AP Calculus AB and BC

Derivative of ln(x+1): Answer and Chain Rule

The derivative of ln of x plus 1 is 1 over x plus 1. The chain rule rule for a logarithm is the inner derivative over the inner function, and here the inner derivative is 1. The domain requires x greater than negative 1.

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

The logarithm chain rule

ddxln⁡u=u′u  ⟹  ddxln⁡(x+1)=1x+1\frac{d}{dx}\ln u = \frac{u'}{u} \implies \frac{d}{dx}\ln(x+1) = \frac{1}{x+1}

The inner derivative is 11, so nothing extra appears on top. With an inner function like x2+1x^{2}+1 the numerator would be 2x2x instead.

Why this function matters

At the origin the function passes through 00 with slope 11, so its tangent line is y=xy = x. That is the source of the approximation ln⁡(1+x)≈x\ln(1+x) \approx x for small xx, and of the standard limit.

lim⁡x→0ln⁡(1+x)x=1\lim_{x \to 0}\frac{\ln(1+x)}{x} = 1

Its Maclaurin series starts x−x22+x33−⋯x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \cdots, converging on −1<x≤1-1 < x \le 1, which is a standard BC result.

Common mistakes

  • Answering 1x\frac{1}{x}. That is the derivative of ln⁡x\ln x; the shift stays in the denominator.
  • Splitting ln⁡(x+1)\ln(x+1) into ln⁡x+ln⁡1\ln x + \ln 1. There is no logarithm rule for a sum inside.
  • Ignoring the domain. The function needs 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(x+1)?

It is 1x+1\frac{1}{x+1}.

Why is it not 1/x?

The chain rule puts the INNER function in the denominator, and the inner function is x+1x+1.

What is its Maclaurin series?

x−x22+x33−⋯x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \cdots, valid for −1<x≤1-1 < x \le 1.