finding the derivative of a definite integral

**tehbrosta** · October 3rd 2008, 01:13 PM

Find the derivative of the following function at x = 1

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

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.

**ThePerfectHacker** · October 3rd 2008, 01:25 PM

$y' = \frac{1 - x + x^2}{1+x+x^2}$ by Fundamental Theorem.

To find $y'(1)$ evaluate that function on RHS at 1.

**CaptainBlack** · October 3rd 2008, 01:26 PM

Originally Posted by tehbrosta

Find the derivative of the following function at x = 1

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

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.

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

**tehbrosta** · October 3rd 2008, 02:34 PM

So when I take the derivative at x=1 I get:

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

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