Answer
$2e^x \sec^2 (2e^x-1)$
Work Step by Step
Our aim is to solve the function $f(x)=\tan (2e^{x}-1)$
We differentiate both sides with respect to $x$.
$f^{\prime}(x)=\dfrac{d}{dx}[\tan (2e^{x}-1)]$
or, $=\sec^2 (2e^x-1) \dfrac{d}{dx} (2e^{x}-1)$
or, $=2e^x \sec^2 (2e^x-1)$
