Derivative of a composed arctan-ln function

Identify the outer function as $f(u) = \arctan(u)$ and the inner function as $u = g(x) = \ln(x)$, so $h(x) = f(g(x))$. Recall that $f'(u) = \dfrac{1}{1+u^2}$ and $g'(x) = \dfrac{1}{x}$.

By the chain rule, $h'(x) = f'(g(x))\cdot g'(x) = \dfrac{1}{1+(\ln x)^2}\cdot \dfrac{1}{x}$. Combining these gives the final simplified derivative.

The outer function is arctangent, since h of x equals arctangent applied to the natural log of x. The natural log of x plays the role of the inner function, being evaluated first. This ordering determines how the chain rule is applied, with arctangent's derivative evaluated at the inner function. Recognizing this structure is essential for correctly setting up the derivative.