Derivative of \( \displaystyle 5 x \ln{\left(2 x - 3 \right)} - 5 x - \frac{15 \ln{\left(2 x - 3 \right)}}{2} \)
Problem 2.447 · hard
Differentiate \( \displaystyle f(x) = 5 x \ln{\left(2 x - 3 \right)} - 5 x - \frac{15 \ln{\left(2 x - 3 \right)}}{2} \).
- \[ \frac{d}{d x} \left(5 x \ln{\left(2 x - 3 \right)} - 5 x - \frac{15 \ln{\left(2 x - 3 \right)}}{2}\right) \]derivativeDifferentiate the entire function.✓ Proved
- \[ = - \frac{d}{d x} 5 x + \frac{d}{d x} 5 x \ln{\left(2 x - 3 \right)} - \frac{d}{d x} \frac{15 \ln{\left(2 x - 3 \right)}}{2} \]sumApply the sum rule for derivatives.✓ Proved
- \[ = - \frac{d}{d x} 5 x + \frac{d}{d x} 5 x \ln{\left(2 x - 3 \right)} - \frac{15 \frac{d}{d x} \ln{\left(2 x - 3 \right)}}{2} \]constant-multiplePull out the constant factor from the third term.✓ Proved
- \[ = - \frac{d}{d x} 5 x + 5 \frac{d}{d x} x \ln{\left(2 x - 3 \right)} - \frac{15 \frac{d}{d x} \ln{\left(2 x - 3 \right)}}{2} \]constant-multiplePull out the constant factor from the first term.✓ Proved
- \[ = 5 x \frac{d}{d x} \ln{\left(2 x - 3 \right)} + 5 \ln{\left(2 x - 3 \right)} \frac{d}{d x} x - \frac{d}{d x} 5 x - \frac{15 \frac{d}{d x} \ln{\left(2 x - 3 \right)}}{2} \]productApply the product rule to the first term.✓ Proved
- \[ = 5 x \frac{d}{d x} \ln{\left(2 x - 3 \right)} + 5 \ln{\left(2 x - 3 \right)} - \frac{d}{d x} 5 x - \frac{15 \frac{d}{d x} \ln{\left(2 x - 3 \right)}}{2} \]derivativeDifferentiate x.✓ Proved
- \[ = 5 x \frac{d}{d x} \ln{\left(2 x - 3 \right)} + 5 \ln{\left(2 x - 3 \right)} - 5 \frac{d}{d x} x - \frac{15 \frac{d}{d x} \ln{\left(2 x - 3 \right)}}{2} \]constant-multiplePull out the constant factor from the second term.✓ Proved
- \[ = 5 x \frac{d}{d x} \ln{\left(2 x - 3 \right)} + 5 \ln{\left(2 x - 3 \right)} - \frac{15 \frac{d}{d x} \ln{\left(2 x - 3 \right)}}{2} - 5 \]derivative productDifferentiate x. Distribute the constant 5 into the parentheses.✓ Proved
- \[ = \frac{5 x \frac{d}{d x} \left(2 x - 3\right)}{2 x - 3} + 5 \ln{\left(2 x - 3 \right)} - \frac{15 \frac{d}{d x} \ln{\left(2 x - 3 \right)}}{2} - 5 \]chainApply the chain rule to log(2x - 3).✓ Proved
- \[ = \frac{10 x}{2 x - 3} + 5 \ln{\left(2 x - 3 \right)} - \frac{15 \frac{d}{d x} \ln{\left(2 x - 3 \right)}}{2} - 5 \]derivative algebra algebraDifferentiate 2x - 3. Simplify the fraction inside the parentheses. Multiply 5*x by 2/(2*x - 3).✓ Proved
- \[ = \frac{10 x}{2 x - 3} + 5 \ln{\left(2 x - 3 \right)} - 5 - \frac{15 \frac{d}{d x} \left(2 x - 3\right)}{2 \left(2 x - 3\right)} \]chainApply the chain rule to the third term's log component.✓ Proved
- \[ = \frac{10 x}{2 x - 3} + 5 \ln{\left(2 x - 3 \right)} - 5 - \frac{15}{2 x - 3} \]derivative algebra algebraDifferentiate 2x - 3. Simplify the fraction inside the parentheses. Multiply (15/2) by 2/(2*x - 3).✓ Proved
- \[ = 5 \ln{\left(2 x - 3 \right)} - 5 + \frac{10 x - 15}{2 x - 3} \]algebra algebraCombine the terms with the common denominator (2*x - 3). Factor 5 out of the numerator (10x - 15).✓ Proved
- \[ = 5 \ln{\left(2 x - 3 \right)} \]algebra simplifyCancel the common factor (2*x - 3) in the fraction. Subtract 5 - 5.✓ Proved
Answer \( 5 \log{\left(2 x - 3 \right)} \)
✓ 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 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 3 = 0 |
| 11 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 3 = 0 |
| 12 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 3 = 0 |
| 13 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 3 = 0 |
| 14 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 3 = 0 |
| 15 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 3 = 0 |
| 16 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 3 = 0 |
| 17 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 3 = 0 |
| 18 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 3 = 0 |
| 19 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 3 = 0 |
| 20 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 3 = 0 |
| 21 | ✓ 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: fail (style) — Step 9 distributes the constant 5 into the parentheses, which is an algebraic simplification, but it was labeled as "product". The correct label for this step is "algebra".qwen3.6:27b-mlx: pass — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. All labels are appropriate for the operations performed.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-09-21 — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. All labels are appropriate for the operations performed.gpt-oss:20b: fail (style) 2026-09-21 — Step 9 distributes the constant 5 into the parentheses, which is an algebraic simplification, but it was labeled as "product". The correct label for this step is "algebra".qwen3.6:27b-mlx: pass 2026-09-20 — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. All labels are appropriate for the operations performed.gpt-oss:20b: fail (style) 2026-09-20 — Step 9 applies a distribution of the constant 5 into the parentheses, which is a simple algebraic simplification, yet it is labeled as "product". The correct label for this step should be "algebra".qwen3.6:27b-mlx: pass 2026-09-20 — The solution correctly applies differentiation rules and algebraic simplifications in a step-by-step manner. Each step changes only one aspect of the expression, and the labels used are appropriate for the operations performed.gpt-oss:20b: pass 2026-09-20
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.