a.) Suppose that f(x)=4x−tanx, −π2<x<π2. Find f′ and f″
\begin{equation} \begin{aligned} f'(x) &= 4 \frac{d}{dx} (x) - \frac{d}{dx} (\tan x)\\ \\ f'(x) &= 4 - \sec ^2 x\\ \\ f''(x) &= \frac{d}{dx} (4) - \frac{d}{dx} (\sec^2 x)\\ \\ f''(x) &= \frac{d}{dx} (4) - \frac{d}{dx} (\sec x)^2\\ \\ f''(x) &= 0 -2 (\sec x) \frac{d}{dx} (\sec x)\\ \\ f''(x) &= - 2 \sec x \sec x \tan x \\ \\ f''(x) &= -2 \sec^2 x \tan x \end{aligned} \end{equation}
b.) Graph f, f', and f''
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