Home›Calculus 1›The chain rule in Leibniz notation›Problem 2.2487

The chain rule in Leibniz notation

Problem 2.2487 · easy

Let \( \displaystyle y = \sqrt{u} \) and \( \displaystyle u = - 2 x^{2} - 4 \). 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
\[ \frac{d}{d x} \left(- 2 x^{2} - 4\right) = - 4 x \]
du/dx.✓ Proved
\[ - 4 x \left. 1 \frac{1}{2 \sqrt{u}} \right|_{\substack{ u=- 2 x^{2} - 4 }} = \frac{\sqrt{2} i x}{\sqrt{x^{2} + 2}} \]
Multiply, and write u in terms of x.✓ Proved

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

✓ 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 and handles the complex arithmetic resulting from the negative argument of the square root. The final expression is algebraically equivalent to the stated answer.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-24 — The solution correctly applies the chain rule and handles the complex arithmetic resulting from the negative argument of the square root. The final expression is algebraically equivalent to the stated answer.
gpt-oss:20b: pass 2026-10-24
qwen3.6:27b-mlx: fail (style) 2026-10-23 — [domain objection, downgraded to style] The solution fails to specify the domain of validity. Since u = -2x^2 - 4 is always negative for real x, sqrt(u) is imaginary, but the problem does not specify working in complex numbers. Even if complex numbers are allowed, the simplification of sqrt(-2(x^2+2)) to i*sqrt(2)*sqrt(x^2+2) assumes a specific branch cut and ignores that the derivative of sqrt(u) involves 1/sqrt(u), which would be 1/(i*sqrt(2)*sqrt(x^2+2)). The final result has a sign/branch ambiguity issue typical of complex square roots, but more importantly, standard calculus problems of this form usually imply real-valued functions where the derivative is undefined. If complex numbers are intended, the step is algebraically correct but contextually ambiguous without stating the branch. However, the most critical error is likely that for real calculus, the function is not differentiable (not even defined) on any real interval. If the context is complex analysis, the notation 'I' is used, but the simplification sqrt(-2) = i*sqrt(2) is standard. Let's look closer. dy/dx = (1/(2*sqrt(-2x^2-4))) * (-4x). sqrt(-2x^2-4) = sqrt(-2(x^2+2)) = i*sqrt(2)*sqrt(x^2+2). So dy/dx = -4x / (2 * i * sqrt(2) * sqrt(x^2+2)) = -2x / (i * sqrt(2) * sqrt(x^2+2)) = -2x * (-i) / (sqrt(2) * sqrt(x^2+2)) = 2ix / (sqrt(2)*sqrt(x^2+2)) = i*sqrt(2)*x / sqrt(x^2+2). The algebra holds for complex numbers. Is it an error? Usually, 'find dy/dx' implies real variables unless specified. If x is real, y is imaginary. The derivative exists in the complex sense. But typically, this is a 'trick' question or an error in problem statement expecting 'undefined'. However, looking at the provided answer, it matches the complex derivation. The issue is whether the solution *should* have pointed out that the function is not real-valued. If the prompt implies standard real calculus, the answer is 'undefined'. If it allows complex, it's correct. Given the 'adversarial' nature, pointing out the domain issue is key. The solution presents a complex result for what looks like a standard real calculus problem without justification. This is misleading/error-prone for a student expecting real analysis.
gpt-oss:20b: pass 2026-10-23
qwen3.6:27b-mlx: pass 2026-10-23 — The solution correctly applies the chain rule and simplifies the complex expression involving the square root of a negative number. The use of 'I' for the imaginary unit is consistent with the result.
gpt-oss:20b: pass 2026-10-23

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 2026-10-24 with SymPy 1.14.0.