Home›Calculus 1›The chain rule in Leibniz notation›Problem 2.2332

The chain rule in Leibniz notation

Problem 2.2332 · easy

Let \( \displaystyle y = \sqrt{u} \) and \( \displaystyle u = \cos{\left(x \right)} \). Find \( \displaystyle \dfrac{dy}{dx} \) using \( \displaystyle \dfrac{dy}{dx} = \dfrac{dy}{du}\dfrac{du}{dx} \).

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

Answer \( \frac{dy}{dx} = - \frac{\sin{\left(x \right)}}{2 \sqrt{\cos{\left(x \right)}}} \)

✓ 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: pass
qwen3.6:27b-mlx: pass

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-18
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: pass 2026-10-16
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly applies the chain rule and performs the necessary substitutions and algebraic simplifications.
gpt-oss:20b: pass 2026-10-15

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 2026-10-18 with SymPy 1.14.0.