Find the derivative of y with respect to x,t,or theta, as appropriate $$y=\int_{\sqrt{x}}^{\sqrt{4x}}\ln(t^2)\,dt$$
I'm having trouble getting started with this, thanks for any help.

First Step
First, we need to recognize to which variables you are supposed to differentiate with respect.  The important thing to realize here is that if you perform a definite integration with respect to one variable, that variable "goes away" after the computation.  Symbolically:
$$\frac{d}{dt}\int_a^b f(t)\,dt = 0$$
Why? Because the result of a definite integral is a constant, and the derivative of a constant is zero! :)
So, it isn't appropriate here to differentiate with respect to $t$. With respect to $\theta$ doesn't make much sense, either--that's not even in the problem!  So, we are looking at differentiating with respect to $x$.
Second Step
We now use a very fun theorem: the fundamental theorem of calculus!  (bad pun, sorry)
The relevant part states that:
$$\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$$
We now make your integral look like this:
$$\begin{align}
   y &= \int_{\sqrt{x}}^{\sqrt{4x}}\ln(t^2)\,dt\\
     & = \int_{\sqrt{x}}^a\ln(t^2)\,dt + \int_{a}^{\sqrt{4x}}\ln(t^2)\,dt\\
     & = -\int_{a}^{\sqrt{x}}\ln(t^2)\,dt + \int_{a}^{\sqrt{4x}}\ln(t^2)\,dt\\
\end{align}$$
Can you now find $\frac{dy}{dx}$?  (Hint: Don't forget the chain rule!)
If you still want some more guidance, just leave a comment.
EDIT:
Note that $y$ is a sum of two integral functions, so you can differentiate both independently.  I'll do one, and leave the other for you:
$$\begin{align}
\frac{d}{dx}\left[\int_{a}^{\sqrt{4x}}\ln(t^2)\,dt\right] &= \left[\ln\left(\sqrt{4x}^2\right)\right]\cdot\frac{d}{dx}\left(\,\sqrt{4x}\right)\\
&=\left[\ln\left(4|x|\right)\right]\left(2\frac{x^{-1/2}}{1/2}\right)\\
&=4x^{-1/2}\ln\left(4|x|\right)
\end{align}$$
