Can someone help me answer this?
Determine F ' (x) and F ' (1) if
F(x) = integral sign lower limit = 4, upper limit = x
4t^10 (2 t^(1) 0 + 8 ln t) ^4 dt
a) F ' (x) =
b) F' (1) =
To determine F'(x) and F'(1) for the given function F(x), we need to use the Fundamental Theorem of Calculus and apply the rules of differentiation.
The Fundamental Theorem of Calculus states that if F(x) = ∫[a to x] f(t) dt, then F'(x) = f(x), where f(x) is the integrand of the original function.
In this case, we have F(x) = ∫[4 to x] 4t^10 (2t^0 + 8ln(t))^4 dt.
To find F'(x), we differentiate the integrand with respect to x. Let's break it down step by step:
1. Start by differentiating the outer function, which is (2t^0 + 8ln(t))^4, using the chain rule. Let's call this function g(t).
g(t) = (2t^0 + 8ln(t))^4.
To differentiate g(t), we need to differentiate the power function first. So, we get:
g'(t) = 4(2t^0 + 8ln(t))^3 * (d/dt)(2t^0 + 8ln(t)).
2. Next, differentiate the inner function, (2t^0 + 8ln(t)), using the sum rule and the derivative of ln(t):
(d/dt)(2t^0 + 8ln(t)) = 0 + 8(1/t) = 8/t.
Plugging this back into g'(t), we have:
g'(t) = 4(2t^0 + 8ln(t))^3 * (8/t).
3. Finally, substitute g'(t) back into the original expression for F'(x):
F'(x) = 4t^10 * g'(t).
F'(x) = 4t^10 * 4(2t^0 + 8ln(t))^3 * (8/t).
Simplifying further, we have:
F'(x) = 16t^10 * (2t^0 + 8ln(t))^3 * (8/t).
Now let's evaluate F'(1) by substituting t = 1 into the expression:
F'(1) = 16(1)^10 * (2(1)^0 + 8ln(1))^3 * (8/1).
Simplifying this further, we have:
F'(1) = 16 * 2^3 * 8.
F'(1) = 2048.
So, the answers are:
a) F'(x) = 16t^10 * (2t^0 + 8ln(t))^3 * (8/t).
b) F'(1) = 2048.