Derivative of \( \displaystyle - \frac{\ln{\left(\tan^{2}{\left(5 x + 1 \right)} + 1 \right)}}{10} + \frac{\ln{\left(\tan{\left(5 x + 1 \right)} \right)}}{5} \)
Problem 2.1417 · hard
Differentiate \( \displaystyle f(x) = - \frac{\ln{\left(\tan^{2}{\left(5 x + 1 \right)} + 1 \right)}}{10} + \frac{\ln{\left(\tan{\left(5 x + 1 \right)} \right)}}{5} \).
- \[ \frac{d}{d x} \left(- \frac{\ln{\left(\tan^{2}{\left(5 x + 1 \right)} + 1 \right)}}{10} + \frac{\ln{\left(\tan{\left(5 x + 1 \right)} \right)}}{5}\right) \]sumDifferentiate the sum of two terms.✓ Proved
- \[ = \frac{d}{d x} \left(- \frac{\ln{\left(\tan^{2}{\left(5 x + 1 \right)} + 1 \right)}}{10}\right) + \frac{d}{d x} \frac{\ln{\left(\tan{\left(5 x + 1 \right)} \right)}}{5} \]constant-multipleFactor out the constants.✓ Proved
- \[ = - \frac{\frac{d}{d x} \ln{\left(\tan^{2}{\left(5 x + 1 \right)} + 1 \right)}}{10} + \frac{\frac{d}{d x} \ln{\left(\tan{\left(5 x + 1 \right)} \right)}}{5} \]constant-multipleMove the constants outside the derivatives.✓ Proved
- \[ = \frac{\frac{d}{d x} \tan{\left(5 x + 1 \right)}}{5 \tan{\left(5 x + 1 \right)}} - \frac{\frac{d}{d x} \left(\tan^{2}{\left(5 x + 1 \right)} + 1\right)}{10 \left(\tan^{2}{\left(5 x + 1 \right)} + 1\right)} \]chainApply the chain rule to the logarithmic functions.✓ Proved
- \[ = \frac{\frac{d}{d x} \tan{\left(5 x + 1 \right)}}{5 \tan{\left(5 x + 1 \right)}} - \frac{\tan{\left(5 x + 1 \right)} \frac{d}{d x} \tan{\left(5 x + 1 \right)}}{5 \left(\tan^{2}{\left(5 x + 1 \right)} + 1\right)} \]powerApply the power rule to the inner function.✓ Proved
- \[ = \frac{\sec^{2}{\left(5 x + 1 \right)}}{\tan{\left(5 x + 1 \right)}} - \frac{\tan{\left(5 x + 1 \right)} \sec^{2}{\left(5 x + 1 \right)}}{\tan^{2}{\left(5 x + 1 \right)} + 1} \]chain algebra algebraApply the chain rule to the tangent function. Multiply the constants together. Simplify the expression by distributing the constants.≈ Checked numerically
- \[ = - \tan{\left(5 x + 1 \right)} + \frac{\sec^{2}{\left(5 x + 1 \right)}}{\tan{\left(5 x + 1 \right)}} \]algebra simplifySubstitute 1 + tan(5*x + 1)**2 with sec(5*x + 1)**2. Cancel the sec(5*x + 1)**2 term in the first fraction.≈ Checked numerically
- \[ = - \tan{\left(5 x + 1 \right)} + \frac{1}{\cos^{2}{\left(5 x + 1 \right)} \tan{\left(5 x + 1 \right)}} \]rewrite algebraRewrite sec(x)**2 as 1/cos(x)**2. Simplify the complex fraction.✓ Proved
- \[ = - \tan{\left(5 x + 1 \right)} + \frac{1}{\sin{\left(5 x + 1 \right)} \cos{\left(5 x + 1 \right)}} \]algebra algebraUse the identity tan(x) = sin(x)/cos(x) to simplify the denominator. Split the product in the denominator.✓ Proved
- \[ = - \tan{\left(5 x + 1 \right)} + \csc{\left(5 x + 1 \right)} \sec{\left(5 x + 1 \right)} \]rewriteRewrite the terms using secant and cosecant.✓ Proved
- \[ = - \tan{\left(5 x + 1 \right)} + \frac{1}{\sin{\left(5 x + 1 \right)} \cos{\left(5 x + 1 \right)}} \]algebra algebraConvert back to sine and cosine for simplification. Combine the fractions.✓ Proved
- \[ = \frac{\tan^{2}{\left(5 x + 1 \right)} + 1}{\tan{\left(5 x + 1 \right)}} - \tan{\left(5 x + 1 \right)} \]algebraUse the identity 1/(sin(x)cos(x)) = (sin(x)^2 + cos(x)^2)/(sin(x)cos(x)) = tan(x) + cot(x) or similar algebraic manipulation.≈ Checked numerically
- \[ = \frac{1}{\tan{\left(5 x + 1 \right)}} \]algebra simplifyDistribute the division by tan(5*x + 1). Cancel the tangent terms.✓ Proved
- \[ = \cot{\left(5 x + 1 \right)} \]simplifyRewrite the cotangent function.✓ Proved
Answer \( \frac{1}{\tan{\left(5 x + 1 \right)}} \)
