Question

Given that F(x) = the integral of the sqrt of (t^2+9) from to 0, calculate F''.

Answer 1

\[F(x) = \int\limits_{0}^{x} \sqrt{t^2+9}\]

Answer 2

there shud be a dt after the integral

Answer 3

F'(x)= \[ \sqrt{t ^{2}=9}\]

then find the derivative of that to find F''(x). you'll need the chain rule

Answer 4

let t = f(x) = x\\[\[\int\limits\limits_{0}^{x} (\sqrt{x^2 + 9} )dx\]\]

by the Fundamental Theorem of Calculus, integrals and derivatives just cancel out, so:

dx/dy \[dx/dy \int\limits\limits_{0}^{x} (\sqrt{x^2 + 9} )dx = \sqrt{x^2 + 9}\]

now just derive that to get F''(x)

Answer 5

im sorry! substitute that t for x in the F'(x)

Answer 6

oh right what tmmss5533 said, you'll have to chain rule the x^2 because it really is f(x)^2

Answer 7

wait so wats the answer???

Answer 8

wow im terrible lol i meant to put x^(2)+9 under that radical

Answer 9

is the answer x/ sqrt of (x^2+9)

Answer 10

Well what's the derivative of x^2? It just 2(x) right?

so 2(x) d/dx(sqrt(x^2 + 9)

whats the derivative of a sqrt(x^2 + 9)? It's gonna be 1/sqrt(x^2 + 9)

so putting them together:

2x/[sqrt(x^2 + 9)]

Answer 11

can u help me with anotehr problem

Answer 12

Hermeezey can you please help me with my last two recent math questions?

Answer 13

lol i asked hermezy first

Answer 14

LOL post your problems and i'll see what i can do

Answer 15

lol i know i meant afterwards

Answer 16

ok i posted my problem

Answer 17

dx/dy ∫ √ ( t² + 9) [ 0, x ]
= √( x² + 9)
Then,
( √( x² + 9) )' = 2x/ 2√( x² + 9) = x/ √( x² + 9)