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!
 A: 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}
A: Note that $t$ is a dummy variable so both integrals are the same.
$$\int_{2}^{x^4} \tan (x^2)dx= \int_{2}^{x^4} \tan (t^2)dt$$
This is a composite function situation. The upper limit of integration is the inner function.
The derivative is found by chain rule, which is derivative of the outer function multiplied by derivative of the inner function. 
Therefore you answer is  $${d\over dx}\int_{2}^{x^4} \tan (t^2)dt= (4x^3)\tan (x^8) $$
A: $${d\over dx}\int_{2}^{x^4} \tan (t^2)dt,$$
Let $$F(u)= \int_2^u \tan (t^2)dt$$
Then you want to compute $F'(x^4)$.
\begin{align}
   F'(x^4) 
   &= \left. {dF \over du} \cdot {du \over dx}\right|_{u=x^4} \\
   &=\tan(x^8) \cdot 4x^3 \\
   &= 4x^3 \tan(x^8)
\end{align}
A: $$I=\int_2^{x^4}\tan(t^2)dt$$
and we want to find $\frac{dI}{dx}$
$$\frac{d}{dx}\int_2^{x^4}\tan(t^2)dt=\frac{d}{dx}\left[x^4\right]*\tan\left((x^4)^2\right)=4x^3\tan(x^8)$$
and we know this from the full version of the Leibniz integral rule:
$$\frac{d}{dx} \left (\int_{a(x)}^{b(x)}f(x,t)\,dt \right) = f\big(x,b(x)\big)\cdot \frac{d}{dx} b(x) - f\big(x,a(x)\big)\cdot \frac{d}{dx} a(x) + \int_{a(x)}^{b(x)}\frac{\partial}{\partial x} f(x,t) \,dt$$
