∫Calc Practice

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 \( \displaystyle f(x) = 3 x \ln{\left(2 x + 1 \right)} - 3 x + \frac{3 \ln{\left(2 x + 1 \right)}}{2} \).
  1. \[ \frac{d}{d x} \left(3 x \ln{\left(2 x + 1 \right)} - 3 x + \frac{3 \ln{\left(2 x + 1 \right)}}{2}\right) \]
    Start with the derivative of the function.✓ Proved
  2. \[ = - \frac{d}{d x} 3 x + \frac{d}{d x} 3 x \ln{\left(2 x + 1 \right)} + \frac{d}{d x} \frac{3 \ln{\left(2 x + 1 \right)}}{2} \]
    sumApply the sum rule.✓ Proved
  3. \[ = \frac{d}{d x} 3 x \ln{\left(2 x + 1 \right)} + \frac{d}{d x} \frac{3 \ln{\left(2 x + 1 \right)}}{2} - 3 \]
    constantThe derivative of 3*x is 3.✓ Proved
  4. \[ = 3 \frac{d}{d x} x \ln{\left(2 x + 1 \right)} + \frac{d}{d x} \frac{3 \ln{\left(2 x + 1 \right)}}{2} - 3 \]
    constant-multipleFactor out the constant 3 from the first term.✓ Proved
  5. \[ = 3 x \frac{d}{d x} \ln{\left(2 x + 1 \right)} + 3 \ln{\left(2 x + 1 \right)} \frac{d}{d x} x + \frac{d}{d x} \frac{3 \ln{\left(2 x + 1 \right)}}{2} - 3 \]
    productApply the product rule to the first term.✓ Proved
  6. \[ = 3 x \frac{d}{d x} \ln{\left(2 x + 1 \right)} + 3 \ln{\left(2 x + 1 \right)} + \frac{d}{d x} \frac{3 \ln{\left(2 x + 1 \right)}}{2} - 3 \]
    derivativeThe derivative of x is 1.✓ Proved
  7. \[ = \frac{3 x \frac{d}{d x} \left(2 x + 1\right)}{2 x + 1} + 3 \ln{\left(2 x + 1 \right)} + \frac{d}{d x} \frac{3 \ln{\left(2 x + 1 \right)}}{2} - 3 \]
    chainApply the chain rule to log(2*x + 1).✓ Proved
  8. \[ = \frac{6 x}{2 x + 1} + 3 \ln{\left(2 x + 1 \right)} + \frac{d}{d x} \frac{3 \ln{\left(2 x + 1 \right)}}{2} - 3 \]
    derivative algebraThe derivative of 2*x + 1 is 2. Simplify the fraction.✓ Proved
  9. \[ = \frac{6 x}{2 x + 1} + 3 \ln{\left(2 x + 1 \right)} + \frac{3 \frac{d}{d x} \ln{\left(2 x + 1 \right)}}{2} - 3 \]
    constant-multipleFactor out 3/2 from the last term.✓ Proved
  10. \[ = \frac{6 x}{2 x + 1} + 3 \ln{\left(2 x + 1 \right)} - 3 + \frac{3}{2 x + 1} \]
    derivative algebra algebraDifferentiate the log term again. Simplify the last term. Distribute the 3.✓ Proved
  11. \[ = 3 \ln{\left(2 x + 1 \right)} - 3 + \frac{6 x + 3}{2 x + 1} \]
    algebra algebraCombine the fractions. Factor the numerator.✓ Proved
  12. \[ = 3 \ln{\left(2 x + 1 \right)} \]
    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
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0line 1 is the problem as stated
2✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
3✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
4✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
5✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
6✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
7✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where 2*x + 1 = 0
8✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where 2*x + 1 = 0
9✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where 2*x + 1 = 0
10✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where 2*x + 1 = 0
11✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where 2*x + 1 = 0
12✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where 2*x + 1 = 0
13✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where 2*x + 1 = 0
14✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where 2*x + 1 = 0
15✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where 2*x + 1 = 0
16✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where 2*x + 1 = 0
17✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
answer✓ Provedsympy 1.14.0final line against the stated answer: simplify(a - b) reduced to 0
log is undefined for non-positive arguments
answer, a second way✓ Provedsympy 1.14.0SymPy differentiated f directly and got the stated answer

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-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 2026-09-27
  • qwen3.6:27b-mlx: pass 2026-09-27
  • gpt-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.