^(x²) is always positive, but the inner rate 2x vanishes at 0, so the product is zero - a flat point on the curve, not a rule failure.
A worked case
For y=(3x−1)^4, set u=3x−1: y′=4u³×3=12(3x−1)³, giving y′(0)=−12. For z=e^(x²), z′=e^(x²)×2x; z′(0)=0 although the outer exponential derivative is never zero. For w=ln(2x+1), x>−1/2, w′=2/(2x+1); w′(1)=2/3. For r=sin((2x+1)²), first set u=(2x+1)²: r′=cos((2x+1)²)×2(2x+1)×2=4(2x+1)cos((2x+1)²). For s=√(5−x²), the original domain is −√5≤x≤√5, but s′=−x/√(5−x²) is finite only in the open interval.
Chain rules for nested functions and restricted inputs
For y=f(u) and u=g(x), dy/dx=f′(g(x))g′(x)
Choose the derivative rule and preserve every coefficient, inner rate and input restriction.
For (3x−1)^4, find the derivative at x=0.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
4(−1)³×3=−12.
Where it goes wrong
- Do not stop after differentiating the outer power or exponential.
- For sin(u²), the cosine input stays u²; it is not replaced by 2u.
- Missing one link in a nested derivative loses a factor.
- A zero inner derivative can give zero total derivative, so dividing by it without checking can lose valid cases.
- A square root can exist at an endpoint without a finite derivative there.
Not true: The derivative of e^(x²) is e^(x²) without an inner-rate factor. The chain rule multiplies by the inner derivative: the derivative of e^(x²) is 2x·e^(x²).
For e^(x²), find the derivative at x=0.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The inner rate is 2x, which is zero at zero.
The chain rule multiplies by the inner derivative.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
True. The chain rule multiplies by the inner derivative: the derivative of e^(x²) is 2x·e^(x²).
Applying it
- Work from the outside inward and record each input change.
- Multiply the factors, then simplify.
- For a short polynomial, expansion gives an independent alternative check.
- Retain every log/root restriction and interpret whether a zero rate comes from the outer function, the inner input or both.
For ln(2x+1), find the derivative at x=1. Give it to 3 significant figures.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The inner rate is 2; 2/(2×1+1)=2/3.
Match each composite to its derivative.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Differentiate the outer function AT the inner expression, then multiply by the inner derivative; a nested chain multiplies every link.
In your course
- AQA A-level Mathematics (7357), specification section G.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Composite derivative: The outer derivative evaluated at the inner output, multiplied by the derivative of that inner input.
- For y=f(u) and u=g(x), dy/dx=f′(g(x))g′(x).
- Watch for: do not stop after differentiating the outer power or exponential.