Find the derivative from x to x^8 (integral) of ln t dt.

f(x) = ∫[x,x^8] ln(t) dt

The 2nd fundamental theorem is really just the chain rule.

f'(x) = ln(x^8)*8x^7 - ln(x)*1/x
= 8lnx*8x^7 - ln * 1/x
= lnx (84x^7 - 1/x)

This works because if

g(t) = ∫lnt dt
g'(t) = ln(t)
So,
f(x) = g(x^8) - g(x)
so, f'(x) = g'(x^8)*8x^7 - g'(x)*1/x
as shown above

To find the derivative of the integral of ln(t) dt from x to x^8, we can use the Fundamental Theorem of Calculus.

According to the Fundamental Theorem of Calculus, if F(x) is the antiderivative of f(x), then the derivative of the integral of f(x) from a to x is simply F(x).

In this case, we need to find the antiderivative of ln(t). The integral of ln(t) is t*ln(t) - t + C, where C is a constant.

Using the Fundamental Theorem of Calculus, the derivative of the integral from x to x^8 of ln(t) dt is equal to the antiderivative evaluated at the upper limit of integration (x^8).

Therefore, the derivative of the integral from x to x^8 of ln(t) dt is (x^8 * ln(x^8) - x^8) - (x * ln(x) - x) = 8x^7 * ln(x^8) - x^8 * ln(x) + x.

So, the derivative is 8x^7 * ln(x^8) - x^8 * ln(x) + x.