# Derivatives of Integrals

• Dec 4th 2008, 06:45 PM
fishguts
Derivatives of Integrals
I am wondering if someone can verify if I did these two derivatives of integrals correctly

$h(y) = \int_2^B \sin^4t (dt) B = 1/y
$

$h^1(y) = -sin^4(1/y)/y^2$

and

$g(y) = \int_A^1 \frac{u^3}{1+ u^2} du A = 1-3x$
$g^1(x) = -\frac{(1-3x)^2}{1+(1-3x)^2} (-3)$
• Dec 5th 2008, 10:11 PM
mr fantastic
Quote:

Originally Posted by fishguts
I am wondering if someone can verify if I did these two derivatives of integrals correctly

$h(y) = \int_2^B \sin^4t (dt) B = 1/y
$

$h^1(y) = -sin^4(1/y)/y^2$

and

$g(y) = \int_A^1 \frac{u^3}{1+ u^2} du A = 1-3x$
$g^1(x) = -\frac{(1-3x)^2}{1+(1-3x)^2} (-3)$

Your answers are correct. The second answer can be simplified.