Question

Medal!! Question Below!!

Answer 1

can you see the attachment?

Answer 2

@Data_LG2

Answer 3

@freckles

Answer 4

hi

Answer 5

Hi :)

Answer 6

\[\text{ Let } F(x)=\int\limits_{1}^{a(x)} g(t) dt \\ \text{ so assume } G'=g \\ \text{ anyways } F(x)=G(t)|_1^{a(x)}=G(a(x))-G(1) \\ \text{ for first term use chain rule } \\ \text{ second term use constant rule } \\ F'(x)=a'(x)G'(a(x))-0 \\ F'(x)=a'(x)g(a(x))\]

Answer 7

you can differentiate your integral in a similar way

Answer 8

notice your g(t) is sqrt(t^2+1)
and your a(x) is 4x

Answer 9

a'(x)=?

Answer 10

wait do you plug it into x?

Answer 11

I was currently just asking for a'(x) given a(x)=4x

Answer 12

\[F(x)=\int\limits_1^{4x} \sqrt{t^2+1} dt \\ F'(x)=a'(x)g(a(x)) \text{ where } g(t)=\sqrt{t^2+1} \text{ and } a(x)=4x\]

Answer 13

you will plug 4x into g(t)=sqrt(t^2+1) but you also need to differentiate (4x)

Answer 14

8x^2+4x - 3/2

Answer 15

the derivative of 4x is 8x^2+4x-3/2?
how did you get that?

Answer 16

Picture the line y=4x
what is the slope of the line y=4x?

Answer 17

the slope is 4

Answer 18

right the derivative of 4x is 4..
\[F(x)=\int\limits\limits_1^{4x} \sqrt{t^2+1} dt \\ F'(x)=a'(x)g(a(x)) \text{ where } g(t)=\sqrt{t^2+1} \text{ and } a(x)=4x \\ F'(x)=(4x)'g(4x)=4g(4x)\]
last thing to do is replace the t in sqrt(t^2+1) with (4x)

Answer 19

\[F'(x)=4 \sqrt{(4t)^2+1}\]

Answer 20

oops that t is suppose to be x

Answer 21

but anyways i think you got it from here

Answer 22

if not i will be back after i eat