Derivative of definite integral I am learning Calculus online and this problem stumped me.  The solution doesn't really make sense either.  I'm starting to think that the user made a typo.  Here is the question to simplify:
$$\frac{d}{dx}\int_{x^2}^1\frac{t^4 + 1}{t^2 + 1}dt$$
if $u = x^2$, then $\frac{du}{dx} = 2x$, which means $dx = \frac{du}{2x}$ So we have,
$$-\frac{d}{\frac{du}{2x}}\int_1^u\frac{t^4 + 1}{t^2 + 1}dt$$
And then when I simplify it,
$$-\frac{d}{du}\int_1^u\frac{t^4 + 1}{t^2 + 1}*(2x)dt$$
But this still doesn't get rid of the t's.  I have no idea how to replace t with x. Also, I am unable to get rid of $dt$, if I could get rid of $dt$ then I would be able to simplify the equation as the derivative of an integral of $f(x)$ is simply $f(x)$
 A: So, to evaluate 
$$
\frac{d}{dx}\int_{x^2}^1\frac{t^4 + 1}{t^2 + 1}dt
$$
it suffices to note that if $F$ is a primitive (unique up to a constant) of $\frac{t^4 + 1}{t^2 + 1}$, then the integral becomes $F(1) - F(x^2)$. Differentiating in turn gives the result 
$$
\frac{d}{dx}\int_{x^2}^1\frac{t^4 + 1}{t^2 + 1}dt = \frac{d}{dx} \left( F(1) - F(x^2) \right) = 0 - F'(x^2) \cdot \frac{d}{d x} (x^2)
$$
which is simply 
$$
- 2x \frac{x^8 + 1}{x^4 + 1} .
$$
A: When you calculate the derivative of $$\int_{u(x)}^{v(x)} f(t) dt,$$
what you must realize is that this is actually a compositum function. You have $3$ functions: $F(u,v) = \int_u^v{f(t) dt}$ and the functions $u(x)$ and $v(x)$, giving you.
$$\int_{u(x)}^{v(x)} f(t) dt = F(u(x),v(x)).$$
The chain rule for such a function is,
\begin{align}
\frac{d}{dx}F(u(x),v(x)) &= \frac{dF}{du}\frac{du}{dx} + \frac{dF}{dv}\frac{dv}{dx} \\&=-f(u(x)) \cdot u'(x) + f(v(x))\cdot v'(x).
\end{align}         
In your case, $f(t)=\frac{t^4+1}{t^2+1}$, $u(x) = x^2$ and $v(x) = 1$.
