Derivative of arctan(e^x) via the chain rule

Recognize $h(x) = f(g(x))$ where $f(u) = \arctan(u)$ and $g(x) = e^x$. Since $f'(u) = \dfrac{1}{1+u^2}$ and $g'(x) = e^x$, the chain rule gives $h'(x) = f'(g(x)) \cdot g'(x)$.

Substituting $u = e^x$ into $f'(u)$ gives $\dfrac{1}{1+(e^x)^2} = \dfrac{1}{1+e^{2x}}$. Multiplying by $g'(x) = e^x$ yields the final result $h'(x) = \dfrac{e^x}{1+e^{2x}}$.

The outer function was arctan of u, since h of x was viewed as f of g of x with f of u equal to arctan of u. The inner function was e to the x, which played the role of u. Recognizing this split is what let the chain rule be applied correctly. So the outer function is simply arctan of u.