Home›Calculus 1›Quotient rule›Problem 2.586

Derivative of \( \displaystyle \frac{\sqrt{2} \sqrt{x}}{2 x + 1} \)

Problem 2.586 · hard Video

Differentiate \( \displaystyle f(x) = \frac{\sqrt{2} \sqrt{x}}{2 x + 1} \).

\[ \frac{d}{d x} \frac{\sqrt{2} \sqrt{x}}{2 x + 1} \]
Start with the derivative of the function.✓ Proved
\[ = \sqrt{2} \frac{d}{d x} \frac{\sqrt{x}}{2 x + 1} \]
constant-multiple rewritePull out the constant factor sqrt(2). Rewrite the denominator using a negative exponent.✓ Proved
\[ = \sqrt{2} \left(\sqrt{x} \frac{d}{d x} \frac{1}{2 x + 1} + \frac{\frac{d}{d x} \sqrt{x}}{2 x + 1}\right) \]
productApply the product rule.✓ Proved
\[ = \sqrt{2} \left(- \frac{\sqrt{x} \frac{d}{d x} \left(2 x + 1\right)}{\left(2 x + 1\right)^{2}} + \frac{\frac{d}{d x} \sqrt{x}}{2 x + 1}\right) \]
chainApply the chain rule to the second term.✓ Proved
\[ = \sqrt{2} \left(- \frac{2 \sqrt{x}}{\left(2 x + 1\right)^{2}} + \frac{\frac{d}{d x} \sqrt{x}}{2 x + 1}\right) \]
derivative constant-multipleDifferentiate the inner function 2x + 1. Simplify the second term's constant.✓ Proved
\[ = \sqrt{2} \left(- \frac{2 \sqrt{x}}{\left(2 x + 1\right)^{2}} + \frac{1}{2 \sqrt{x} \left(2 x + 1\right)}\right) \]
derivative algebraDifferentiate sqrt(x). Simplify the expression.✓ Proved
\[ = \frac{\sqrt{2} \left(1 - 2 x\right)}{2 \sqrt{x} \left(2 x + 1\right)^{2}} \]
algebra algebra algebra simplifyFind a common denominator. Simplify the numerator. Combine like terms. Final simplified form.✓ Proved

Answer \( \frac{\sqrt{2} \left(\frac{1}{2} - x\right)}{\sqrt{x} \left(2 x + 1\right)^{2}} \)

Top: the graph of f = sqrt(2)·sqrt(x)/(2x + 1). A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = -2sqrt(2)·sqrt(x)/(2x + 1)² + sqrt(2)/(2sqrt(x)·(2x + 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	line 1 is the problem as stated
2	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where 2*x + 1 = 0
3	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where 2*x + 1 = 0
4	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where 2*x + 1 = 0
5	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where 2*x + 1 = 0
6	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where 2*x + 1 = 0
7	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where 2*x + 1 = 0
8	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where 2*x + 1 = 0 undefined where x = 0
9	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where x = 0 undefined where 2*x + 1 = 0
10	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where x = 0 undefined where 2*x + 1 = 0
11	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where 2*x + 1 = 0 undefined where x = 0
12	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where 2*x + 1 = 0 undefined where x = 0
13	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where 2*x + 1 = 0 undefined where x = 0
answer	✓ Proved	sympy 1.14.0	final line against the stated answer: simplify(a - b) reduced to 0 undefined where 2*x + 1 = 0 undefined where x = 0
answer, a second way	✓ Proved	sympy 1.14.0	SymPy differentiated f directly and got the stated answer

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass — The solution correctly applies differentiation rules and algebraic simplifications in single-step increments. The labels accurately reflect the operations performed.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly applies differentiation rules and algebraic simplifications in single-step increments. The labels accurately reflect the operations performed.
gpt-oss:20b: pass 2026-09-26
qwen3.6:27b-mlx: pass 2026-09-21 — The solution correctly applies the constant multiple, rewrite, product, and chain rules in distinct steps. The algebraic simplification steps are valid and lead to the correct final answer.
gpt-oss:20b: pass 2026-09-21
qwen3.6:27b-mlx: pass 2026-09-21
gpt-oss:20b: pass 2026-09-21

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 gemma4:26b, checked 2026-09-26 with SymPy 1.14.0.