Home›Calculus 1›Derivatives of trigonometric functions›Problem 2.998

Derivative of \( \displaystyle \ln{\left(\tan{\left(x + 1 \right)} + \sec{\left(x + 1 \right)} \right)} \)

Problem 2.998 · hard Beautiful

Differentiate \( \displaystyle f(x) = \ln{\left(\tan{\left(x + 1 \right)} + \sec{\left(x + 1 \right)} \right)} \).

\[ \frac{d}{d x} \ln{\left(\tan{\left(x + 1 \right)} + \sec{\left(x + 1 \right)} \right)} \]
derivativeStart with the derivative of the function.✓ Proved
\[ = \frac{\frac{d}{d x} \left(\tan{\left(x + 1 \right)} + \sec{\left(x + 1 \right)}\right)}{\tan{\left(x + 1 \right)} + \sec{\left(x + 1 \right)}} \]
chainApply the chain rule for the logarithm.✓ Proved
\[ = \frac{\frac{d}{d x} \tan{\left(x + 1 \right)} + \frac{d}{d x} \sec{\left(x + 1 \right)}}{\tan{\left(x + 1 \right)} + \sec{\left(x + 1 \right)}} \]
sumApply the sum rule to the inner expression.✓ Proved
\[ = \frac{\tan{\left(x + 1 \right)} \sec{\left(x + 1 \right)} \frac{d}{d x} \left(x + 1\right) + \sec^{2}{\left(x + 1 \right)} \frac{d}{d x} \left(x + 1\right)}{\tan{\left(x + 1 \right)} + \sec{\left(x + 1 \right)}} \]
trigDifferentiate the trigonometric functions.≈ Checked numerically
\[ = \frac{\tan{\left(x + 1 \right)} \sec{\left(x + 1 \right)} + \sec^{2}{\left(x + 1 \right)}}{\tan{\left(x + 1 \right)} + \sec{\left(x + 1 \right)}} \]
chain constant algebraDifferentiate the inner linear terms. Simplify the derivative of the inner terms. Multiply the fraction by the numerator.✓ Proved
\[ = \sec{\left(x + 1 \right)} \]
algebra simplifyFactor out sec(x + 1) from the numerator. Cancel the common term in the numerator and denominator.✓ Proved

Answer \( \frac{1}{\cos{\left(x + 1 \right)}} \)

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.

✓ 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 undefined where tan(x + 1) + sec(x + 1) = 0
3	✓ 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
4	≈ Checked numerically	sympy 1.14.0	sympy 1.14.0: simplify left (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
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	✓ 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
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 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: pass
qwen3.6:27b-mlx: pass — The solution correctly applies the chain rule, sum rule, and trigonometric derivatives in a step-by-step manner. Each step changes only one aspect of the expression, and the labels accurately reflect the rules applied.

Every verdict on record (4)

qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly applies the chain rule, sum rule, and trigonometric derivatives in a step-by-step manner. Each step changes only one aspect of the expression, and the labels accurately reflect the rules applied.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly applies the chain rule, sum rule, and trigonometric derivatives in separate steps. The algebraic simplification and cancellation are handled properly with appropriate labels.
gpt-oss:20b: pass 2026-09-27

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 2026-09-27 with SymPy 1.14.0.