Derivative of ln(x + sqrt(x^2+1)): Answer and Proof

The cancellation that cleans it up

Set $u=x+\sqrt{x^{2}+1}$, so the outer derivative is $\frac{1}{u}$. The inner derivative needs its own chain rule on the root, where the inside $x^{2}+1$ contributes a factor of $2x$ that cancels the $2$ underneath.

Putting those two terms over the common denominator $\sqrt{x^{2}+1}$ is the whole trick, because the numerator that appears is $u$ itself. Multiply by $\frac{1}{u}$ and the bracket disappears.

Why every real number is in the domain

A logarithm needs a positive input, so the question is whether $x+\sqrt{x^{2}+1}$ can ever fail to be positive. It cannot. For $x\geq 0$ both pieces are non-negative and the root is strictly positive. For $x<0$, compare sizes: $\sqrt{x^{2}+1}>\sqrt{x^{2}}=|x|$, so the root always outweighs the negative term.

That is why the sample point $x=-1$ is legitimate here, unlike on a page about $\ln x$. The derivative $\frac{1}{\sqrt{x^{2}+1}}$ is positive for every $x$, so the function increases across the whole real line, with its steepest slope of $1$ at the origin.

The function is the inverse hyperbolic sine, written $\sinh^{-1}x$, and $\frac{1}{\sqrt{x^{2}+1}}$ is the standard formula for its derivative. Compare $\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^{2}}}$: one sign inside the root separates the two results.

The mistakes students make

The size of the expression tempts students into shortcuts that are not legal.

• Answering $\frac{1}{x+\sqrt{x^{2}+1}}$ and stopping. That is only the outer derivative, and it is precisely the factor that gets cancelled once the inner derivative is written down.
• Differentiating the root as $\frac{1}{2\sqrt{x^{2}+1}}$ and forgetting the $2x$ from inside. The correct term is $\frac{x}{\sqrt{x^{2}+1}}$, and without it the cancellation never happens.
• Splitting the logarithm as $\ln x+\ln\sqrt{x^{2}+1}$. The log of a sum is not the sum of logs, and the split also throws away the negative half of the domain.