AP Calculus AB and BC

Derivative of ln x / x^2: Quotient Rule

The derivative of ln x over x squared is 1 minus 2 ln x, all over x cubed. Setting the numerator to zero gives ln x equal to one half, so the function has a maximum at x equals the square root of e.

ddx[ln⁡xx2]=1−2ln⁡xx3\frac{d}{dx}\left[\frac{\ln x}{x^{2}}\right] = \frac{1 - 2\ln x}{x^{3}}

Quotient rule, then cancel

1x⋅x2−ln⁡x⋅2xx4=x−2xln⁡xx4=1−2ln⁡xx3\frac{\frac{1}{x}\cdot x^{2} - \ln x\cdot 2x}{x^{4}} = \frac{x - 2x\ln x}{x^{4}} = \frac{1 - 2\ln x}{x^{3}}

The maximum at the square root of e

For x>0x > 0 the denominator is positive, so the sign is that of 1−2ln⁡x1 - 2\ln x. That is positive until ln⁡x=12\ln x = \frac{1}{2}, giving a maximum at x=e≈1.649x = \sqrt{e} \approx 1.649 with value 12e\frac{1}{2e}.

Common mistakes

  • Failing to cancel the shared factor of xx and leaving x4x^{4} underneath.
  • Solving 1−2ln⁡x=01 - 2\ln x = 0 as x=12x = \frac{1}{2} rather than e1/2e^{1/2}.

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

What is the derivative of ln x / x^2?

It is 1−2ln⁡xx3\frac{1 - 2\ln x}{x^{3}}.

Where is the maximum?

At x=ex = \sqrt{e}, with value 12e\frac{1}{2e}.