Derivative of \( \displaystyle \left(2 x - 1\right)^{\frac{1}{2 x - 1}} \)
Problem 2.549 · hard
Differentiate \( \displaystyle f(x) = \left(2 x - 1\right)^{\frac{1}{2 x - 1}} \).
- \[ \frac{d}{d x} \left(2 x - 1\right)^{\frac{1}{2 x - 1}} \]derivativeStart with the derivative of the function.✓ Proved
- \[ = \frac{d}{d x} e^{\frac{\ln{\left(2 x - 1 \right)}}{2 x - 1}} \]rewriteRewrite the power using the exponential and logarithm.≈ Checked numerically
- \[ = e^{\frac{\ln{\left(2 x - 1 \right)}}{2 x - 1}} \frac{d}{d x} \frac{\ln{\left(2 x - 1 \right)}}{2 x - 1} \]chainApply the chain rule to the exponential function.✓ Proved
- \[ = \left(\ln{\left(2 x - 1 \right)} \frac{d}{d x} \frac{1}{2 x - 1} + \frac{\frac{d}{d x} \ln{\left(2 x - 1 \right)}}{2 x - 1}\right) e^{\frac{\ln{\left(2 x - 1 \right)}}{2 x - 1}} \]productApply the product rule to the inner expression.✓ Proved
- \[ = \left(\ln{\left(2 x - 1 \right)} \frac{d}{d x} \frac{1}{2 x - 1} + \frac{\frac{d}{d x} \left(2 x - 1\right)}{\left(2 x - 1\right)^{2}}\right) e^{\frac{\ln{\left(2 x - 1 \right)}}{2 x - 1}} \]chainApply the chain rule to the logarithmic term.✓ Proved
- \[ = \left(\ln{\left(2 x - 1 \right)} \frac{d}{d x} \frac{1}{2 x - 1} + \frac{2}{\left(2 x - 1\right)^{2}}\right) e^{\frac{\ln{\left(2 x - 1 \right)}}{2 x - 1}} \]derivativeDifferentiate the linear term 2x - 1.✓ Proved
- \[ = \left(- \frac{\ln{\left(2 x - 1 \right)} \frac{d}{d x} \left(2 x - 1\right)}{\left(2 x - 1\right)^{2}} + \frac{2}{\left(2 x - 1\right)^{2}}\right) e^{\frac{\ln{\left(2 x - 1 \right)}}{2 x - 1}} \]chainApply the chain rule to the reciprocal term.✓ Proved
- \[ = \left(- \frac{2 \ln{\left(2 x - 1 \right)}}{\left(2 x - 1\right)^{2}} + \frac{2}{\left(2 x - 1\right)^{2}}\right) e^{\frac{\ln{\left(2 x - 1 \right)}}{2 x - 1}} \]derivative algebraDifferentiate the linear term 2x - 1. Simplify the expression inside the parentheses.✓ Proved
- \[ = \frac{\left(2 - 2 \ln{\left(2 x - 1 \right)}\right) e^{\frac{\ln{\left(2 x - 1 \right)}}{2 x - 1}}}{\left(2 x - 1\right)^{2}} \]algebraFactor out common terms.✓ Proved
- \[ = \frac{\left(2 - 2 \ln{\left(2 x - 1 \right)}\right) \left(2 x - 1\right)^{\frac{1}{2 x - 1}}}{\left(2 x - 1\right)^{2}} \]simplifyConvert the exponential back to the original power form.≈ Checked numerically
Answer \( 2 \left(1 - \log{\left(2 x - 1 \right)}\right) \left(2 x - 1\right)^{-2 + \frac{1}{2 x - 1}} \)
✓ Nihil obstat Lines: 10 proved, 2 checked numerically. 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 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left 2*((1 - log(2*x - 1))*(2*x - 1)**(1/(2*x - 1)) + (log(2*x - 1) - 1)*exp(log(2*x - 1)/(2*x - 1)))/(2*x - 1)**2; numeric agreement only, at 24 of 24 sampled points undefined where 2*x - 1 = 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 undefined where 2*x - 1 = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 11 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left 2*((1 - log(2*x - 1))*exp(log(2*x - 1)/(2*x - 1)) + (2*x - 1)**(1/(2*x - 1))*(log(2*x - 1) - 1))/(2*x - 1)**2; numeric agreement only, at 24 of 24 sampled points log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 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 undefined where 2*x - 1 = 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
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-21gpt-oss:20b: pass 2026-09-21qwen3.6:27b-mlx: pass 2026-09-21gpt-oss:20b: pass 2026-09-21
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.