If Upper F(x) = int_1^x (f(t)) d t where f(t) = int_1^(t^2) (sqrt(2 + u^2) divided by u) d u find (Upeer F) prime prime (2)

Answer 1

To find the second derivative of Upper F(x), we will first find the first derivative and then take its derivative again.

Given: Upper F(x) = int_1^x f(t) dt

Let's find the first derivative of Upper F(x):

Upper F'(x) = f(x)

Since f(t) = int_1^(t^2) (sqrt(2 + u^2) / u) du, we need to find the derivative of f(t):

f'(t) = d/dt (int_1^(t^2) (sqrt(2 + u^2) / u) du)

To find f'(t), we will use the Leibniz rule:

f'(t) = (d/dt t^2) (sqrt(2 + (t^2)^2) / t^2) - (d/dt 1) (sqrt(2 + 1^2) / 1)

f'(t) = (2t) (sqrt(2 + t^4) / t^2) - 0

f'(t) = 2 (sqrt(2 + t^4) / t)

Therefore, Upper F'(x) = f(x) = 2 (sqrt(2 + x^4) / x)

Now, to find the second derivative:

Upper F''(x) = f'(x) = 2 (sqrt(2 + x^4) / x)

Now, to find (Upper F)''(2), we substitute x = 2 into Upper F''(x):

(Upper F)''(2) = f'(2) = 2 (sqrt(2 + 2^4) / 2)

(Upper F)''(2) = 2 (sqrt(18) / 2)

(Upper F)''(2) = sqrt(18)

Therefore, the second derivative of Upper F at x = 2 is sqrt(18).