∫Calc Practice

Derivative of \( \displaystyle \left(x + 2\right)^{\frac{1}{x + 2}} \)

Problem 2.494 · hard

Differentiate \( \displaystyle f(x) = \left(x + 2\right)^{\frac{1}{x + 2}} \).
  1. \[ \frac{d}{d x} \left(x + 2\right)^{\frac{1}{x + 2}} \]
    derivativeStart with the derivative of the function.✓ Proved
  2. \[ = \frac{d}{d x} e^{\frac{\ln{\left(x + 2 \right)}}{x + 2}} \]
    rewriteRewrite the base and exponent using the exponential and logarithm.≈ Checked numerically
  3. \[ = e^{\frac{\ln{\left(x + 2 \right)}}{x + 2}} \frac{d}{d x} \frac{\ln{\left(x + 2 \right)}}{x + 2} \]
    chainApply the chain rule for the exponential function.✓ Proved
  4. \[ = \left(\ln{\left(x + 2 \right)} \frac{d}{d x} \frac{1}{x + 2} + \frac{\frac{d}{d x} \ln{\left(x + 2 \right)}}{x + 2}\right) e^{\frac{\ln{\left(x + 2 \right)}}{x + 2}} \]
    productApply the product rule to the inner expression.✓ Proved
  5. \[ = \left(- \frac{\ln{\left(x + 2 \right)}}{\left(x + 2\right)^{2}} + \frac{1}{\left(x + 2\right)^{2}}\right) e^{\frac{\ln{\left(x + 2 \right)}}{x + 2}} \]
    derivative algebraDifferentiate the individual terms. Combine the terms inside the parentheses.✓ Proved
  6. \[ = \frac{\left(1 - \ln{\left(x + 2 \right)}\right) e^{\frac{\ln{\left(x + 2 \right)}}{x + 2}}}{\left(x + 2\right)^{2}} \]
    algebraFactor out the common denominator.✓ Proved
  7. \[ = \frac{\left(1 - \ln{\left(x + 2 \right)}\right) \left(x + 2\right)^{\frac{1}{x + 2}}}{\left(x + 2\right)^{2}} \]
    simplifySubstitute the original expression back for the exponential term.≈ Checked numerically
Answer \( \left(1 - \log{\left(x + 2 \right)}\right) \left(x + 2\right)^{-2 + \frac{1}{x + 2}} \)

✓ Nihil obstat Lines: 7 proved, 2 checked numerically. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.

The full receipt
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0line 1 is the problem as stated
2≈ Checked numericallysympy 1.14.0sympy 1.14.0: simplify left ((1 - log(x + 2))*(x + 2)**(1/(x + 2)) + (log(x + 2) - 1)*exp(log(x + 2)/(x + 2)))/(x + 2)**2; numeric agreement only, at 24 of 24 sampled points
undefined where x + 2 = 0
log is undefined for non-positive arguments
3✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where x + 2 = 0
4✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where x + 2 = 0
5✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where x + 2 = 0
6✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where x + 2 = 0
7✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where x + 2 = 0
8≈ Checked numericallysympy 1.14.0sympy 1.14.0: simplify left ((1 - log(x + 2))*exp(log(x + 2)/(x + 2)) + (x + 2)**(1/(x + 2))*(log(x + 2) - 1))/(x + 2)**2; numeric agreement only, at 24 of 24 sampled points
log is undefined for non-positive arguments
undefined where x + 2 = 0
answer✓ Provedsympy 1.14.0final line against the stated answer: simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where x + 2 = 0
answer, a second way✓ Provedsympy 1.14.0SymPy differentiated f directly and got the stated answer

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass
Every verdict on record (6)
  • qwen3.6:27b-mlx: pass 2026-09-21
  • gpt-oss:20b: pass 2026-09-21
  • qwen3.6:27b-mlx: pass 2026-09-20
  • gpt-oss:20b: pass 2026-09-20
  • qwen3.6:27b-mlx: pass 2026-09-20
  • gpt-oss:20b: pass 2026-09-20

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.