# finding the derivative of a definite integral

• October 3rd 2008, 01:13 PM
tehbrosta
finding the derivative of a definite integral
Find the derivative of the following function at x = 1

http://img512.imageshack.us/img512/8052/58866883iz3.jpg

Based on the Fundamental Theorem of Calculus, Part I would this be correct?

http://img296.imageshack.us/img296/6778/3attemptjo2.jpg
http://img296.imageshack.us/img296/3...jpg/1/w232.png
Or would it become something like

$(-1+2t)/(1+2t)$

thanks, my teacher to so horrible. I might actually be able to learn something this way. (Nod)
• October 3rd 2008, 01:25 PM
ThePerfectHacker
$y' = \frac{1 - x + x^2}{1+x+x^2}$ by Fundamental Theorem.

To find $y'(1)$ evaluate that function on RHS at 1.
• October 3rd 2008, 01:26 PM
CaptainBlack
Quote:

Originally Posted by tehbrosta
Find the derivative of the following function at x = 1

http://img512.imageshack.us/img512/8052/58866883iz3.jpg

Based on the Fundamental Theorem of Calculus, Part I would this be correct?

http://img296.imageshack.us/img296/6778/3attemptjo2.jpg
http://img296.imageshack.us/img296/3...jpg/1/w232.png
Or would it become something like

$(-1+2t)/(1+2t)$

thanks, my teacher to so horrible. I might actually be able to learn something this way. (Nod)

If

$y(x)=\int_0^x f(t) dt$

then the fundamental theorem tell us that:

$y'(x)=f(x)$

so:

$y'(1)=f(1)$

RonL
• October 3rd 2008, 02:34 PM
tehbrosta
So when I take the derivative at x=1 I get:

$y'(1) = (1-(1)+1^2)/(1+1+1^2) = 1/3$?