Derivative of \( \displaystyle 3 \ln{\left(\tan{\left(x - 1 \right)} + \sec{\left(x - 1 \right)} \right)} \)
Problem 2.1167 · hard Beautiful Video
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. How we verify
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 of the inner linear functions, which is not a single valid rule application.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) * Derivative(f(g(x)), x), which is circular and dimensionally wrong (multiplying the derivative of the composite function by the inner derivative again). It should have directly substituted the derivatives of tan and sec, or applied the chain rule to get sec^2(x-1)*1 and sec(x-1)tan(x-1)*1.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-03 — 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) * Derivative(f(g(x)), x), which is circular and dimensionally wrong (multiplying the derivative of the composite function by the inner derivative again). It should have directly substituted the derivatives of tan and sec, or applied the chain rule to get sec^2(x-1)*1 and sec(x-1)tan(x-1)*1.gpt-oss:20b: fail (error) 2026-10-03 — Step 5 applies two rules at once: it uses the chain rule for each inner function and also multiplies the derivatives of the inner linear functions, which is not a single valid rule application.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 2026-10-03 with SymPy 1.14.0.