## Calculus 10th Edition

$g''(\dfrac{\pi}{6})=32\sqrt{3}.$
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.