Single Variable Calculus, Chapter 5, Review Exercises, Section Review Exercises, Problem 36

Determine the derivative of the function $\displaystyle f(x) = \int^{\sin x}_1 \frac{1-t^2}{1+t^4} dt$ using the properties of integral.
$\displaystyle \frac{d}{dx} \int^{\sin x}_1 \frac{1-t^2}{1+t^4} dt = \frac{d}{dx} \left( \int^u_1 \frac{1-t^2}{1+t^4} dt\right)$

$\begin{equation} \begin{aligned} g'(x) &= \frac{d}{du} \left( \int^u_1 \frac{1-t^2}{1+t4} dt \right)\\ \\ g'(x) &= \frac{d}{du} \left( \int^u_1 \frac{1-t^2}{1+t^4} dt \right) \frac{du}{dx}\\ \\ g'(x) &= \frac{1 - u^2}{1+u^4} \frac{du}{dx}\\ \\ g'(x) &= \frac{1(\sin x)^2}{1+(\sin x)^4} \cdot \cos x\\ \\ g'(x) &= \frac{\left( 1 - \sin^2 x\right)(\cos x)}{1 + \sin^4 x} && \Longleftarrow \text{ (Apply Pythagorean Identity } \sin^2 \theta + \cos^2 \theta = 1)\\ \\ g'(x) &= \frac{\left( \cos^2 x \right)(\cos x)}{1 + \sin^4 x}\\ \\ g'(x) &= \frac{\cos^3 x}{1 + \sin^4 x} \end{aligned} \end{equation}$