Derivative of \( \displaystyle \left(1 + \frac{1}{5 x}\right)^{5 x} \)
Problem 2.516 · hard Video
Differentiate \( \displaystyle f(x) = \left(1 + \frac{1}{5 x}\right)^{5 x} \).
- \[ \frac{d}{d x} \left(1 + \frac{1}{5 x}\right)^{5 x} \]rewriteRewrite the base and exponent using the exponential and logarithmic functions.✓ Proved
- \[ = \frac{d}{d x} e^{5 x \ln{\left(1 + \frac{1}{5 x} \right)}} \]chainApply the chain rule for the exponential function.✓ Checked independently
- \[ = e^{5 x \ln{\left(1 + \frac{1}{5 x} \right)}} \frac{d}{d x} 5 x \ln{\left(1 + \frac{1}{5 x} \right)} \]productApply the product rule to the exponent.✓ Proved
- \[ = \left(5 x \frac{d}{d x} \ln{\left(1 + \frac{1}{5 x} \right)} + \ln{\left(1 + \frac{1}{5 x} \right)} \frac{d}{d x} 5 x\right) e^{5 x \ln{\left(1 + \frac{1}{5 x} \right)}} \]derivativeDifferentiate both parts of the product.✓ Proved
- \[ = \left(5 x \frac{d}{d x} \ln{\left(1 + \frac{1}{5 x} \right)} + 5 \ln{\left(1 + \frac{1}{5 x} \right)}\right) e^{5 x \ln{\left(1 + \frac{1}{5 x} \right)}} \]derivativeThe derivative of 5*x is 5.✓ Proved
- \[ = \left(\frac{5 x \frac{d}{d x} \frac{1}{5 x}}{1 + \frac{1}{5 x}} + 5 \ln{\left(1 + \frac{1}{5 x} \right)}\right) e^{5 x \ln{\left(1 + \frac{1}{5 x} \right)}} \]derivativeApply the chain rule to the logarithm.✓ Proved
- \[ = \left(5 \ln{\left(1 + \frac{1}{5 x} \right)} - \frac{1}{x \left(1 + \frac{1}{5 x}\right)}\right) e^{5 x \ln{\left(1 + \frac{1}{5 x} \right)}} \]derivative algebra algebra algebraThe derivative of 1/(5*x) is -1/(5*x^2). Simplify the double negative. Cancel 5*x in the numerator and denominator. Distribute x into the denominator.✓ Proved
- \[ = \left(5 \ln{\left(1 + \frac{1}{5 x} \right)} - \frac{1}{x + \frac{1}{5}}\right) e^{5 x \ln{\left(1 + \frac{1}{5 x} \right)}} \]algebraDistribute x into the denominator term.✓ Proved
- \[ = \left(5 \ln{\left(1 + \frac{1}{5 x} \right)} - \frac{5}{5 x + 1}\right) e^{5 x \ln{\left(1 + \frac{1}{5 x} \right)}} \]algebraMultiply numerator and denominator by 5 to clear the fraction.✓ Proved
- \[ = \left(1 + \frac{1}{5 x}\right)^{5 x} \left(5 \ln{\left(1 + \frac{1}{5 x} \right)} - \frac{5}{5 x + 1}\right) \]simplifyConvert the exponential form back to the original power form.✓ Checked independently
Answer \( \left(1 + \frac{1}{5 x}\right)^{5 x} \left(5 \log{\left(1 + \frac{1}{5 x} \right)} - \frac{1}{x \left(1 + \frac{1}{5 x}\right)}\right) \)
Lines: 12 proved, 2 checked by independent solution. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see the receipt. 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 | ✓ Checked independently | claude-sonnet-5-5 script, run by sympy 1.14.0, sympy 1.14.0 | sympy 1.14.0: simplify left 5*(((5*x + 1)/(5*x))**(5*x)*((5*x + 1)*log((5*x + 1)/(5*x)) - 1) + (-(5*x + 1)*log((5*x + 1)/(5*x)) + 1)*exp(5*x*log((5*x + 1)/(5*x))))/(5*x + 1); numeric agreement only, at 24 of 24 sampled points claude-sonnet-5-5 script, run by sympy 1.14.0: log of both inner functions agree for positive p (b**e = exp(e*log b)); derivatives of equal functions agree; diff residual: 0 ( checks/2.516/line2-acbc612b.py)undefined where x = 0 log is undefined for non-positive arguments |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x = 0 undefined where 1 + 1/(5*x) = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 1 + 1/(5*x) = 0 undefined where x = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 1 + 1/(5*x) = 0 undefined where x = 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 1 + 1/(5*x) = 0 undefined where x = 0 |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 1 + 1/(5*x) = 0 undefined where x = 0 |
| 11 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 1 + 1/(5*x) = 0 undefined where x = 0 undefined where x + 1/5 = 0 |
| 12 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x = 0 undefined where x + 1/5 = 0 undefined where 5*x + 1 = 0 |
