Derivative of \( \displaystyle - 3 x + \left(3 x + \frac{6}{5}\right) \ln{\left(5 x + 2 \right)} \)
Problem 2.2415 · hard Video
Differentiate \( \displaystyle f(x) = - 3 x + \left(3 x + \frac{6}{5}\right) \ln{\left(5 x + 2 \right)} \).
- \[ \frac{d}{d x} \left(- 3 x + \left(3 x + \frac{6}{5}\right) \ln{\left(5 x + 2 \right)}\right) \]derivativeStart with the derivative of the function.✓ Proved
- \[ = \frac{d}{d x} \left(- 3 x\right) + \frac{d}{d x} \left(3 x + \frac{6}{5}\right) \ln{\left(5 x + 2 \right)} \]sumApply the sum rule.✓ Proved
- \[ = \frac{d}{d x} \left(- 3 x\right) + \frac{d}{d x} 3 x \ln{\left(5 x + 2 \right)} + \frac{d}{d x} \frac{6 \ln{\left(5 x + 2 \right)}}{5} \]productApply the product rule to the second term.✓ Proved
- \[ = 3 x \frac{d}{d x} \ln{\left(5 x + 2 \right)} + \ln{\left(5 x + 2 \right)} \frac{d}{d x} 3 x + \frac{d}{d x} \left(- 3 x\right) + \frac{d}{d x} \frac{6 \ln{\left(5 x + 2 \right)}}{5} \]productApply the product rule to the term 3*x*log(5*x + 2).✓ Proved
- \[ = 3 x \frac{d}{d x} \ln{\left(5 x + 2 \right)} + \ln{\left(5 x + 2 \right)} \frac{d}{d x} 3 x + \frac{d}{d x} \left(- 3 x\right) + \frac{6 \frac{d}{d x} \ln{\left(5 x + 2 \right)}}{5} \]constant-multipleApply the constant multiple rule to the last term.✓ Proved
- \[ = 3 x \frac{d}{d x} \ln{\left(5 x + 2 \right)} + 3 \ln{\left(5 x + 2 \right)} + \frac{d}{d x} \left(- 3 x\right) + \frac{6 \frac{d}{d x} \ln{\left(5 x + 2 \right)}}{5} \]derivativeDifferentiate the term 3*x.✓ Proved
- \[ = \frac{3 x \frac{d}{d x} \left(5 x + 2\right)}{5 x + 2} + 3 \ln{\left(5 x + 2 \right)} + \frac{d}{d x} \left(- 3 x\right) + \frac{6 \frac{d}{d x} \ln{\left(5 x + 2 \right)}}{5} \]chainApply the chain rule to log(5*x + 2).✓ Proved
- \[ = \frac{15 x}{5 x + 2} + 3 \ln{\left(5 x + 2 \right)} + \frac{d}{d x} \left(- 3 x\right) + \frac{6 \frac{d}{d x} \ln{\left(5 x + 2 \right)}}{5} \]derivative algebraDifferentiate the inner function 5*x + 2. Simplify the term 3*x*5/(5*x + 2).✓ Proved
- \[ = \frac{15 x}{5 x + 2} + 3 \ln{\left(5 x + 2 \right)} + \frac{d}{d x} \left(- 3 x\right) + \frac{6 \frac{d}{d x} \left(5 x + 2\right)}{5 \left(5 x + 2\right)} \]algebraDistribute the 3*x into the fraction.✓ Proved
- \[ = \frac{15 x}{5 x + 2} + 3 \ln{\left(5 x + 2 \right)} + \frac{d}{d x} \left(- 3 x\right) + \frac{6}{5 x + 2} \]derivative algebraDifferentiate the inner function 5*x + 2 again. Simplify (6/5)*5/(5*x + 2).✓ Proved
- \[ = \frac{15 x}{5 x + 2} + 3 \ln{\left(5 x + 2 \right)} - 3 + \frac{6}{5 x + 2} \]derivativeDifferentiate the term -3*x.✓ Proved
- \[ = 3 \ln{\left(5 x + 2 \right)} - 3 + \frac{15 x + 6}{5 x + 2} \]algebra algebraCombine the fractions. Factor out 3 from the numerator.✓ Proved
- \[ = 3 \ln{\left(5 x + 2 \right)} \]simplify simplifyCancel the common factor (5*x + 2). Combine the constant terms.✓ Proved
Answer \( 3 \ln{\left(5 x + 2 \right)} \)
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. A reviewer's objection to the explanation stands; see the receipt. 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 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 5*x + 2 = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 5*x + 2 = 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 5*x + 2 = 0 |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 5*x + 2 = 0 |
| 11 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 5*x + 2 = 0 |
| 12 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 5*x + 2 = 0 |
| 13 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 5*x + 2 = 0 |
