Derivative of \( \displaystyle 5 x \ln{\left(2 x + 2 \right)} - 5 x + 5 \ln{\left(x + 1 \right)} \)
Problem 2.2392 · hard Video
Differentiate \( \displaystyle f(x) = 5 x \ln{\left(2 x + 2 \right)} - 5 x + 5 \ln{\left(x + 1 \right)} \).
- \[ \frac{d}{d x} \left(5 x \ln{\left(2 x + 2 \right)} - 5 x + 5 \ln{\left(x + 1 \right)}\right) \]derivativeStart with the derivative of the function.✓ Proved
- \[ = - \frac{d}{d x} 5 x + \frac{d}{d x} 5 x \ln{\left(2 x + 2 \right)} + \frac{d}{d x} 5 \ln{\left(x + 1 \right)} \]sumApply the sum rule.✓ Proved
- \[ = \frac{d}{d x} 5 x \ln{\left(2 x + 2 \right)} + \frac{d}{d x} 5 \ln{\left(x + 1 \right)} - 5 \]constantDifferentiate the linear term.✓ Proved
- \[ = 5 \frac{d}{d x} x \ln{\left(2 x + 2 \right)} + 5 \frac{d}{d x} \ln{\left(x + 1 \right)} - 5 \]constant-multipleFactor out the constant 5.✓ Proved
- \[ = 5 x \frac{d}{d x} \ln{\left(2 x + 2 \right)} + 5 \ln{\left(2 x + 2 \right)} \frac{d}{d x} x + 5 \frac{d}{d x} \ln{\left(x + 1 \right)} - 5 \]productApply the product rule to the first term.✓ Proved
- \[ = \frac{5 x \frac{d}{d x} \left(2 x + 2\right)}{2 x + 2} + 5 \ln{\left(2 x + 2 \right)} + 5 \frac{d}{d x} \ln{\left(x + 1 \right)} - 5 \]chainApply the chain rule to the logarithm.✓ Proved
- \[ = \frac{10 x}{2 x + 2} + 5 \ln{\left(2 x + 2 \right)} + 5 \frac{d}{d x} \ln{\left(x + 1 \right)} - 5 \]derivativeDifferentiate the inner function 2x + 2.✓ Proved
- \[ = \frac{5 x}{x + 1} + 5 \ln{\left(2 x + 2 \right)} + 5 \frac{d}{d x} \ln{\left(x + 1 \right)} - 5 \]algebra algebraSimplify the fraction 2/(2x + 2) to 1/(x + 1). Distribute the 5.✓ Proved
- \[ = \frac{5 x}{x + 1} + 5 \ln{\left(2 x + 2 \right)} - 5 + \frac{5}{x + 1} \]logarithmic algebraDifferentiate the logarithm log(x + 1). Remove parentheses.✓ Proved
- \[ = 5 \ln{\left(2 x + 2 \right)} - 5 + \frac{5 x + 5}{x + 1} \]algebra algebraCombine the fractions with the same denominator. Factor out 5 from the numerator.✓ Proved
- \[ = 5 \ln{\left(2 x + 2 \right)} \]algebra simplifySimplify the fraction (x + 1)/(x + 1) to 1. Final simplification.✓ Proved
Answer \( 5 \ln{\left(2 x + 2 \right)} \)
✓ 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. How we verify
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 undefined where 2*x + 2 = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x + 2 = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x + 2 = 0 undefined where x + 1 = 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 1 = 0 |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 1 = 0 |
| 11 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 1 = 0 |
| 12 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 1 = 0 |
| 13 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 1 = 0 |
| 14 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 1 = 0 |
| 15 | ✓ 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 and algebraic simplifications in single-step increments. All labels are appropriate and the logic is sound.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly applies differentiation rules and algebraic simplifications in single-step increments. All labels are appropriate and the logic is sound.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The labels used are appropriate for the operations performed, and the algebraic simplifications are valid.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly applies differentiation rules and algebraic simplifications in a step-by-step manner, adhering to the one-change-per-step constraint.gpt-oss:20b: pass 2026-10-15
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 2026-10-18 with SymPy 1.14.0.