Problem involving first fundamental theorem of calculus help

If F(x) = ∫ (from 1 to x) f(t)dt, where f(t)=∫ (from 1 to t^2) (√(1+u^2))du/u , find F"(x).

I'm so confused, I do not get what to plug in where. Help?

Let G(t) be an antiderivative of f(t), so G'(t) = f(t).

Then by the fundamental theorem of calculus,
F(x) = ∫ (from 1 to x) f(t)dt = G(x) - G(1)

So F'(x) = G'(x) - 0 = G'(x) = f(x) = ∫ (from 1 to x^2) (√(1+u^2))du/u.

Let H(u) be an antiderivative of (√(1+u^2))/u, so H'(u) = (√(1+u^2))/u.

Then by the fundamental theorem of calculus,
F'(x) = ∫ (from 1 to x^2) (√(1+u^2))du/u = H(x^2) - H(1).

So from the chain rule,
F''(x) = [(d/dx)(x^2)]H'(x^2) - 0 = 2xH'(x^2) = 2x(√(1+x^4))/(x^2) = 2(√(1+x^4))/x.

Note: the second answerer almost had the solution correct, but replaced u by x instead of by x^2 in the denominator, in the second line of the solution.

If F(x)=Integral{1 to x of f(t)} then the fundamental theorem of calculus says that F '(x)=f(x) and so F ''(x)=f '(x). If you compute the derivative of f(x) w.r.t x you have F''(x).

Hope this helps

F(x) = ∫ (from 1 to x) {f(t)} dt, where f(t) = ∫ (from 1 to t^2) {√(1+u^2)/u} du

Need to find F '' (x). First find F ' (x).

By the FTC, F ' (x) = f(x) = ∫ (from 1 to x^2) {√(1+u^2)/u} du

Let G(x) = ∫ (from 1 to x) {√(1+u^2)/u} du.
Then G(x^2) = f(x) = F ' (x)

F '' (x) = f ' (x) = G ' (x^2) = 2x * √(1+(x^2)^2)/x^2

F '' (x) = (2/x) √(1 + x^4)

F'(t) = f(t)

Let g(u) = (√(1+u^2))/u
f(t)=t^2

F''(t) = g(f(u)) f'(u) = g(t^2) (2t) = 2t (√(1+t^2))/t
F''(x) = 2x (√(1+x^2))/x

F'(x) = f(x)

F"(x) = f'(x) = √(1 + (x^2)^2)/x • 2x = 2 √(1 + x^4)