Chain, product, quotient and implicit differentiation
Введено| English | Значение |
|---|---|
| chain rule/tʃeɪn ruːl/ | правило дифференцирования сложной функции |
Does the inside expression change too?
- A cost curve is a power of a changing expression. Differentiating the outer power alone misses the rate of its input.
- This lesson studies chain rule 链式法则: The rule that multiplies the outer derivative by the inner derivative for a composite function.
The method
- For y=f(g(x)), y prime=f prime(g(x))g prime(x).
- For uv, differentiate to u prime v+uv prime.
- For u/v, use (u prime v-uv prime)/v², where v≠0.
Why is the derivative of x·eˣ not simply 1·eˣ?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Each factor varies, so both contribute: with u=x and v=eˣ, the product rule gives u′v+uv′=eˣ+xeˣ. Multiplying the separate derivatives drops the uv′ term.
A worked case
For y=(3x+1)^4, y prime=4(3x+1)^3×3=12(3x+1)^3. At x=0, the gradient is 12. For x²+y²=25, differentiate implicitly: 2x+2y y prime=0, so y prime=-x/y when y≠0.
Chain, product, quotient and implicit differentiation
For y=f(g(x)), y prime=f prime(g(x))g prime(x)
Compare the model with the worked case and explain one change.
Find the derivative of (3x+1)^4 at x=0.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Chain rule gives 4(3x+1)³×3; at zero this is 12.
Where it goes wrong
- A derivative of a product is not the product of the derivatives.
- In implicit differentiation, every differentiated function of y brings a dy/dx factor.
- A quotient's denominator is squared.
Not true: The derivative of u(x)v(x) is u prime times v prime. The product rule is (uv)′=u′v+uv′: for x·x the derivative is 2x, not 1×1=1.
For x²+y²=25, find dy/dx at (3,4).
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Implicit differentiation gives 2x+2y y prime=0, so y prime=-3/4.
The derivative of u(x)v(x) is u prime times v prime.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The product rule is (uv)′=u′v+uv′: for x·x the derivative is 2x, not 1×1=1.
Applying it
- Choose a useful form before differentiating: expanding a short polynomial may be simpler.
- For related rates, write the relation in symbols, differentiate with respect to time, then substitute measured values.
For y=xe^x, find dy/dx at x=0.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Product rule gives e^x+xe^x; at zero this is 1.
Match each rule to the situation it handles.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Composites chain the outer and inner derivatives; products add u′v+uv′; implicit relations bring a dy/dx factor for every differentiated y term.
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.
- Chain rule: The rule that multiplies the outer derivative by the inner derivative for a composite function.
- For y=f(g(x)), y prime=f prime(g(x))g prime(x).
- Watch for: A derivative of a product is not the product of the derivatives.