Question

Calculus
Find the second derivative of f(x) = sec(pix)

Answer 1

\[f(x) = \sec (\pi x)\]\[f'(x) = \pi \sec (\pi x) \tan (\pi x)\]\[f''(x) = \pi \left[\sec (\pi x) (\pi \sec^{2}(\pi x)) + \pi \sec(\pi x) \tan (\pi x) \tan (\pi x) \right]\]\[f''(x) = \pi^{2} \left[\sec^{3}(\pi x)+ \sec(\pi x)\tan^{2}(\pi x) \right]\]\[f''(x) = \pi^{2}\sec(\pi x) \left[\sec^{2}(\pi x)+ \tan^{2}(\pi x) \right]\]

Answer 2

Also, you can check it with wolfram alpha: http://www.wolframalpha.com/input/?i=second+derivative+of+sec%28pi+x%29

Answer 3

\[\frac{d}{dx}\sec(\pi x)=\frac{d}{dx}(\pi x)*\frac{d}{dx}\sec(x)|_{x=\pi x}=\pi*\left[ \sec(x)\tan(x) \right]_{x=\pi x}=\pi \sec(\pi x) \tan(\pi x)\]
Repeat for the second derivative but you have a product rule.

Answer 4

why did you put pi on the side for the second row?

Answer 5

I pulled out pi as a constant from the first derivative to make it easier for me when using the product rule.

Answer 6

but you still have it in the middle

Answer 7

That's because when I differentiated the sec(pi x)tan(pi x) from the first derivative, the chain rule created another pi

Answer 8

ah i see thanks

Answer 9

sec πx = 1/cos πx
(1/cos πx) ' = - ( - π sin πx)/ cos² πx)
= + π tan πx / cos πx)

Answer 10

f"(x) = π² ( sec² πx cosπx + tanπx sinπx) / cos²πx