Home›Calculus 1›The chain rule in Leibniz notation›Problem 2.2438

The chain rule in Leibniz notation

Problem 2.2438 · easy

Let \( \displaystyle y = \ln{\left(u \right)} \) and \( \displaystyle u = x^{2} + 1 \). Find \( \displaystyle \dfrac{dy}{dx} \) using \( \displaystyle \dfrac{dy}{dx} = \dfrac{dy}{du}\dfrac{du}{dx} \).

\[ \frac{d}{d u} \ln{\left(u \right)} = \frac{1}{u} \]
dy/du.✓ Proved
\[ \frac{d}{d x} \left(x^{2} + 1\right) = 2 x \]
du/dx.✓ Proved
\[ 2 x \left. 1 \frac{1}{u} \right|_{\substack{ u=x^{2} + 1 }} = \frac{2 x}{x^{2} + 1} \]
Multiply, and write u in terms of x.✓ Proved

Answer \( \frac{dy}{dx} = \frac{2 x}{x^{2} + 1} \)

✓ 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: pass
qwen3.6:27b-mlx: pass — The solution correctly applies the chain rule, computes the derivatives, and substitutes back to express the result in terms of x.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-19 — The solution correctly applies the chain rule, computes the derivatives, and substitutes back to express the result in terms of x.
gpt-oss:20b: pass 2026-10-19
qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly applies the chain rule and substitutes the inner function into the derivative of the outer function.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly applies the chain rule and substitutes the inner function back into the derivative of the outer function.
gpt-oss:20b: pass 2026-10-16

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 2026-10-19 with SymPy 1.14.0.