| 14 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 5*x + 2 = 0 |
| 15 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 5*x + 2 = 0 |
| 16 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 5*x + 2 = 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: fail (error) — Step 3 incorrectly applies the sum rule to the product (3*x+6/5)*log(5*x+2); it should use the product rule on the entire product, not split it into two terms. This violates the rule‑granularity requirement and leads to an incorrect derivative.qwen3.6:27b-mlx: fail (error) — Step 3 incorrectly labels the expansion of a product into a sum as 'product'; the product rule is applied in Step 4. Step 3 is actually an algebraic expansion (distributive property) or a 'sum' rule application after rewriting the term, but 'product' is the wrong label for splitting a product into a sum of derivatives. More critically, Step 10 is labeled 'algebra' but performs the chain rule expansion on the last term, which is a differentiation step, not just algebra. Actually, looking closer at Step 10: it changes `Derivative(log(5*x + 2), x)` to `(1/(5*x + 2))*(Derivative(5*x + 2, x))`. This is the chain rule. The label 'algebra' is incorrect for a differentiation step. Step 3 is also problematic: it splits `(3*x + 6/5)*log(...)` into `3*x*log(...) + (6/5)*log(...)`. This is algebraic distribution, not the product rule. The label 'product' is wrong. However, the prompt says 'A computer algebra system has already confirmed that consecutive lines are algebraically equal'. Step 3 changes the structure from one derivative term to two. This is applying the linearity of the derivative (sum rule) to the expanded product. But the label is 'product'. The product rule is `d(uv) = u'dv + udv'`. Step 3 does `d(u+v) = du + dv` effectively on the expanded form? No, Step 3 is `d((A+B)C) -> d(AC) + d(BC)`. This is valid algebra inside the derivative operator, but the label 'product' usually refers to the differentiation rule. If it's just algebraic manipulation inside the derivative, it should be 'algebra' or 'rewrite'. If it's applying the sum rule to the terms inside, it should be 'sum'. Calling it 'product' is misleading/wrong. Let's look at Step 10 again. Line 9: `... + (6/5)*Derivative(log(5*x + 2), x)` Line 10: `... + (6/5)*(1/(5*x + 2))*(Derivative(5*x + 2, x))` This step applies the chain rule to `log(5*x+2)`. The label is 'algebra'. This is a clear error. Differentiation is not algebra. Also Step 3: Line 2: `Derivative((3*x + 6/5)*log(5*x + 2), x)` Line 3: `Derivative(3*x*log(5*x + 2), x) + Derivative((6/5)*log(5*x + 2), x)` This step distributes the derivative over the sum inside the product? No, it expands the product `(A+B)C` to `AC + BC` inside the derivative. This is an algebraic step (distribution). The label 'product' is incorrect. It should be 'algebra' or 'rewrite'. So we have two labeling errors. Step 10 is a differentiation step labeled 'algebra'. Step 3 is an algebraic step labeled 'product'. The verdict is fail.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-19 — Step 3 incorrectly labels the expansion of a product into a sum as 'product'; the product rule is applied in Step 4. Step 3 is actually an algebraic expansion (distributive property) or a 'sum' rule application after rewriting the term, but 'product' is the wrong label for splitting a product into a sum of derivatives. More critically, Step 10 is labeled 'algebra' but performs the chain rule expansion on the last term, which is a differentiation step, not just algebra. Actually, looking closer at Step 10: it changes `Derivative(log(5*x + 2), x)` to `(1/(5*x + 2))*(Derivative(5*x + 2, x))`. This is the chain rule. The label 'algebra' is incorrect for a differentiation step. Step 3 is also problematic: it splits `(3*x + 6/5)*log(...)