Derivative of \( \displaystyle \ln{\left(e^{2 x - 1} + 1 \right)} \)
Problem 2.424 · hard
Differentiate \( \displaystyle f(x) = \ln{\left(e^{2 x - 1} + 1 \right)} \).
- \[ \frac{d}{d x} \ln{\left(e^{2 x - 1} + 1 \right)} \]Start with the derivative of the function.✓ Proved
- \[ = \frac{\frac{d}{d x} \left(e^{2 x - 1} + 1\right)}{e^{2 x - 1} + 1} \]chainApply the chain rule for the logarithm.✓ Proved
- \[ = \frac{\frac{d}{d x} 1 + \frac{d}{d x} e^{2 x - 1}}{e^{2 x - 1} + 1} \]sumDifferentiate the sum inside the parentheses.✓ Proved
- \[ = \frac{\frac{d}{d x} e^{2 x - 1}}{e^{2 x - 1} + 1} \]constant algebraThe derivative of the constant 1 is 0. Simplify the expression.✓ Proved
- \[ = \frac{e^{2 x - 1} \frac{d}{d x} \left(2 x - 1\right)}{e^{2 x - 1} + 1} \]exponentialApply the chain rule for the exponential function.✓ Proved
- \[ = \frac{2 e^{2 x - 1}}{e^{2 x - 1} + 1} \]derivative algebraDifferentiate the linear function 2*x - 1. Simplify the final expression.✓ Proved
Answer \( \frac{2 e^{2 x}}{e^{2 x} + e} \)
✓ 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 undefined where exp(2*x - 1) + 1 = 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where exp(2*x - 1) + 1 = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where exp(2*x - 1) + 1 = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where exp(2*x - 1) + 1 = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where exp(2*x - 1) + 1 = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where exp(2*x - 1) + 1 = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where exp(2*x - 1) + 1 = 0 |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 undefined where exp(2*x) + E = 0 |
| 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 the chain rule, sum rule, and derivative rules for exponential and linear functions. The final simplification is algebraically correct and matches the stated answer.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-09-21 — The solution correctly applies the chain rule, sum rule, and derivative rules for exponential and linear functions. The final simplification is algebraically correct and matches the stated answer.gpt-oss:20b: pass 2026-09-21qwen3.6:27b-mlx: pass 2026-09-20gpt-oss:20b: pass 2026-09-20qwen3.6:27b-mlx: pass 2026-09-20 — The solution correctly applies the chain rule, sum rule, and derivative rules for exponential and linear functions. Each step isolates a single rule application, and the labels are accurate.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.