Home›Calculus 1›Chain rule›Problem 2.230

Derivative of \( \displaystyle x \ln{\left(5 x + 2 \right)} - x + \frac{2 \ln{\left(5 x + 2 \right)}}{5} \)

Problem 2.230 · hard Beautiful

Differentiate \( \displaystyle f(x) = x \ln{\left(5 x + 2 \right)} - x + \frac{2 \ln{\left(5 x + 2 \right)}}{5} \).

\[ \frac{d}{d x} \left(x \ln{\left(5 x + 2 \right)} - x + \frac{2 \ln{\left(5 x + 2 \right)}}{5}\right) \]
sumStart with the derivative of the entire function.✓ Proved
\[ = - \frac{d}{d x} x + \frac{d}{d x} x \ln{\left(5 x + 2 \right)} + \frac{d}{d x} \frac{2 \ln{\left(5 x + 2 \right)}}{5} \]
sumDistribute the derivative over the sum.✓ Proved
\[ = \frac{d}{d x} x \ln{\left(5 x + 2 \right)} + \frac{d}{d x} \frac{2 \ln{\left(5 x + 2 \right)}}{5} - 1 \]
constantThe derivative of x is 1.✓ Proved
\[ = \frac{d}{d x} x \ln{\left(5 x + 2 \right)} + \frac{2 \frac{d}{d x} \ln{\left(5 x + 2 \right)}}{5} - 1 \]
constant-multiplePull out the constant factor 2/5.✓ Proved
\[ = x \frac{d}{d x} \ln{\left(5 x + 2 \right)} + \ln{\left(5 x + 2 \right)} \frac{d}{d x} x + \frac{2 \frac{d}{d x} \ln{\left(5 x + 2 \right)}}{5} - 1 \]
productApply the product rule to the first term.✓ Proved
\[ = x \frac{d}{d x} \ln{\left(5 x + 2 \right)} + \ln{\left(5 x + 2 \right)} + \frac{2 \frac{d}{d x} \ln{\left(5 x + 2 \right)}}{5} - 1 \]
constant algebraThe derivative of x is 1. Simplify the multiplication by 1.✓ Proved
\[ = \left(x + \frac{2}{5}\right) \frac{d}{d x} \ln{\left(5 x + 2 \right)} + \ln{\left(5 x + 2 \right)} - 1 \]
algebraGroup the terms involving the derivative of the logarithm.✓ Proved
\[ = \frac{\left(x + \frac{2}{5}\right) \frac{d}{d x} \left(5 x + 2\right)}{5 x + 2} + \ln{\left(5 x + 2 \right)} - 1 \]
logarithmicApply the chain rule for the derivative of the logarithm.✓ Proved
\[ = \ln{\left(5 x + 2 \right)} \]
derivative algebra simplify simplifyThe derivative of 5*x + 2 is 5. Distribute the 5 into the parentheses. Simplify the fraction. Final simplification.✓ Proved

Answer \( \log{\left(5 x + 2 \right)} \)

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. The reviewers disagree about how one step is explained; every verdict is in the receipt.

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 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
12	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 log is undefined for non-positive arguments
13	✓ 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: pass
deepseek-r1:70b: pass
qwen3.6:27b-mlx: pass — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint and using valid labels from the fixed vocabulary.

Every verdict on record (12)

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 and using valid labels from the fixed vocabulary.
gpt-oss:20b: pass 2026-09-20
qwen3.6:27b-mlx: fail (error) 2026-09-20 — Step 9 is labeled 'logarithmic' but the note explicitly states 'Apply the chain rule'. The step applies the chain rule to the logarithm, so the label should be 'chain'. Furthermore, the rule 'logarithmic' is not in the fixed vocabulary list provided in the instructions (which lists 'logarithmic' as a valid label, but the note contradicts the label by citing the chain rule, and typically 'logarithmic' refers to the base derivative d/dx log(u) = 1/u * u', which is a combination of logarithmic and chain rules. However, looking at the vocabulary list: 'logarithmic' IS in the list. The issue is that Step 9 applies the chain rule to the log function. The derivative of log(u) is (1/u)*u'. This step writes out the u' part. Is this 'logarithmic' or 'chain'? Usually, 'logarithmic' covers d/dx log(x) = 1/x. Applying it to a composite function requires 'chain'. The note says 'Apply the chain rule'. The label is 'logarithmic'. This is a mismatch. If the step is applying the chain rule, the label should be 'chain'. If the label is 'logarithmic', the note is misleading. Given the strict contract, a label must name the rule applied. The rule applied is the chain rule (differentiating the outer log function with respect to its argument, then multiplying by the derivative of the argument). Wait, actually, the standard decomposition is: d/dx log(g(x)) = (d/dg log(g)) * g'. The step shows Derivative(log(5x+2), x) -> Derivative(5x+2, x) * (1/(5x+2)). This is the Chain Rule. The label 'logarithmic' is incorrect for this specific step because the core operation being performed is the chain rule expansion. The 'logarithmic' rule would typically be d/dx log(x) = 1/x. Here we have a composite. So 'chain' is the correct label. The label 'logarithmic' is a defect.
gpt-oss:20b: pass 2026-09-20
qwen3.6:27b-mlx: pass 2026-09-19 — 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.
deepseek-r1:70b: pass 2026-09-19
gpt-oss:20b: pass 2026-09-19
qwen3.6:27b-mlx: pass 2026-09-18 — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint and using valid labels from the fixed vocabulary.
deepseek-r1:70b: pass 2026-09-18
gpt-oss:20b: pass 2026-09-18
gpt-oss:20b: fail 2026-09-17 — Step 3 (and similarly step 6) incorrectly labels the rule as 'constant' when differentiating x; the correct rule is the derivative of the identity function. The solution also omits the domain restriction x>-2/5 for the logarithm.
deepseek-r1:70b: pass 2026-09-17

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.