Derivative of an inverse function
Problem 2.1548 · medium
Let \( \displaystyle f(x) = x^{3} + x^{2} + 2 x \), which is one-to-one. Find \( \displaystyle (f^{-1})'(-8) \).
- (f⁻¹)′(a) = 1 / f′(f⁻¹(a)). First find the x with f(x) = a; a small integer is worth trying.Reviewed
- \[ \left. x^{3} + x^{2} + 2 x \right|_{\substack{ x=-2 }} = -8 \]f(-2) = -8, so f⁻¹(-8) = -2.✓ Proved
- \[ \frac{d}{d x} \left(x^{3} + x^{2} + 2 x\right) = 3 x^{2} + 2 x + 2 \]Differentiate f.✓ Proved
- \[ \frac{1}{\left. 3 x^{2} + 2 x + 2 \right|_{\substack{ x=-2 }}} = \frac{1}{10} \]Take the reciprocal of f′ there.✓ Proved
Answer \( (f^{-1})'(-8) = \frac{1}{10} \)
✓ Nihil obstat Lines: 3 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | inverted f numerically at a ± 1e-10 and took the difference quotient |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-04gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: pass 2026-10-04gpt-oss:20b: pass 2026-10-04
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/inverse_derivative, checked 2026-10-04 with SymPy 1.14.0.