AP Calculus AB and BC

Derivative of 1/ln x: Chain Rule

The derivative of 1 over ln x is negative 1 over x times ln x squared. Rewrite the expression as ln x to the negative one, then the power rule and chain rule give the answer without needing the quotient rule.

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

Rewrite as a negative power

ddx(ln⁡x)−1=−(ln⁡x)−2⋅1x=−1x(ln⁡x)2\frac{d}{dx}(\ln x)^{-1} = -(\ln x)^{-2}\cdot\frac{1}{x} = -\frac{1}{x(\ln x)^{2}}

The minus comes from the exponent −1-1 coming down, and the 1x\frac{1}{x} is the chain rule.

Domain and behaviour

The function needs x>0x > 0 and x≠1x \neq 1, since ln⁡1=0\ln 1 = 0 makes it undefined. The derivative is negative wherever it exists, so the function is decreasing on each side of x=1x = 1.

Common mistakes

  • Answering −1(ln⁡x)2-\frac{1}{(\ln x)^{2}} and dropping the chain rule factor 1x\frac{1}{x}.
  • Answering ln⁡x\ln x or −ln⁡x-\ln x, inverting the wrong thing.
  • Missing that x=1x = 1 is excluded.

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 1/ln x?

It is −1x(ln⁡x)2-\frac{1}{x(\ln x)^{2}}.

What is the domain?

x>0x > 0 with x≠1x \neq 1, since ln⁡1=0\ln 1 = 0.