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 apply the Fundamental Theorem of Calculus and then differentiate the integrand with respect to x.
a) F'(x):
According to the Fundamental Theorem of Calculus, to find the derivative of a function defined as the integral of another function, we can differentiate the integrand and evaluate it at x.
In this case, the integrand is 4t^10(2t^10 + 8ln(t))^4 dt. So, we need to differentiate 4t^10(2t^10 + 8ln(t))^4 with respect to t.
To differentiate this expression, we can use the chain rule. The derivative of t^n with respect to t is n*t^(n-1). Additionally, the derivative of ln(t) with respect to t is 1/t.
Applying the chain rule, we differentiate 4t^10(2t^10 + 8ln(t))^4 as follows:
d/dt [4t^10(2t^10 + 8ln(t))^4]
= 4(t^10)'(2t^10 + 8ln(t))^4 + 4t^10(2t^10 + 8ln(t))^3 * (2t^10 + 8ln(t))'
= 4(10t^9)(2t^10 + 8ln(t))^4 + 4t^10(2t^10 + 8ln(t))^3 * (2(10t^9) + 8(1/t))
= 80t^9(2t^10 + 8ln(t))^4 + 4t^10(2t^10 + 8ln(t))^3 * (20t^9 + 8/t)
= 80t^9(2t^10 + 8ln(t))^4 + 4t^10(2t^10 + 8ln(t))^3 * (20t^10 + 8)/t
= 80t^9(2t^10 + 8ln(t))^4 + 4t^10(2t^10 + 8ln(t))^3 * (20t^10 + 8)/t
Now that we have the derivative of the integrand, F'(x) is given by that expression.
Therefore, the answer to a) is: F'(x) = 80x^9(2x^10 + 8ln(x))^4 + 4x^10(2x^10 + 8ln(x))^3 * (20x^10 + 8)/x
b) F'(1):
To find F'(1), we substitute x = 1 into the expression we obtained in part a).
F'(1) = 80(1)^9(2(1)^10 + 8ln(1))^4 + 4(1)^10(2(1)^10 + 8ln(1))^3 * (20(1)^10 + 8)/(1)
= 80(1)(2 + 0)^4 + 4(1)(2 + 0)^3 * (20 + 8)
= 80(2)^4 + 4(2)^3 * 28
= 80 * 16 + 4 * 8 * 28
= 1280 + 896
= 2176
Therefore, the answer to b) is: F'(1) = 2176.