Derivative of \( \displaystyle 3 \ln{\left(\tan{\left(x - 1 \right)} + \sec{\left(x - 1 \right)} \right)} \)
Problem 2.1167 · hard Beautiful
Differentiate \( \displaystyle f(x) = 3 \ln{\left(\tan{\left(x - 1 \right)} + \sec{\left(x - 1 \right)} \right)} \).
- \[ \frac{d}{d x} 3 \ln{\left(\tan{\left(x - 1 \right)} + \sec{\left(x - 1 \right)} \right)} \]derivativeStart with the derivative of the function.✓ Proved
- \[ = 3 \frac{d}{d x} \ln{\left(\tan{\left(x - 1 \right)} + \sec{\left(x - 1 \right)} \right)} \]constant-multiplePull out the constant factor.✓ Proved
- \[ = \frac{3 \frac{d}{d x} \left(\tan{\left(x - 1 \right)} + \sec{\left(x - 1 \right)}\right)}{\tan{\left(x - 1 \right)} + \sec{\left(x - 1 \right)}} \]chainApply the chain rule for the logarithm.✓ Proved
- \[ = \frac{3 \left(\frac{d}{d x} \tan{\left(x - 1 \right)} + \frac{d}{d x} \sec{\left(x - 1 \right)}\right)}{\tan{\left(x - 1 \right)} + \sec{\left(x - 1 \right)}} \]sumDifferentiate the sum inside the parentheses.✓ Proved
- \[ = \frac{3 \left(\frac{d}{d x} \left(x - 1\right) \frac{d}{d x} \tan{\left(x - 1 \right)} + \frac{d}{d x} \left(x - 1\right) \frac{d}{d x} \sec{\left(x - 1 \right)}\right)}{\tan{\left(x - 1 \right)} + \sec{\left(x - 1 \right)}} \]chainApply the chain rule to each term.✓ Proved
- \[ = \frac{3 \left(\frac{d}{d x} \tan{\left(x - 1 \right)} + \frac{d}{d x} \sec{\left(x - 1 \right)}\right)}{\tan{\left(x - 1 \right)} + \sec{\left(x - 1 \right)}} \]derivativeDifferentiate the inner linear functions.✓ Proved
- \[ = \frac{3 \left(\tan{\left(x - 1 \right)} \sec{\left(x - 1 \right)} + \sec^{2}{\left(x - 1 \right)}\right)}{\tan{\left(x - 1 \right)} + \sec{\left(x - 1 \right)}} \]trigCompute the derivatives of tangent and secant.≈ Checked numerically
- \[ = \frac{3 \tan{\left(x - 1 \right)} \sec{\left(x - 1 \right)} + 3 \sec^{2}{\left(x - 1 \right)}}{\tan{\left(x - 1 \right)} + \sec{\left(x - 1 \right)}} \]algebraCombine the terms into a single fraction.✓ Proved
- \[ = 3 \sec{\left(x - 1 \right)} \]algebra simplifyFactor out sec(x - 1) from the numerator. Cancel the common factor in the numerator and denominator.✓ Proved
Answer \( \frac{3}{\cos{\left(x - 1 \right)}} \)
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.
Lines: 10 proved, 1 checked numerically. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see 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 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(x - 1) + sec(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(x - 1) + sec(x - 1) = 0 |
| 5 | ✓ 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(x - 1) + sec(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(x - 1) + sec(x - 1) = 0 |
| 7 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left 3*(tan(x - 1)**2 - sec(x - 1)**2 + 1)/(tan(x - 1) + sec(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(x - 1) + sec(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(x - 1) + sec(x - 1) = 0 |
| 9 | ✓ 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(x - 1) + sec(x - 1) = 0 |
| 10 | ✓ 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(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: fail (error) — Step 5 applies two rules at once: it uses the chain rule for each inner function and also multiplies the derivatives, effectively applying a product rule. The label "chain" is therefore incorrect for that step.qwen3.6:27b-mlx: fail (error) — Step 5 is mathematically incorrect. The chain rule states that d/dx[f(g(x))] = f'(g(x)) * g'(x). The step incorrectly writes this as g'(x) * d/dx[f(g(x))], which results in a recursive definition rather than a simplification. Specifically, it replaces Derivative(tan(x-1), x) with Derivative(x-1, x)*Derivative(tan(x-1), x), which is circular and does not evaluate the derivative of the outer function.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-09-29 — Step 5 is mathematically incorrect. The chain rule states that d/dx[f(g(x))] = f'(g(x)) * g'(x). The step incorrectly writes this as g'(x) * d/dx[f(g(x))], which results in a recursive definition rather than a simplification. Specifically, it replaces Derivative(tan(x-1), x) with Derivative(x-1, x)*Derivative(tan(x-1), x), which is circular and does not evaluate the derivative of the outer function.gpt-oss:20b: fail (error) 2026-09-29 — Step 5 applies two rules at once: it uses the chain rule for each inner function and also multiplies the derivatives, effectively applying a product rule. The label "chain" is therefore incorrect for that step.qwen3.6:27b-mlx: fail (error) 2026-09-29 — Step 5 incorrectly applies the chain rule by multiplying by Derivative(x - 1, x) while retaining the unevaluated Derivative(tan(x - 1), x) term, resulting in a double application of the chain rule logic. Step 6 then attempts to 'differentiate' the inner function again, which is logically incoherent given the state of the expression.gpt-oss:20b: fail (error) 2026-09-29 — Step 5 incorrectly applies the chain rule to the sum inside the logarithm. It multiplies each inner derivative by Derivative(x-1,x), effectively treating it as a product, whereas the correct application should simply replace Derivative(tan(x-1)+sec(x-1),x) with Derivative(tan(x-1),x)+Derivative(sec(x-1),x). This introduces an extra factor of 1 that is not justified by any rule in the allowed vocabulary.
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 2026-09-29 with SymPy 1.14.0.