✓ Nihil obstat Lines: 19 proved, 3 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 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments tan has poles at odd multiples of pi/2 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments tan has poles at odd multiples of pi/2 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments tan has poles at odd multiples of pi/2 undefined where tan(5*x + 1)**2 + 1 = 0 undefined where tan(5*x + 1) = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(5*x + 1)**2 + 1 = 0 undefined where tan(5*x + 1) = 0 |
| 6 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (tan(5*x + 1)**2 - sec(5*x + 1)**2 + 1)/(tan(5*x + 1)**3 + tan(5*x + 1)); numeric agreement only, at 24 of 24 sampled points tan has poles at odd multiples of pi/2 undefined where tan(5*x + 1)**2 + 1 = 0 undefined where tan(5*x + 1) = 0 sec has poles at odd multiples of pi/2 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where tan(5*x + 1) = 0 undefined where tan(5*x + 1)**2 + 1 = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where tan(5*x + 1) = 0 undefined where tan(5*x + 1)**2 + 1 = 0 |
| 9 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (tan(5*x + 1)**2 - sec(5*x + 1)**2 + 1)*tan(5*x + 1)/(tan(5*x + 1)**2 + 1); numeric agreement only, at 24 of 24 sampled points tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where tan(5*x + 1) = 0 undefined where tan(5*x + 1)**2 + 1 = 0 |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where tan(5*x + 1) = 0 |
| 11 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where tan(5*x + 1) = 0 undefined where cos(5*x + 1) = 0 |
| 12 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(5*x + 1) = 0 undefined where cos(5*x + 1) = 0 |
| 13 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(5*x + 1) = 0 undefined where cos(5*x + 1) = 0 undefined where sin(5*x + 1) = 0 |
| 14 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where sin(5*x + 1) = 0 undefined where cos(5*x + 1) = 0 |
| 15 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where sin(5*x + 1) = 0 undefined where cos(5*x + 1) = 0 sec has poles at odd multiples of pi/2 csc has poles at multiples of pi |
| 16 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 csc has poles at multiples of pi undefined where sin(5*x + 1) = 0 undefined where cos(5*x + 1) = 0 |
| 17 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where sin(5*x + 1) = 0 undefined where cos(5*x + 1) = 0 |
| 18 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left -tan(5*x + 1) - 1/tan(5*x + 1) + 1/(sin(5*x + 1)*cos(5*x + 1)); numeric agreement only, at 24 of 24 sampled points tan has poles at odd multiples of pi/2 undefined where sin(5*x + 1) = 0 undefined where cos(5*x + 1) = 0 undefined where tan(5*x + 1) = 0 |
| 19 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(5*x + 1) = 0 |
| 20 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(5*x + 1) = 0 |
| 21 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(5*x + 1) = 0 cot has poles at multiples of pi |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(5*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 — The solution correctly applies differentiation rules and simplifies the expression to the stated answer. Each step adheres to the single-rule constraint and uses valid labels from the fixed vocabulary.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-03 — The solution correctly applies differentiation rules and simplifies the expression to the stated answer. Each step adheres to the single-rule constraint and uses valid labels from the fixed vocabulary.gpt-oss:20b: pass 2026-10-03qwen3.6:27b-mlx: pass 2026-10-03 — The solution correctly applies differentiation rules and simplifies the expression to the stated answer. Each step adheres to the one-change-per-step constraint and uses valid labels from the fixed vocabulary.gpt-oss:20b: pass 2026-10-03
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 2026-10-03 with SymPy 1.14.0.