Is $\exists x(P(x) \to Q(x)) \equiv (\exists xP(x) \to \exists xQ(x))$? My intuition is that this statement is false and here is my proof. 
$\exists x(P(x) \to Q(x))$
$\exists x(\lnot P(x) \lor Q(x))$ using logical equivalence.
$\exists x\lnot P(x) \lor \exists x Q(x)$ using distributive properties of $\exists$ over $\lor$.
Assuming Q(x) is always false, we simply need 1 such x where $\lnot P(x)$ is true and that makes the entire statement true. The other expression can be transformed into this:
$\exists xP(x) \to \exists xQ(x)$
$\lnot\exists xP(x) \lor \exists xQ(x)$ using logical equivalence.
$\forall x\lnot P(x) \lor \exists xQ(x)$
If we assume Q(x) is always false, the only thing that makes this true is for all $\lnot P(x)$ to be true. For example, the domain could be all positive integers and $P(x) = x \le 10$. To be complete, let's assume $Q(x) = x \le -10$. 
Therefore, the lefthand side of $\exists x(P(x) \to Q(x))$ is true and the righthand side of $(\exists xP(x) \to \exists xQ(x))$ is false.
I'm pretty sure I'm right but I would like another set of eyes to see if I did anything stupid.
 A: You know two things, as Jonny pointed out:


*

*$\exists x(P(x)\to Q(x))\equiv \neg\forall xP(x)\lor\exists x Q(x)\equiv\exists x\neg P(x)\lor\exists xQ(x)$

*$\exists xP(x)\to \exists xQ(x)\equiv\forall x\neg P(x)\lor\exists x Q(x)$


Hence, in our attempt to create a counterexample, we are really concerned with whether or not $\exists x\neg P(x)\equiv\forall x\neg P(x)$. This should be pretty easy to invalidate.
Counterexample: Let the universe be the natural numbers. Let $Q(x)$ and $P(x)$ be the following:


*

*$Q(x) : x$ is both prime and not prime.

*$P(x) : x$ is not an even prime number.


Well, $Q(x)$ is clearly always false. Now, $\exists x\neg P(x)$ is true because there does exist an even prime number, namely $2$. However, $\forall x\neg P(x)$ is clearly not true, for $2$ is the only even prime number. Thus, we can see that $\exists x\neg P(x) \not\equiv\forall\neg P(x)$, thus effectively showing that $$\exists x(P(x)\to Q(x))\not\equiv \exists xP(x)\to \exists xQ(x).$$
A: I would leave out the assuming Qx is always false and stick with the logical equivalence moves.
The first statement is equivalent to $\neg\forall x P(x) \lor \exists x Q(x)$, while the second statement is equivalent to $\forall x \neg P(x) \lor \exists x Q(x)$
From here all you need is a counter example... recall that statements are logically equivalent if their truth is evaluated to be the same in every model.  So we need a structure where one statement is true and the other is false. Perhaps we are in the universe of natural numbers and let P mean x is prime and Q means x is the largest number.   
A: Your counterexample is fine.
You might be interested in another slightly more formal approach, written in the style of Dijkstra-Scholten and Gries-Schneider (see also Dijkstra's EWD1300), to find all counterexamples.
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\followsfrom}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$Let's calculate when the two sides are equivalent:
$$\calc
\langle \exists x :: P(x) \then Q(x) \rangle \;\equiv\; (\langle \exists x :: P(x) \rangle \then \langle \exists x :: Q(x) \rangle)
\op\equiv\hints{write $\;\phi \then \psi\;$ as $\;\lnot \phi \lor \psi\;$, twice}\hint{-- to make manipulations easier}
\langle \exists x :: \lnot P(x) \lor Q(x) \rangle \;\equiv\; \lnot \langle \exists x :: P(x) \rangle \lor \langle \exists x :: Q(x) \rangle
\op\equiv\hints{$\;\exists\;$ distributes over $\;\lor\;$}\hint{-- because $\;\exists\;$ is a kind of 'repeated disjunction'}
\langle \exists x :: \lnot P(x) \rangle \lor \langle \exists :: Q(x) \rangle \;\equiv\; \lnot \langle \exists x :: P(x) \rangle \lor \langle \exists x :: Q(x) \rangle
\op\equiv\hint{$\;\lor\;$ distributes over $\;\equiv\;$ -- to simplify}
(\langle \exists x :: \lnot P(x) \rangle \equiv \lnot \langle \exists x :: P(x) \rangle) \;\lor\; \langle \exists x :: Q(x) \rangle
\op\equiv\hints{DeMorgan on right hand side of $\;\equiv\;$}\hint{-- making both sides more alike}
(\langle \exists x :: \lnot P(x) \rangle \equiv \langle \forall x :: \lnot P(x) \rangle) \;\lor\; \langle \exists x :: Q(x) \rangle
\op\equiv\hint{split $\;\equiv\;$ into $\;\then\;$ and $\;\followsfrom\;$; $\;\forall\then\exists\;$}
(\langle \exists x :: \lnot P(x) \rangle \then \langle \forall x :: \lnot P(x) \rangle) \;\lor\; \langle \exists x :: Q(x) \rangle
\op\equiv\hint{write $\;\phi \then \psi\;$ as $\;\lnot \phi \lor \psi\;$}
\tag{*} \lnot \langle \exists x :: \lnot P(x) \rangle \;\lor\; \langle \forall x :: \lnot P(x) \rangle \;\lor\; \langle \exists x :: Q(x) \rangle
\endcalc$$
So you have a counterexample whenever the negation of $\ref *$ is true:
$$
\tag{$\lnot$*}
\langle \exists x :: \lnot P(x) \rangle \;\land\; \langle \exists x :: P(x) \rangle \;\land\; \langle \forall x :: \lnot Q(x) \rangle
$$
In other words, a counterexample is any $\;P(x)\;$ which is true for some $\;x\;$ and false for some other, together with any $\;Q(x)\;$ which is always false.
