Inverse derivatives at corresponding points
Введено| English | Значение |
|---|---|
| inverse derivative/ɪnˈvɜːs dɪˈrɪvətɪv/ | производная обратной функции |
How do we reverse its local rate without taking the reciprocal of its output?
- A calibration converts length to a reading.
- This lesson studies inverse derivative 反函数导数: The rate of the inverse output with respect to its input, evaluated at the corresponding original point.
The method
- Let f be one-to-one on a stated interval and g=f⁻¹.
- The identity f(g(v))=v gives f′(g(v))g′(v)=1.
- When the derivatives exist and f′(g(v))≠0, g′(v)=1/f′(g(v)).
- The inverse is evaluated at the original output v, not automatically at the same original input x.
- The reciprocal 1/f(x) is a different function.
For f(x)=x²+1 on x≥0, why does g′(10) use f′(3), not f′(10)?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The inverse at v=10 corresponds to the original input 3, where f′(3)=6. Using f′(v) directly would give 1/20 - the classic wrong-point error.
A worked case
For f(x)=x²+1 on x≥0, g(v)=√(v−1), v≥1. At v=10, g(10)=3 and f′(3)=6, so g′(10)=1/6. On x≤0 the inverse branch is −√(v−1), with g(10)=−3 and g′(10)=−1/6. For f=e^x, g=ln v: f′(g(v))=e^(lnv)=v, hence g′=1/v for v>0. For f=sin x on −π/2≤x≤π/2, g=arcsin v. Within −1<v<1, cos(g(v))=√(1−v²)>0, so g′=1/√(1−v²). The nonnegative cosine follows from the restricted sine branch; it is not an arbitrary sign choice.
Inverse derivatives at corresponding points
Let f be one-to-one on a stated interval and g=f⁻¹
Identify the appropriate changing variable and validate the derivative denominator or local branch.
For the positive quadratic branch, find g′(10). Give it to 3 significant figures.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The corresponding positive input is 3 and f′(3)=6, giving 1/6.
Where it goes wrong
- Do not replace the corresponding-point denominator f′(g(v)) by f′(v).
- One-to-one behaviour alone does not ensure a finite inverse gradient: x³ is invertible, but its derivative is zero at zero and cube root has no finite derivative there.
- Root/arcsine inverse endpoints can also lack finite gradients.
- Different inverse branches can give different signs.
Not true: The derivative of an inverse is 1/f′(v), evaluated at the inverse input v. The derivative of the inverse at y is 1/f′(x), where x is the input with f(x)=y.
For g(v)=ln v, find g′(4).
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The exponential derivative at ln4 is 4, so its inverse rate is 1/4.
The derivative of the inverse at y is 1/f′(x), where x is the input with f(x)=y.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
True. The derivative of the inverse at y is 1/f′(x), where x is the input with f(x)=y.
Applying it
- Restrict the original domain, find the inverse input/output pair, differentiate f and evaluate its derivative at the recovered original input.
- Then take the reciprocal only if that finite gradient is nonzero.
- Swap horizontal and vertical units for the inverse rate.
- Check the result by differentiating an explicit inverse when one is available.
For the negative quadratic branch, find g′(10). Give it to 3 significant figures.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The corresponding negative input is −3 and f′(−3)=−6, giving −1/6.
Match each inverse-derivative fact to its statement.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Different branches give different signs: on x≤0 the same v=10 corresponds to -3 and g′(10)=-1/6.
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.
- Inverse derivative: The rate of the inverse output with respect to its input, evaluated at the corresponding original point.
- Let f be one-to-one on a stated interval and g=f⁻¹.
- Watch for: do not replace the corresponding-point denominator f′(g(v)) by f′(v).