Question

If f(x) = the integral from -2 to x^4 of sqrt(t^2+8)dt then f'(x)=?
I will write this as an equation as well below

Answer 1

use main theorem of calculus again. Just substitute t^2 by (x^4)^2

Answer 2

so answer should be sqr(x^8+8)

Answer 3

If
\[f(x)=\int\limits_{-2}^{x^4} \sqrt{t^2+8}dt \]
f'(x)?

Answer 4

For some reason it is saying that \[\sqrt{x^8+8}\] is the wrong answer

Answer 5

sqr(x^8+8) isn't the right answer