| 13 | ✓ Checked independently | claude-sonnet-5-5 script, run by sympy 1.14.0, sympy 1.14.0 | sympy 1.14.0: simplify left 5*(((5*x + 1)/(5*x))**(5*x)*(-(5*x + 1)*log((5*x + 1)/(5*x)) + 1) + ((5*x + 1)*log((5*x + 1)/(5*x)) - 1)*exp(5*x*log((5*x + 1)/(5*x))))/(5*x + 1); numeric agreement only, at 24 of 24 sampled points claude-sonnet-5-5 script, run by sympy 1.14.0: exp(5p log base) = base**(5p) for positive base, remaining bracket identical ( checks/2.516/line13-3b8cd1c9.py)log is undefined for non-positive arguments undefined where 5*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 log is undefined for non-positive arguments undefined where 1 + 1/(5*x) = 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: fail (error) — Step 3 incorrectly labels the derivative of exp(u) as a product rule; it should be a chain rule. Step 4 mislabels the product rule applied to 5*x*log(...) as a derivative. Step 11 incorrectly simplifies 1/(x*(1+1/(5*x))) to 1/(x+1/5); the correct simplification is 5/(5*x+1).qwen3.6:27b-mlx: pass
Senior review claude-sonnet-5-5, 2026-10-09: fail — The math is correct and the final answer matches. However, line 2 is labelled 'chain' when it is really the rewrite a**b = exp(b*log a), and line 3 is labelled 'product' when it is the chain rule for exp(u). The labels are shifted one step, so a student would learn which rule is used where incorrectly.
gpt-oss:20b: uphold — The objection to step 3 is right: it differentiates exp(u) to exp(u)*u', which is the chain rule, so the 'product' label names a rule that step did not apply (step 2 is likewise labelled 'chain' when it is only the rewrite a**b = exp(b*log a)). The objections to steps 4 and 11 are dismissed below.gpt-oss:20b: dismiss — Step 4 does apply the product rule, but the label 'derivative' is in the vocabulary and the step legitimately differentiates, so either label is acceptable. The objection that it is mislabelled is not a real defect.gpt-oss:20b: dismiss — Step 11 is correct, because x*(1+1/(5x)) = x + 1/5 exactly and SymPy proved the line. It is not required to equal 5/(5x+1) at that point, since line 12 makes that change next.
Every verdict on record (7)
claude-sonnet-5-5: fail (misleading) 2026-10-09 — The math is correct and the final answer matches. However, line 2 is labelled 'chain' when it is really the rewrite a**b = exp(b*log a), and line 3 is labelled 'product' when it is the chain rule for exp(u). The labels are shifted one step, so a student would learn which rule is used where incorrectly.qwen3.6:27b-mlx: pass 2026-09-21gpt-oss:20b: fail (error) 2026-09-21 — Step 3 incorrectly labels the derivative of exp(u) as a product rule; it should be a chain rule. Step 4 mislabels the product rule applied to 5*x*log(...) as a derivative. Step 11 incorrectly simplifies 1/(x*(1+1/(5*x))) to 1/(x+1/5); the correct simplification is 5/(5*x+1).qwen3.6:27b-mlx: fail (error) 2026-09-20 — Step 1 applies the rewrite rule but leaves the outer derivative operator intact, which violates the 'one thing per step' constraint (it should be a separate step or the label should reflect the differentiation). More critically, Step 4 is labeled 'derivative' but performs the product rule expansion, which is a distinct rule from 'derivative' (which is reserved for basic derivatives like d/dx(x)). The product rule application should be labeled 'product'.gpt-oss:20b: fail (error) 2026-09-20 — Step 3 incorrectly labels the application of the chain rule for the exponential as a product rule, and step 4 applies a product rule to a single term. These steps combine multiple rules and mislabel the rule used, violating the contract.qwen3.6:27b-mlx: fail (error) 2026-09-20 — Step 1 is labeled 'rewrite' but performs the differentiation setup (converting to exp/log form) which is not a differentiation rule itself, and more critically, Step 2 is labeled 'chain' but the step shown is just the result of the rewrite from Step 1, not the application of the chain rule (which happens in Step 3). The label 'chain' in Step 2 is incorrect because the chain rule is not applied in that specific transition; the transition is purely algebraic rewriting. Furthermore, Step 3 is labeled 'product' but applies the chain rule to the outer exponential function. The label should be 'chain'. Step 4 is labeled 'derivative' but applies the product rule. The labels are consistently swapped or misapplied regarding which rule is being invoked at which stage.gpt-oss:20b: fail (error) 2026-09-20 — Step 3 incorrectly labels a chain‑rule derivative of exp(g) as a product rule, and step 4 repeats the same mistake. Moreover, steps 11–12 algebraically transform 1/(x*(1+1/(5x))) into 1/(x+1/5), which is not equivalent. These errors invalidate the solution.
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.