Calculating the derivatative

Let $y=2^{t^2}$.
Taking logs,
$\log(y) = t^2 \log(2)$,
Taking derivatives on both sides,
$\dfrac{y'}{y} = 2t \log(2)$,
so that 
$y' = 2ty \log(2) = 2t2^{t^2} \log(2) = t2^{t^2+1} \log(2)$.

On the other hand, let $z = \tan^2(t^2)$.
$z' = 2\tan(t^2)\cdot\sec^2(t^2)\cdot 2t$.

Therefore, for $f(t) = 2^{t^2}\cdot \tan^2(t^2)$,
$f'(t) = t2^{t^2+1} \log(2)\cdot \tan^2(t^2) + 2^{t^2}\cdot 2\tan(t^2)\cdot\sec^2(t^2)\cdot 2t$
$f'(t) = t2^{t^2+1} \log(2)\cdot \tan^2(t^2) + 2^{t^2+2}\cdot \tan(t^2)\cdot\sec^2(t^2)\cdot t$