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

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

Answer 2

Sorry
It's x^8

Answer 3

\[\sqrt{x^8+8}\]

Answer 4

i'd go with

\[4x^3 \times \sqrt{x^8 + 8}\]

Answer 5

sub x in f(t), and apply the chain rule to x^4

Answer 6

thank you so much for the help it is much appreciated :)

Answer 7

Sorry, I'm wrong, I forgot to account for the upper limit being x^4 not just x