Tangent lines
Problem 2.726 · medium
Find an equation of the tangent line to \( \displaystyle y = x e^{2 x} \) at \( \displaystyle x = 2 \).
- The tangent line passes through the point (a, f(a)) and has slope f'(a).
- \[ \left. x e^{2 x} \right|_{\substack{ x=2 }} = 2 e^{4} \]The point of tangency.✓ Proved
- \[ \frac{d}{d x} x e^{2 x} = \left(2 x + 1\right) e^{2 x} \]Differentiate.✓ Proved
- \[ \left. \left(2 x + 1\right) e^{2 x} \right|_{\substack{ x=2 }} = 5 e^{4} \]The slope at the point.✓ Proved
- \[ \left(5 x - 10\right) e^{4} + 2 e^{4} = 5 x e^{4} - 8 e^{4} \]Point-slope form, then simplify.✓ Proved
Answer \( y = 5 x e^{4} - 8 e^{4} \)
Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 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 |
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 2026-09-26 with SymPy 1.14.0.