need to figure this out

Question

Calculate the derivative

 

ddx∫e8xx4ln(t)dtddx∫x4e8xlntdt

 

using Part 2 of the Fundamental Theorem of Calculus.

 

Enclose arguments of functions in parentheses. For example, sin(2x)sin2x.

 

ddx∫e8xx4ln(t)dt=

Answer 1

Let f(x) = ∫_{(from x^4 to e^(8x))} ln(t)dt = ∫_{(x^4 to 1)} ln(t)dt + ∫_{(1 to e^(8x))} ln(t)dt

= -∫_{(1 to x^4)} ln(t)dt +∫_{(1 to e^(8x))} ln(t)dt

So, f'(x) = - ln(x⁴)(x⁴)' + ln(e^8x)(e^8x)' = - 4(lnx)(4x³) + 8x(8e^8x) = -16x³lnx + 64xe^8x

Note: If f(x) = ∫_{(a to r(x))} g(t)dt, then f'(x) = g(r(x))(r'(x))

Answer 2

d/dx(f(x) = 8x·8e^8x- 4ln(x³)4x³ = 64xe^8x - 48x³ln(IxI)