Derivative of \( \displaystyle 3 x \ln{\left(2 x + 1 \right)} - 3 x + \frac{3 \ln{\left(2 x + 1 \right)}}{2} \)
Problem 2.994 · hard
Differentiate \( \displaystyle f(x) = 3 x \ln{\left(2 x + 1 \right)} - 3 x + \frac{3 \ln{\left(2 x + 1 \right)}}{2} \).
- \[ \frac{d}{d x} \left(3 x \ln{\left(2 x + 1 \right)} - 3 x + \frac{3 \ln{\left(2 x + 1 \right)}}{2}\right) \]Start with the derivative of the function.✓ Proved
- \[ = - \frac{d}{d x} 3 x + \frac{d}{d x} 3 x \ln{\left(2 x + 1 \right)} + \frac{d}{d x} \frac{3 \ln{\left(2 x + 1 \right)}}{2} \]sumApply the sum rule.✓ Proved
- \[ = \frac{d}{d x} 3 x \ln{\left(2 x + 1 \right)} + \frac{d}{d x} \frac{3 \ln{\left(2 x + 1 \right)}}{2} - 3 \]constantThe derivative of 3*x is 3.✓ Proved
- \[ = 3 \frac{d}{d x} x \ln{\left(2 x + 1 \right)} + \frac{d}{d x} \frac{3 \ln{\left(2 x + 1 \right)}}{2} - 3 \]constant-multipleFactor out the constant 3 from the first term.✓ Proved
- \[ = 3 x \frac{d}{d x} \ln{\left(2 x + 1 \right)} + 3 \ln{\left(2 x + 1 \right)} \frac{d}{d x} x + \frac{d}{d x} \frac{3 \ln{\left(2 x + 1 \right)}}{2} - 3 \]productApply the product rule to the first term.✓ Proved
- \[ = 3 x \frac{d}{d x} \ln{\left(2 x + 1 \right)} + 3 \ln{\left(2 x + 1 \right)} + \frac{d}{d x} \frac{3 \ln{\left(2 x + 1 \right)}}{2} - 3 \]derivativeThe derivative of x is 1.✓ Proved
- \[ = \frac{3 x \frac{d}{d x} \left(2 x + 1\right)}{2 x + 1} + 3 \ln{\left(2 x + 1 \right)} + \frac{d}{d x} \frac{3 \ln{\left(2 x + 1 \right)}}{2} - 3 \]chainApply the chain rule to log(2*x + 1).✓ Proved
- \[ = \frac{6 x}{2 x + 1} + 3 \ln{\left(2 x + 1 \right)} + \frac{d}{d x} \frac{3 \ln{\left(2 x + 1 \right)}}{2} - 3 \]derivative algebraThe derivative of 2*x + 1 is 2. Simplify the fraction.✓ Proved
- \[ = \frac{6 x}{2 x + 1} + 3 \ln{\left(2 x + 1 \right)} + \frac{3 \frac{d}{d x} \ln{\left(2 x + 1 \right)}}{2} - 3 \]constant-multipleFactor out 3/2 from the last term.✓ Proved
- \[ = \frac{6 x}{2 x + 1} + 3 \ln{\left(2 x + 1 \right)} - 3 + \frac{3}{2 x + 1} \]derivative algebra algebraDifferentiate the log term again. Simplify the last term. Distribute the 3.✓ Proved
- \[ = 3 \ln{\left(2 x + 1 \right)} - 3 + \frac{6 x + 3}{2 x + 1} \]algebra algebraCombine the fractions. Factor the numerator.✓ Proved
- \[ = 3 \ln{\left(2 x + 1 \right)} \]algebra simplifyCancel the (2*x + 1) terms. Final simplification.✓ Proved
Answer \( 3 \ln{\left(2 x + 1 \right)} \)
✓ 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 2*x + 1 = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x + 1 = 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x + 1 = 0 |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x + 1 = 0 |
| 11 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x + 1 = 0 |
| 12 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x + 1 = 0 |
| 13 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x + 1 = 0 |
| 14 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x + 1 = 0 |
| 15 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x + 1 = 0 |
| 16 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x + 1 = 0 |
| 17 | ✓ 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 single-rule-per-step constraint. All labels are appropriate and the algebraic simplifications are sound.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly applies differentiation rules step-by-step, adhering to the single-rule-per-step constraint. All labels are appropriate and the algebraic simplifications are sound.gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: pass 2026-09-27gpt-oss:20b: pass 2026-09-27
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 2026-09-27 with SymPy 1.14.0.