Answer
$g''(\dfrac{\pi}{6})=32\sqrt{3}.$
Work Step by Step
Using the Chain Rule:
$g'(t)=2\sec^2{2t}$
Using the Chain Rule:
$u=\sec{2t}$; $\dfrac{du}{dt}=2\sec{2t}\tan{2t}$
$g(u)=2u^2;\dfrac{d}{du}g'(u)=4u$
$\dfrac{d}{dt}g'(t)=\dfrac{d}{du}g'(u)\times\dfrac{du}{dt}=8\sec^2{2t}\tan{2t}.$
$g''(\dfrac{\pi}{6})=8\sec^2{\dfrac{2\pi}{6}}\tan{\dfrac{2\pi}{6}}=32\sqrt{3}.$
A computer algebra system was used to verify these results.
