# What is the derivative of f(x)=tan^-1(e^x) ?

By Chain Rule, we can find $f ' \left(x\right) = \frac{{e}^{x}}{1 + {e}^{2 x}}$.
Note: $\left[{\tan}^{- 1} \left(x\right)\right] ' = \frac{1}{1 + {x}^{2}}$.
$f ' \left(x\right) = \frac{1}{1 + {\left({e}^{x}\right)}^{2}} \cdot {e}^{x} = \frac{{e}^{x}}{1 + {e}^{2 x}}$