Question

find F'(x) when F(x)= integral from 4 to x^2 (2 sqrt(1+t^3) dt)

Answer 1

\[F(x)=\int\limits_4^{x^2} g(t) dt=G(t)|_4^{x^2}=G(x^2)-G(4) \\ \text{so we have } F(x)=G(x^2)-G(4) \\ \text{ differentiate both sides }\]

Answer 2

\[F'(x)=(x^2)'g(x^2)-0 \text{ by chain rule and constant rule }\]

Answer 3

@aawowaa are you there?

Answer 4

+

Answer 5

yes, I'm going over it.

Answer 6

you understand that g(t) is 2sqrt(1+t^3)

Answer 7

And that I was using G(t) to represent the antiderivative of g(t)
that is G'=g

Answer 8

yes, I understand

Answer 9

so now I plug in 2sqrt(1+t^3) back?

Answer 10

g(t)=2sqrt(1+t^3)
so
g(x^2)=?"

Answer 11

=2sqrt(1+x^6)

Answer 12

ok so that is what you replace g(x^2) with

Answer 13

don't forget to find (x^2)' which is next to g(x^2)

Answer 14

thank you I got it

Answer 15

np