# calculus

Let F(x)= the integral from 0 to 2x of tan(t^2) dt. Use your calculator to find F″(1)

By applying the fundamental theorem of calculus, I got the derivative of the integral (F'(x)) to be 2tan(2x^2)

When I take the derivative to find F''(x) I get 8x sec^2(2x^2). When I plug 1 in for x, I don't get any of the answer choices, but I don't know where I went wrong in my evaluations. I just need to know my mistake

1. actually, F' = 2tan((2x)^2) = 2tan(4x^2)
F' = 2sec^2(4x^2)(8x) = 16x sec^2(4x^2)

posted by Steve

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