Derivative of \( \displaystyle - \frac{5 x^{2}}{4} - 5 x + \left(\frac{5 x^{2}}{2} + 10 x + 10\right) \ln{\left(x + 2 \right)} \)
Problem 2.1909 · hard
Differentiate \( \displaystyle f(x) = - \frac{5 x^{2}}{4} - 5 x + \left(\frac{5 x^{2}}{2} + 10 x + 10\right) \ln{\left(x + 2 \right)} \).
- \[ \frac{d}{d x} \left(- \frac{5 x^{2}}{4} - 5 x + \left(\frac{5 x^{2}}{2} + 10 x + 10\right) \ln{\left(x + 2 \right)}\right) \]derivativeStart with the derivative of the function.✓ Proved
- \[ = \frac{d}{d x} \left(- 5 x\right) + \frac{d}{d x} \left(- \frac{5 x^{2}}{4}\right) + \frac{d}{d x} \left(\frac{5 x^{2}}{2} + 10 x + 10\right) \ln{\left(x + 2 \right)} \]sumApply the sum rule.✓ Proved
- \[ = \left(\frac{5 x^{2}}{2} + 10 x + 10\right) \frac{d}{d x} \ln{\left(x + 2 \right)} + \ln{\left(x + 2 \right)} \frac{d}{d x} \left(\frac{5 x^{2}}{2} + 10 x + 10\right) + \frac{d}{d x} \left(- 5 x\right) + \frac{d}{d x} \left(- \frac{5 x^{2}}{4}\right) \]productApply the product rule to the third term.✓ Proved
- \[ = \left(5 x + 10\right) \ln{\left(x + 2 \right)} + \left(\frac{5 x^{2}}{2} + 10 x + 10\right) \frac{d}{d x} \ln{\left(x + 2 \right)} + \frac{d}{d x} \left(- 5 x\right) + \frac{d}{d x} \left(- \frac{5 x^{2}}{4}\right) \]derivativeDifferentiate the polynomial and logarithmic terms.✓ Proved
- \[ = - \frac{5 x}{2} + \left(5 x + 10\right) \ln{\left(x + 2 \right)} + \left(\frac{5 x^{2}}{2} + 10 x + 10\right) \frac{d}{d x} \ln{\left(x + 2 \right)} + \frac{d}{d x} \left(- 5 x\right) \]derivativeDifferentiate the first term.✓ Proved
- \[ = - \frac{5 x}{2} + \left(5 x + 10\right) \ln{\left(x + 2 \right)} + \left(\frac{5 x^{2}}{2} + 10 x + 10\right) \frac{d}{d x} \ln{\left(x + 2 \right)} - 5 \]derivativeDifferentiate the second term.✓ Proved
- \[ = - \frac{5 x}{2} + \left(5 x + 10\right) \ln{\left(x + 2 \right)} - 5 + \frac{\frac{5 x^{2}}{2} + 10 x + 10}{x + 2} \]logarithmic algebraDifferentiate the natural logarithm. Factor out 5 from the numerator.✓ Proved
- \[ = \left(5 x + 10\right) \ln{\left(x + 2 \right)} \]algebra algebra algebra simplifyRewrite the quadratic term as a square. Simplify the fraction. Distribute the 5/2. Combine all terms.✓ Proved
Answer \( 5 \left(x + 2\right) \ln{\left(x + 2 \right)} \)
✓ Nihil obstat 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. Reviewers found nothing wrong with the explanation.
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 log is undefined for non-positive arguments |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 2 = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 2 = 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 2 = 0 |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 11 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 12 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 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: pass — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The algebraic simplifications in the final steps are valid and clearly labeled.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-09qwen3.6:27b-mlx: pass 2026-10-09 — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The algebraic simplifications in the final steps are valid and clearly labeled.gpt-oss:20b: pass 2026-10-09qwen3.6:27b-mlx: pass 2026-10-09
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 2026-10-09 with SymPy 1.14.0.