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) =
f(x) = ∫[4,x] 4t^10(2t+8lnt)^4 dt
Don't know what the 0 is doing in there; if I got it wrong, fix it and modify the solution.
Using Leibnitz's Rule for differentiation under the integral sign,
f'(x) = 4x^10 (2x+8lnx)^4
so,
f'(1) = 4(2+0)^4 = 64