` into `3*x*log(...) + (6/5)*log(...)`. This is algebraic distribution, not the product rule. The label 'product' is wrong. However, the prompt says 'A computer algebra system has already confirmed that consecutive lines are algebraically equal'. Step 3 changes the structure from one derivative term to two. This is applying the linearity of the derivative (sum rule) to the expanded product. But the label is 'product'. The product rule is `d(uv) = u'dv + udv'`. Step 3 does `d(u+v) = du + dv` effectively on the expanded form? No, Step 3 is `d((A+B)C) -> d(AC) + d(BC)`. This is valid algebra inside the derivative operator, but the label 'product' usually refers to the differentiation rule. If it's just algebraic manipulation inside the derivative, it should be 'algebra' or 'rewrite'. If it's applying the sum rule to the terms inside, it should be 'sum'. Calling it 'product' is misleading/wrong. Let's look at Step 10 again. Line 9: `... + (6/5)*Derivative(log(5*x + 2), x)` Line 10: `... + (6/5)*(1/(5*x + 2))*(Derivative(5*x + 2, x))` This step applies the chain rule to `log(5*x+2)`. The label is 'algebra'. This is a clear error. Differentiation is not algebra. Also Step 3: Line 2: `Derivative((3*x + 6/5)*log(5*x + 2), x)` Line 3: `Derivative(3*x*log(5*x + 2), x) + Derivative((6/5)*log(5*x + 2), x)` This step distributes the derivative over the sum inside the product? No, it expands the product `(A+B)C` to `AC + BC` inside the derivative. This is an algebraic step (distribution). The label 'product' is incorrect. It should be 'algebra' or 'rewrite'. So we have two labeling errors. Step 10 is a differentiation step labeled 'algebra'. Step 3 is an algebraic step labeled 'product'. The verdict is fail.gpt-oss:20b: fail (error) 2026-10-19 — Step 3 incorrectly applies the sum rule to the product (3*x+6/5)*log(5*x+2); it should use the product rule on the entire product, not split it into two terms. This violates the rule‑granularity requirement and leads to an incorrect derivative.qwen3.6:27b-mlx: fail (error) 2026-10-16 — Step 3 incorrectly applies the product rule to split a sum of terms; the transition from step 2 to step 3 is a distributive algebraic expansion (or sum rule), not a product rule application. Additionally, Step 10 is labeled 'algebra' but performs a differentiation (chain rule) on the last term, which is a logical error in labeling and process.gpt-oss:20b: fail (error) 2026-10-16 — Step 3 incorrectly applies the product rule. It expands (3*x + 6/5)*log(5*x+2) into 3*x*log + (6/5)*log, which is an algebraic distribution, not a product‑rule application. The label should be "algebra", not "product".qwen3.6:27b-mlx: fail (error) 2026-10-16 — Step 3 is labeled 'product' but performs a 'sum' expansion (distributing the derivative over the sum inside the term). Step 10 is labeled 'algebra' but performs a 'chain' rule application on the last term, which was not yet differentiated in the previous step.gpt-oss:20b: fail (error) 2026-10-16 — Step 3 incorrectly labels an algebraic distributive expansion as a product rule. The step applies only the distributive law, not the product rule, and therefore violates the rule‑granularity requirement.
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 2026-10-19 with SymPy 1.14.0.