Home›Calculus 1›Tangent lines›Problem 2.730

Tangent lines

Problem 2.730 · medium Video

Find an equation of the tangent line to \( \displaystyle y = x + \sin{\left(x \right)} \) at \( \displaystyle x = \frac{\pi}{2} \).

The tangent line passes through the point (a, f(a)) and has slope f'(a).
Reviewed
\[ \left. x + \sin{\left(x \right)} \right|_{\substack{ x=\frac{\pi}{2} }} = 1 + \frac{\pi}{2} \]
The point of tangency.✓ Proved
\[ \frac{d}{d x} \left(x + \sin{\left(x \right)}\right) = \cos{\left(x \right)} + 1 \]
Differentiate.✓ Proved
\[ \left. \cos{\left(x \right)} + 1 \right|_{\substack{ x=\frac{\pi}{2} }} = 1 \]
The slope at the point.✓ Proved
\[ x + 1 \]
Point-slope form, then simplify.

Answer \( y = x + 1 \)

The graph of y = x + sin(x) near x = pi/2. A secant line through the point (pi/2, 1 + pi/2) and a second point on the curve turns into the tangent line as the second point slides in; its slope settles at 1, giving y = x + 1.

Lines: 3 proved, 1 reviewed, 1 not checked. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Reviewed	claude-sonnet-5-5	claude-sonnet-5-5: Line 1 is a sentence stating the standard tangent-line setup: the line passes through (a, f(a)) with slope f'(a). It is true and is not an algebraic equivalence, so no script applies.
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
4	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
5	Not checked	—	both sides are the same expression: the line states a value, and the working that produced it is not on the line
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	the line meets the curve at x = a, and its slope matches a central difference quotient of f there

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The final equation y = x + 1 is incorrect. The tangent line passes through (pi/2, 1 + pi/2) with slope 1, so the equation should be y = x + 1. Wait, let me re-calculate. Point: (pi/2, pi/2 + 1). Slope: 1. Equation: y - (pi/2 + 1) = 1 * (x - pi/2) => y = x - pi/2 + pi/2 + 1 => y = x + 1. The math is actually correct. Let me re-read the prompt carefully. 'Stated answer: Eq(y, x + 1)'. The solution derives y = x + 1. Is there a subtle error? Step 5 says 'Point-slope form, then simplify' and shows 'x + 1, x + 1'. This is not an equation, it's just an expression. The step label implies an equation was formed, but the content is just the RHS. However, usually 'proved' equations are checked by CAS. If the CAS checked 'x+1 == x+1', that's a tautology, not the line equation. The line equation is y = x + 1. The step 5 content 'x + 1, x + 1' likely represents the simplified RHS, but it fails to state the full equation y = ... explicitly in the equation line, or rather, the 'proved' tag suggests the CAS verified the algebraic manipulation. If the CAS verified that the point-slope form simplifies to x+1, that's fine. But does it verify it equals y? No. The step is misleadingly labeled or incomplete. However, looking at standard formats, often the LHS is implicit or the 'equation' is just the expression for y. Let's look closer. Step 5: 'x + 1, x + 1'. This looks like a substitution result or identity. It does not show 'y = x + 1'. If the final answer is y = x + 1, the derivation step should show the equation. The sentence 'Point-slope form, then simplify' is correct in intent, but the equation line is malformed or incomplete as an equation. It's just an expression. A student might not know where 'y' went. This is a 'style' or 'misleading' issue. But wait, is the answer actually correct? y = x + sin(x). y' = 1 + cos(x). At pi/2, y = pi/2 + 1. y' = 1. Line: y - (pi/2 + 1) = 1(x - pi/2) => y = x + 1. Yes, the answer is correct. The issue is purely presentation of step 5. It lists 'x + 1, x + 1' as a proved equation. This is likely a tautology check from the CAS, not the derivation of the line equation. The step fails to explicitly state y = x + 1. This is misleading.

Senior review claude-sonnet-5-5, 2026-10-12: pass — The point (pi/2, 1 + pi/2), the derivative cos(x) + 1, and the slope 1 are all correct. The point-slope line y - (1 + pi/2) = x - pi/2 simplifies to y = x + 1, matching the stated answer.

qwen3.6:27b-mlx: dismiss — The objection recomputes and confirms the answer y = x + 1 is correct; the point (pi/2, 1 + pi/2), slope 1 and line x + 1 all check out, and the complaint about the missing 'y =' in step 5 concerns display only, not a false statement or wrong rule.

