Derivative of \( \displaystyle \frac{\operatorname{acos}{\left(x - 3 \right)}}{x - 3} \)
Problem 2.448 · hard
Differentiate \( \displaystyle f(x) = \frac{\operatorname{acos}{\left(x - 3 \right)}}{x - 3} \).
- \[ \frac{d}{d x} \frac{\operatorname{acos}{\left(x - 3 \right)}}{x - 3} \]derivativeStart with the derivative of the function.✓ Proved
- \[ = \operatorname{acos}{\left(x - 3 \right)} \frac{d}{d x} \frac{1}{x - 3} + \frac{\frac{d}{d x} \operatorname{acos}{\left(x - 3 \right)}}{x - 3} \]quotientApply the quotient rule.✓ Proved
- \[ = \operatorname{acos}{\left(x - 3 \right)} \frac{d}{d x} \frac{1}{x - 3} - \frac{1}{\sqrt{1 - \left(x - 3\right)^{2}} \left(x - 3\right)} \]derivativeDifferentiate the numerator using the chain rule.✓ Proved
- \[ = - \frac{\operatorname{acos}{\left(x - 3 \right)}}{\left(x - 3\right)^{2}} - \frac{1}{\sqrt{1 - \left(x - 3\right)^{2}} \left(x - 3\right)} \]derivative algebraDifferentiate the denominator part. Simplify the terms.✓ Proved
- \[ = \frac{- \operatorname{acos}{\left(x - 3 \right)} - \frac{x - 3}{\sqrt{1 - \left(x - 3\right)^{2}}}}{\left(x - 3\right)^{2}} \]algebra simplifyCombine the fractions under a common denominator. Final simplification.✓ Proved
Answer \( - \frac{\operatorname{acos}{\left(x - 3 \right)}}{\left(x - 3\right)^{2}} - \frac{1}{\sqrt{1 - \left(x - 3\right)^{2}} \left(x - 3\right)} \)
Every line of this solution was proved by the computer algebra system SymPy. 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 | ✓ Proved | sympy 1.14.0 | line 1 is the problem as stated |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 acos is real only on [-1, 1] undefined where x - 3 = 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 acos is real only on [-1, 1] undefined where x - 3 = 0 undefined where 1 - (x - 3)**2 = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 acos is real only on [-1, 1] undefined where 1 - (x - 3)**2 = 0 undefined where x - 3 = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 acos is real only on [-1, 1] undefined where x - 3 = 0 undefined where 1 - (x - 3)**2 = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 acos is real only on [-1, 1] undefined where x - 3 = 0 undefined where 1 - (x - 3)**2 = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 acos is real only on [-1, 1] undefined where 1 - (x - 3)**2 = 0 undefined where x - 3 = 0 |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 acos is real only on [-1, 1] undefined where x - 3 = 0 undefined where 1 - (x - 3)**2 = 0 |
| answer, a second way | ✓ Proved | sympy 1.14.0 | SymPy differentiated f directly and got the stated answer |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — Step 2 incorrectly applies the quotient rule to the expression, which is a product of acos(x-3) and 1/(x-3). The quotient rule was applied as if the function were acos(x-3) divided by (x-3), but the structure in Step 1 is a product, so the product rule should have been used (or the quotient rule applied to the original fraction form, but Step 2's RHS is the product rule expansion, not the quotient rule expansion for u/v where u=acos and v=x-3 would be (u'v - uv')/v^2). Specifically, Step 2 writes the result as u'/v + u*v', which is the product rule for u * (1/v), but labels it 'quotient'. More critically, the label 'quotient' is incorrect for the algebraic manipulation shown, which is the product rule applied to acos(x-3) * (x-3)^-1. If the solver intended to use the quotient rule on acos(x-3)/(x-3), the result should be [ (x-3) * d/dx(acos(x-3)) - acos(x-3) * 1 ] / (x-3)^2. The step shown is [ d/dx(acos(x-3)) / (x-3) + acos(x-3) * d/dx(1/(x-3)) ], which is the product rule. Thus, the label 'quotient' is wrong for the operation performed.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-09-21 — Step 2 incorrectly applies the quotient rule to the expression, which is a product of acos(x-3) and 1/(x-3). The quotient rule was applied as if the function were acos(x-3) divided by (x-3), but the structure in Step 1 is a product, so the product rule should have been used (or the quotient rule applied to the original fraction form, but Step 2's RHS is the product rule expansion, not the quotient rule expansion for u/v where u=acos and v=x-3 would be (u'v - uv')/v^2). Specifically, Step 2 writes the result as u'/v + u*v', which is the product rule for u * (1/v), but labels it 'quotient'. More critically, the label 'quotient' is incorrect for the algebraic manipulation shown, which is the product rule applied to acos(x-3) * (x-3)^-1. If the solver intended to use the quotient rule on acos(x-3)/(x-3), the result should be [ (x-3) * d/dx(acos(x-3)) - acos(x-3) * 1 ] / (x-3)^2. The step shown is [ d/dx(acos(x-3)) / (x-3) + acos(x-3) * d/dx(1/(x-3)) ], which is the product rule. Thus, the label 'quotient' is wrong for the operation performed.gpt-oss:20b: pass 2026-09-21qwen3.6:27b-mlx: pass 2026-09-20 — The solution correctly applies the quotient rule, followed by differentiation of the numerator and denominator terms, and simplifies the result. All steps adhere to the one-change-per-step rule and use valid labels from the fixed vocabulary.gpt-oss:20b: fail (error) 2026-09-20 — Step 2 incorrectly applies the quotient rule; it actually uses the product rule on u·(1/v). Step 6 also incorrectly combines the two fractions: the first term has denominator (x‑3)·√(1‑(x‑3)²) while the second has (x‑3)², so the common denominator should be (x‑3)²·√(1‑(x‑3)²), not (x‑3)². These errors violate the one‑rule‑per‑step rule and produce an incorrect final expression.qwen3.6:27b-mlx: fail (error) 2026-09-20 — Step 2 applies the quotient rule but omits the square of the denominator in the first term, resulting in an algebraically incorrect expression. Additionally, Step 3 labels the application of the chain rule for acos as 'derivative', which is imprecise given the vocabulary distinction, but the algebraic error in Step 2 is the primary defect.gpt-oss:20b: fail (error) 2026-09-20 — Step 2 incorrectly applies the quotient rule. The derivative of a quotient u/v should be (u'v - uv')/v^2, but the solution uses u'/v + u·v', which is not equivalent. This is a clear mathematical error.
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
gemma4:26b, checked 2026-09-26 with SymPy 1.14.0.