## Calculus: Early Transcendentals 8th Edition

$\displaystyle y'=\frac{-\sec t \tan t}{2\sqrt{1-\sec t}}$ $\displaystyle y''=\frac{\sec t((\sec t -2)\tan^2 t +2\sec^3 t - 2\sec^2 t)}{4(1-\sec t)^{3/2}}$ OR $\displaystyle y''=\frac{\sec t(3\sec^3 t-4\sec^2 t-\sec t +2)}{4(1-\sec t)^{3/2}}$
$y'=\frac{d\sqrt{1-\sec{t}}}{dt}$ Using the chain rule: $y'=\frac{d\sqrt{1-\sec{t}}}{d(1-\sec t)} \times \frac{d(1-\sec t)}{dt}$ $=\frac{1}{2\sqrt{1-\sec t}} \times -\sec t \tan t$ $=\frac{-\sec t \tan t}{2\sqrt{1-\sec t}}$ $y'' = \frac{d\frac{-\sec t \tan t}{2\sqrt{1-\sec t}}}{dt}$ Using the quotient rule: $y''=\frac{-((\sec t \tan t) \times \tan t + \sec t \times (\sec^2 t)) \times (2\sqrt{1-\sec t}) -(-\sec t \tan t)\times (2 \times \frac{-\sec t \tan t}{2\sqrt{1-\sec t}})}{(2\sqrt{1-\sec t})^2}$ $=\frac{-2(\sec t \tan^2 t + \sec^3 t)\sqrt{1-\sec t} +(\frac{-2\sec^2 t \tan^2 t}{2\sqrt{1-\sec t}})}{4(1-\sec t)}$ $=\frac{-2(\sec t \tan^2 t + \sec^3 t)\frac{1-\sec t}{\sqrt{1-\sec t}} -(\frac{\sec^2 t \tan^2 t}{\sqrt{1-\sec t}})}{4(1-\sec t)}$ $=\frac{-2(\sec t \tan^2 t + \sec^3 t)({1-\sec t}) -(\sec^2 t \tan^2 t)}{4(1-\sec t)^{3/2}}$ $=\frac{-2(\sec t \tan^2 t-\sec^2 t \tan^2 t+\sec^3 t-\sec^4 t)-\sec^2 t \tan^2 t}{4(1-\sec t)^{3/2}}$ $=\frac{-2\sec t \tan^2 t+2\sec^2 t \tan^2 t-2\sec^3 t+2\sec^4 t-\sec^2 t \tan^2 t}{4(1-\sec t)^{3/2}}$ $=\frac{-2\sec t \tan^2 t+\sec^2 t \tan^2 t-2\sec^3 t+2\sec^4 t}{4(1-\sec t)^{3/2}}$ $=\frac{\sec t((\sec t -2)\tan^2 t +2\sec^3 t - 2\sec^2 t)}{4(1-\sec t)^{3/2}}$ Alternatively, this can be written with only sec (using the identity $\tan^2 t+1=\sec^2 t$): $=\frac{\sec t(3\sec^3 t-4\sec^2 t-\sec t +2)}{4(1-\sec t)^{3/2}}$