Every verdict on record (7)

claude-sonnet-5-5: pass 2026-10-12 — The point (pi/2, 1 + pi/2), the derivative cos(x) + 1, and the slope 1 are all correct. The point-slope line y - (1 + pi/2) = x - pi/2 simplifies to y = x + 1, matching the stated answer.
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The final equation y = x + 1 is incorrect. The tangent line passes through (pi/2, 1 + pi/2) with slope 1, so the equation should be y = x + 1. Wait, let me re-calculate. Point: (pi/2, pi/2 + 1). Slope: 1. Equation: y - (pi/2 + 1) = 1 * (x - pi/2) => y = x - pi/2 + pi/2 + 1 => y = x + 1. The math is actually correct. Let me re-read the prompt carefully. 'Stated answer: Eq(y, x + 1)'. The solution derives y = x + 1. Is there a subtle error? Step 5 says 'Point-slope form, then simplify' and shows 'x + 1, x + 1'. This is not an equation, it's just an expression. The step label implies an equation was formed, but the content is just the RHS. However, usually 'proved' equations are checked by CAS. If the CAS checked 'x+1 == x+1', that's a tautology, not the line equation. The line equation is y = x + 1. The step 5 content 'x + 1, x + 1' likely represents the simplified RHS, but it fails to state the full equation y = ... explicitly in the equation line, or rather, the 'proved' tag suggests the CAS verified the algebraic manipulation. If the CAS verified that the point-slope form simplifies to x+1, that's fine. But does it verify it equals y? No. The step is misleadingly labeled or incomplete. However, looking at standard formats, often the LHS is implicit or the 'equation' is just the expression for y. Let's look closer. Step 5: 'x + 1, x + 1'. This looks like a substitution result or identity. It does not show 'y = x + 1'. If the final answer is y = x + 1, the derivation step should show the equation. The sentence 'Point-slope form, then simplify' is correct in intent, but the equation line is malformed or incomplete as an equation. It's just an expression. A student might not know where 'y' went. This is a 'style' or 'misleading' issue. But wait, is the answer actually correct? y = x + sin(x). y' = 1 + cos(x). At pi/2, y = pi/2 + 1. y' = 1. Line: y - (pi/2 + 1) = 1(x - pi/2) => y = x + 1. Yes, the answer is correct. The issue is purely presentation of step 5. It lists 'x + 1, x + 1' as a proved equation. This is likely a tautology check from the CAS, not the derivation of the line equation. The step fails to explicitly state y = x + 1. This is misleading.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: fail (error) 2026-09-27 — The final equation x + 1 is incorrect. The tangent line passes through (pi/2, 1 + pi/2) with slope 1, so the equation should be y = x + 1. The solution fails to include the y-variable in the final equation, resulting in a tautology rather than the line equation.
gpt-oss:20b: fail (misleading) 2026-09-27 — The final step claims the point‑slope form simplifies to "x + 1, x + 1"; it omits the dependent variable and does not explicitly state the equation y = x + 1, which could mislead a student about the correct form of the tangent line.
qwen3.6:27b-mlx: fail (error) 2026-09-26 — The final equation is incorrect. The tangent line passes through (pi/2, 1 + pi/2) with slope 1, so the equation should be y = x + 1. The solution claims the result is y = x + 1, but line 5 shows 'x + 1, x + 1' which implies the equation is y = x + 1? Wait, let's re-read carefully. Line 5 says 'x + 1, x + 1'. This is likely a typo in the representation of the equation y = x + 1. However, the stated answer is 'Eq(y, x + 1)'. Let's check the math. Point: (pi/2, 1 + pi/2). Slope: 1. Equation: y - (1 + pi/2) = 1 * (x - pi/2) => y = x - pi/2 + 1 + pi/2 => y = x + 1. The math is correct. The issue is line 5. It says 'x + 1, x + 1'. This is not a standard equation format. It looks like it's trying to say y = x + 1 but wrote x + 1 on both sides or something similar. Actually, looking at the pattern of previous lines, line 2 was 'Subs(...), result', line 3 'Derivative(...), result', line 4 'Subs(...), result'. Line 5 is 'x + 1, x + 1'. This doesn't follow the pattern of an equation derivation step clearly. It seems to imply the final expression is x+1, but misses the 'y='. The stated answer is Eq(y, x+1). The solution text in line 5 is ambiguous/incorrectly formatted as an equation step. It should explicitly state y = x + 1. As written, 'x + 1, x + 1' is not a valid equation line and fails to clearly derive y = x + 1 from the point-slope form. It's a 'style' or 'misleading' error because it doesn't show the equality y = ... clearly.
gpt-oss:20b: pass 2026-09-26

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 generator:structured/tangent_line, checked 2026-10-13 with SymPy 1.14.0.