AP Calculus AB and BC

Derivative of ln(ln x): Nested Chain Rule

The derivative of ln of ln x is 1 over x times ln x. The outer logarithm contributes 1 over its inner function, and the inner logarithm contributes 1 over x. The domain requires x greater than 1, since ln x must itself be positive.

ddx[ln⁡(ln⁡x)]=1xln⁡x\frac{d}{dx}\left[\ln\left(\ln x\right)\right] = \frac{1}{x\ln x}

Two layers

ddxln⁡u=u′u,u=ln⁡x,u′=1x\frac{d}{dx}\ln u = \frac{u'}{u}, \quad u = \ln x, \quad u' = \frac{1}{x}
ddxln⁡(ln⁡x)=1/xln⁡x=1xln⁡x\frac{d}{dx}\ln(\ln x) = \frac{1/x}{\ln x} = \frac{1}{x\ln x}

The domain is stricter than it looks

The outer logarithm needs its input positive, so ln⁡x>0\ln x > 0, which means x>1x > 1. It is not enough for xx to be positive.

Read backwards this derivative is the antiderivative fact ∫dxxln⁡x=ln⁡∣ln⁡x∣+C\int \frac{dx}{x\ln x} = \ln\left|\ln x\right| + C, which is where it earns most of its exam marks.

Common mistakes

  • Answering 1ln⁡x\frac{1}{\ln x} and forgetting the inner derivative.
  • Answering 1x\frac{1}{x}, differentiating only one layer.
  • Giving the domain as x>0x > 0. It is x>1x > 1.

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(ln x)?

It is 1xln⁡x\frac{1}{x\ln x}.

What is the domain?

x>1x > 1, because the outer logarithm needs ln⁡x\ln x to be positive.

What is the related antiderivative?

∫dxxln⁡x=ln⁡∣ln⁡x∣+C\int \frac{dx}{x\ln x} = \ln\left|\ln x\right| + C.