Question

HELP! Need help on Fundamental Theorem of Calculus!

Find the derivative of the following function...

F(x)= from square root x to 1 (t^2)/(3+4t^4)dt

using the appropriate form of the Fundamental Theorem of Calculus.

Answer 1

the derivative of the integral is the integrand
replace \(t\) by \(x\) in the integrand and you are done

Answer 2

oh no sorry it is
\[\int_{\sqrt{x}}^1\frac{t^2}{3+4t^4}dt\]right?

Answer 3

yes that is it

Answer 4

in that case you need the chain rule
first write it as \[-\int_1^{\sqrt{x}}\frac{t^2}{3+4t^4}dt\]
then replace \(t\) by \(\sqrt{x}\) and finally, because of the chain rule, multiply it by \(\frac{1}{2\sqrt{x}}\)

Answer 5

okay so where did the 1/(2*sqrt(x)) come from?

Answer 6

i know we have to inverse the integral to have x on top which makes the value negative

Answer 7

oh wait! is it because the derivative of square root x becomes (1/2)x^(-1/2)?