If f(x) The integral from 4 to 2x of: the squareroot of (t^2 + 2). What is F' (2).
Would it be 0 because f (2) = 0? and the derivative of 0 is 0?
Or would it 2 x the squareroot of 12?
To find F'(2), we need to find the derivative of f(x) with respect to x and then evaluate it at x = 2.
To find the derivative of f(x), we can use the Fundamental Theorem of Calculus. According to the theorem, if f(x) is defined as the integral from a constant value to x of a function g(t), then the derivative of f(x) is equal to g(x).
In this case, f(x) is defined as the integral from 4 to 2x of √(t^2 + 2). Therefore, we need to find the derivative of √(t^2 + 2) with respect to t, and then evaluate it at t = 2x.
Differentiating √(t^2 + 2) with respect to t, we can use the chain rule. Let's call √(t^2 + 2) as u.
u = √(t^2 + 2)
Now, differentiate u with respect to t:
du/dt = (1/2)(t^2 + 2)^(-1/2) * 2t
= t / √(t^2 + 2)
Now, substitute t = 2x into the derivative:
du/dx = (2x) / √((2x)^2 + 2)
= (2x) / √(4x^2 + 2)
= (2x) / √2(2x^2 + 1)
Therefore, the derivative of f(x) is (2x) / √(2(2x^2 + 1)).
To find F'(2), we evaluate the derivative at x = 2:
F'(2) = (2(2)) / √(2(2(2^2) + 1))
= 4 / √(2(8) + 1)
= 4 / √(16 + 1)
= 4 / √17
So, F'(2) is equal to 4 divided by the square root of 17.