Taking the derivative of an integral with x in the function and in the bound I know how you would do this: 
$${d\over dx}\int_2^{x^4} \tan (t^2)\,dt,$$ but how would you do this: $${d\over dx}\int_2^{x^4} \tan (x^2)\,dx.$$ I am confused on the argument $x$ being in the integral and as the bound of the integral.
Edit: Since everyone is solving in terms of $t$ as the inside, I will assume that it is in terms of $t$, not $x$. I was confused because I was unsure if there were another method to solve the second integral, or if it was just a misprint. I will assume that this is a misprint and carry on. Thanks!

Using Leibniz Formula
\begin{equation}
 \frac{d}{dx} \left (\int_{a(x)}^{b(x)}f(x,t)\,dt \right) = f\big(x,b(x)\big)\cdot \frac{d}{dx} b(x) - f\big(x,a(x)\big)\cdot \frac{d}{dx} a(x) + \int_{a(x)}^{b(x)}\frac{\partial}{\partial x} f(x,t) \,dt.
\end{equation}
where 
\begin{equation}
 a(x) = 2
\end{equation}
and
\begin{equation}
 b(x) = x^4
\end{equation}
and
\begin{equation}
 f(x,t) = \tan t^2
\end{equation}
You get
\begin{equation}
 f(x,b(x)) = \tan x^8
\end{equation}
\begin{equation}
 \frac{d}{dx}b(x) = 4x^3
\end{equation}
Also
\begin{equation}
 f(x,a(x)) = \tan 2^2 = \tan 4
\end{equation}
but
\begin{equation}
 \frac{d}{dx}a(x) = 0
\end{equation}
and also notice that
\begin{equation}
 \frac{\partial}{\partial x} f(x,t) 
 =
 \frac{\partial}{\partial x} \tan t^2 
 =
 0 
\end{equation}
So, we get that
\begin{equation}
 {d\over dx}\int_{2}^{x^4} \tan (t^2)dt,
 =
4x^3 \tan x^8
\end{equation}
