The chain rule in Leibniz notation
Problem 2.2222 · easy
Let \( \displaystyle y = e^{u} \) and \( \displaystyle u = x^{3} - x \). Find \( \displaystyle \dfrac{dy}{dx} \) using \( \displaystyle \dfrac{dy}{dx} = \dfrac{dy}{du}\dfrac{du}{dx} \).
- \[ \frac{d}{d u} e^{u} = e^{u} \]dy/du.✓ Proved
- \[ \frac{d}{d x} \left(x^{3} - x\right) = 3 x^{2} - 1 \]du/dx.✓ Proved
- \[ \left(3 x^{2} - 1\right) \left. e^{u} \right|_{\substack{ u=x^{3} - x }} = \left(3 x^{2} - 1\right) e^{x \left(x^{2} - 1\right)} \]Multiply, and write u in terms of x.✓ Proved
Answer \( \frac{dy}{dx} = \left(3 x^{2} - 1\right) e^{x \left(x^{2} - 1\right)} \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ 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 | a central difference of y(u(x)) at three points |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly applies the chain rule and substitutes the inner function into the outer derivative. The algebraic simplification is correct.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly applies the chain rule and substitutes the inner function into the outer derivative. The algebraic simplification is correct.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly applies the chain rule, computes the derivatives of the inner and outer functions, and substitutes back to express the result in terms of x.gpt-oss:20b: pass 2026-10-14qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly applies the chain rule, computes the derivatives accurately, and substitutes back to express the result in terms of x.gpt-oss:20b: pass 2026-10-14
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/chain_rule_leibniz, checked 2026-10-17 with SymPy 1.14.0.