Answer
$g'(x) = \frac{1-(sin~x)^2}{1+(sin~x)^4}\cdot cos~x=\frac{\cos ^3x}{1+(\sin x)^4}$
Work Step by Step
The function $f(t) = \frac{1-t^2}{1+t^4}$ is continuous for all $t$
Let $h(x) = sin~x$
Then $h'(x) = cos~x$
$h'(x)$ is continuous for all values of $x$
According to the Substitution Rule:
$g(x) = \int_{1}^{sin~x} \frac{1-t^2}{1+t^4}~dt = \int_{\pi/2}^{x} \frac{1-(sin~u)^2}{1+(sin~u)^4}\cdot cos~u~du$
The function $~~\frac{1-(sin~u)^2}{1+(sin~u)^4}\cdot cos~u~~$ is continuous for all $u$
According to the Fundamental Theorem of Calculus (Part 1):
$g'(x) = \frac{1-(sin~x)^2}{1+(sin~x)^4}\cdot cos~x$
