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.
 A: Hint:
The Fundamental Theorem of Calculus states:
$\frac{d}{dx}\int_{a}^{x}f(t)\,dt = f(x)$
Can your problem be broken down into parts that look like this?
A: 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:
$$\begin{align}
\frac{d}{dx}\left[\int_{a}^{\sqrt{4x}}\ln(t^2)\,dt\right] &= \left[\ln\left(\sqrt{4x}^2\right)\right]\cdot\frac{d}{dx}\left(\,\sqrt{4x}\right)\\
&=\left[\ln\left(4|x|\right)\right]\left(2\frac{x^{-1/2}}{1/2}\right)\\
&=4x^{-1/2}\ln\left(4|x|\right)
\end{align}$$
