Product and quotient rules with retained exclusions
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| quotient rule/ˈkwəʊʃənt ruːl/ | quy tắc thương |
Why does multiplying their separate derivatives miss most of the rate?
- Both factors in a changing area can vary.
- This lesson studies quotient rule 商法则: The rule for differentiating a ratio of two changing functions when the denominator is nonzero.
The method
- For differentiable u,v, (uv)′=u′v+uv′.
- For v≠0, (u/v)′=(u′v−uv′)/v².
- The product rule has two contributions: change in either factor while the other provides its current value.
- The quotient rule uses the derivative of the numerator times the denominator, minus the numerator times the derivative of the denominator, all over v².
For w=(x²-1)/(x-1), why is the simplified derivative w′=1 not valid at x=1?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The factor (x-1) cancels to x+1 only for x≠1. The derivative of the simplified form shares that exclusion - the hole in the graph is not repaired by algebra.
A worked case
For y=x²e^x, y′=2xe^x+x²e^x=e^x(x²+2x), so y′(1)=3e. For z=(x²+1)/(x−1), x≠1, z′=[2x(x−1)−(x²+1)]/(x−1)²=(x²−2x−1)/(x−1)²; at x=2 it is −1. Polynomial division gives z=x+1+2/(x−1), hence z′=1−2/(x−1)², an independent check. For w=(x²−1)/(x−1), x≠1, cancellation gives w=x+1 and w′=1 only on that original domain. The missing point at x=1 is not restored; its continuous extension would be a different function.
Product and quotient rules with retained exclusions
For differentiable u,v, (uv)′=u′v+uv′
Choose the derivative rule and preserve every coefficient, inner rate and input restriction.
For x²e^x, find its derivative at x=0.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
e⁰(0²+2×0)=0.
Where it goes wrong
- Multiplying u′v′ is not the derivative of uv.
- Reversing the quotient numerator flips its sign.
- The squared denominator does not eliminate the original pole or hole.
- Simplifying before differentiating can help, but it must preserve exclusions.
- A stationary quotient point must also lie in the original domain.
Not true: Cancelling a factor from a quotient restores every excluded input. Cancelling does not change the original domain; the excluded inputs stay excluded.
For the worked z, find z′(2).
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
(4−4−1)/(2−1)²=−1.
Cancelling does not change the original domain.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
True. Cancelling does not change the original domain; the excluded inputs stay excluded.
Applying it
- Name u,v and write u′,v′ before substitution.
- Keep brackets around both quotient numerator terms until expansion is complete.
- Factor a product derivative only after collecting both contributions.
- Compare an expanded or divided form when available and substitute into the original domain before evaluating or solving y′=0.
For w=(x²−1)/(x−1), find w′(2).
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Cancellation gives x+1 only for x≠1; its derivative at the allowed input 2 is 1.
Match each rule to its formula.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The quotient numerator subtracts in the order u′v-uv′; reversing it flips the sign, and the squared denominator never removes the original exclusions.
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.
- Quotient rule: The rule for differentiating a ratio of two changing functions when the denominator is nonzero.
- For differentiable u,v, (uv)′=u′v+uv′.
- Watch for: multiplying u′v′ is not the derivative of uv.