Second derivative of tan(x) evaluated at a point

Start with $f(x) = \tan(x)$, so $f'(x) = \sec^2(x)$. To find $f''(x)$, write $f'(x) = (\sec x)^2$ and apply the chain rule: $\frac{d}{dx}[(\sec x)^2] = 2\sec(x)\cdot \frac{d}{dx}[\sec x] = 2\sec(x)\sec(x)\tan(x)$. This gives $f''(x) = 2\sec^2(x)\tan(x)$.

Now evaluate at $x = \pi/4$: $\sec(\pi/4) = \sqrt{2}$, so $\sec^2(\pi/4) = 2$, and $\tan(\pi/4) = 1$. Substituting, $f''(\pi/4) = 2(2)(1) = 4$.