Derivative of \( \displaystyle \frac{\ln{\left(\tan{\left(5 x - 1 \right)} + \sec{\left(5 x - 1 \right)} \right)}}{5} \)
Problem 2.690 · hard
Differentiate \( \displaystyle f(x) = \frac{\ln{\left(\tan{\left(5 x - 1 \right)} + \sec{\left(5 x - 1 \right)} \right)}}{5} \).
- \[ \frac{d}{d x} \frac{\ln{\left(\tan{\left(5 x - 1 \right)} + \sec{\left(5 x - 1 \right)} \right)}}{5} \]derivativeStart with the derivative of the function.✓ Proved
- \[ = \frac{\frac{d}{d x} \ln{\left(\tan{\left(5 x - 1 \right)} + \sec{\left(5 x - 1 \right)} \right)}}{5} \]constant-multipleFactor out the constant 1/5.✓ Proved
- \[ = \frac{\frac{d}{d x} \left(\tan{\left(5 x - 1 \right)} + \sec{\left(5 x - 1 \right)}\right)}{5 \left(\tan{\left(5 x - 1 \right)} + \sec{\left(5 x - 1 \right)}\right)} \]chainApply the chain rule to the logarithm.✓ Proved
- \[ = \frac{\frac{d}{d x} \tan{\left(5 x - 1 \right)} + \frac{d}{d x} \sec{\left(5 x - 1 \right)}}{5 \left(\tan{\left(5 x - 1 \right)} + \sec{\left(5 x - 1 \right)}\right)} \]sumDifferentiate the sum inside the parentheses.✓ Proved
- \[ = \frac{5 \tan{\left(5 x - 1 \right)} \sec{\left(5 x - 1 \right)} + 5 \sec^{2}{\left(5 x - 1 \right)}}{5 \left(\tan{\left(5 x - 1 \right)} + \sec{\left(5 x - 1 \right)}\right)} \]chainApply the chain rule to both tangent and secant terms.≈ Checked numerically
- \[ = \frac{\tan{\left(5 x - 1 \right)} \sec{\left(5 x - 1 \right)} + \sec^{2}{\left(5 x - 1 \right)}}{\tan{\left(5 x - 1 \right)} + \sec{\left(5 x - 1 \right)}} \]constant-multiple simplifyFactor out the common constant 5. Cancel the 1/5 and 5.✓ Proved
- \[ = \sec{\left(5 x - 1 \right)} \]algebra simplifyFactor out sec(5*x - 1) from the numerator. Cancel the common factor in the numerator and denominator.✓ Proved
Answer \( \frac{1}{\cos{\left(5 x - 1 \right)}} \)
Mind the domain. The answer is also defined at points where f(x) is not. Substituting there gives a number that is not a slope of f.
✓ Nihil obstat Lines: 9 proved, 1 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 sec 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 sec has poles at odd multiples of pi/2 undefined where tan(5*x - 1) + sec(5*x - 1) = 0 |
| 4 | ✓ 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) + sec(5*x - 1) = 0 |
| 5 | ≈ 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) + sec(5*x - 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) + sec(5*x - 1) = 0 |
| 6 | ✓ 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) + sec(5*x - 1) = 0 |
| 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) + sec(5*x - 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) + sec(5*x - 1) = 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 sec has poles at odd multiples of pi/2 |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 undefined where cos(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
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-21gpt-oss:20b: pass 2026-09-21qwen3.6:27b-mlx: pass 2026-09-21 — The solution correctly applies differentiation rules and algebraic simplifications. Each step changes only one aspect of the expression and uses valid labels from the fixed vocabulary.gpt-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.