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Derivative rules from a table of values

Problem 2.1799 · easy

Use the table to find \( \displaystyle h'(2) \) for \( \displaystyle h(x) = f{\left(g(x) \right)} \). \[ \begin{array}{c|cccc} x & f(x) & f'(x) & g(x) & g'(x) \\ \hline 2 & 1 & -1 & 1 & 5 \\ 1 & -3 & 1 & \cdot & \cdot\end{array} \]

Chain rule: h′(a) = f′(g(a))·g′(a).
\[ 1 \cdot 5 = 5 \]
Read the values from the table and substitute.✓ Proved

Answer \( h'(2) = 5 \)

Lines: 1 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.

The full receipt

Line	Status	Checked by	Detail
1	Not checked	—	a sentence; read, not computed
2	✓ 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	differentiated concrete polynomials built to match the table

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The solution fails to explicitly evaluate g(2) to determine the correct argument for f'. While the final numerical answer is coincidentally correct because g(2)=1 and f'(1)=1, the reasoning skips the critical step of identifying that f' must be evaluated at x=1, not x=2. A student might incorrectly assume f'(g(2)) means f'(2) or fail to look up the value in the second row of the table.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-07
qwen3.6:27b-mlx: fail (error) 2026-10-07 — The solution fails to explicitly evaluate g(2) to determine the correct argument for f'. While the final numerical answer is coincidentally correct because g(2)=1 and f'(1)=1, the reasoning skips the critical step of identifying that f' must be evaluated at x=1, not x=2. A student might incorrectly assume f'(g(2)) means f'(2) or fail to look up the value in the second row of the table.
qwen3.6:27b-mlx: fail (error) 2026-10-07 — The solution substitutes f'(2) instead of f'(g(2)). Since g(2) = 1, the correct term is f'(1), which is 1, not -1. The setup incorrectly identifies the inner function's output value for the outer derivative.
gpt-oss:20b: pass 2026-10-07

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/derivative_from_table, checked 2026-10-07 with SymPy 1.14.0.