Chain rule with a square root and a natural log

Write $h(x) = u^{1/2}$ where $u = \ln(x+1)$. By the chain rule, $h'(x) = \frac{1}{2} u^{-1/2} \cdot \frac{du}{dx}$. Since $u = \ln(x+1)$, another application of the chain rule gives $\frac{du}{dx} = \frac{1}{x+1}$.

Combining these, $h'(x) = \dfrac{1}{2\sqrt{\ln(x+1)}} \cdot \dfrac{1}{x+1}$. Now substitute $x = 1$: $\ln(1+1) = \ln 2$ and $x+1 = 2$, so $h'(1) = \dfrac{1}{2\sqrt{\ln 2}\cdot 2} = \dfrac{1}{4\sqrt{\ln 2}}$.

The outer function is the square root. We wrote h of x as u to the one half power where u equals the natural log of x plus one, so the outer layer applies square root to whatever the inner function produces. The inner function, natural log of x plus one, is applied first, then the square root is applied last.