## Calculus: Early Transcendentals 8th Edition

$$f''(\pi/4)=3\sqrt 2$$
$$f(t)=\sec t$$ - First derivative: $$f'(t)=\sec t\tan t$$ - Second derivative: $$f''(t)=(\sec t\tan t)'$$ Apply the Product Rule: $$f''(t)=(\sec t)'\tan t+\sec t(\tan t)'$$ $$f''(t)=\sec t\tan t\tan t+\sec t\sec^2 t$$ $$f''(t)=\sec t\tan^2 t+\sec^3 t$$ We have $\sec(\pi/4)=\sqrt 2$ and $\tan(\pi/4)=1$. So, $$f''(\pi/4)=\sec(\pi/4)\tan^2(\pi/4)+\sec^3(\pi/4)$$ $$f''(\pi/4)=\sqrt 2\times1^2+(\sqrt2)^3$$ $$f''(\pi/4)=\sqrt2+2\sqrt2=3\sqrt 2$$