Differentiation under integral sign- Multivariable case problem Let $f_{\theta}(x,y)=f(x\cos \theta-y\sin \theta,x\sin\theta+y\cos\theta)$, where $f\in C^2(\Bbb{R}^2)$(Is the range necessarily $\Bbb{R}^2$? This is quite ambiguous.) a function with a bounded support. Show that ${d\over d\theta}\int\int_{\Bbb{R}\times (0,\infty)}f_{\theta}(x,y)dxdy=-\int_{\Bbb{R}}f_{\theta}(x,0)dx$. 
Hint: $({\partial \over \partial \theta}+y{\partial \over \partial x}-x{\partial \over \partial y})f_{\theta}(x,y)=0$. 
While I did prove the hint, I had no idea what to do with the above integral. I've gone through the Lecturer's notes and there was no evidence to differentiation under the integral sign. I am pretty sure I am not expected to think of such theorem and prove it while trying to show that identity above, assuming the theorem was never mentioned in the course, on the other hand. Therefore I could really use your help approaching it, or alternately your conjecture on whether or not the Theorem is necessary and was simply overlooked by me.
 A: Define two new variables as below. We can see that it is rotational transformation.
$$\begin{cases} 
x' = x \cos \theta - y \sin \theta\\ 
y' = x \sin \theta + y \cos \theta
\end{cases}$$
The function $f_{\theta}$ can then be written as as below. It is a function independent to $\theta$.
$$f_{\theta}(x',y')=f(x',y')$$
The Jacobian determinant is unit as shown below. 
$$det \left[ 
\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}
 \right]=1$$
So we get, $$dx'dy'=dxdy$$
$${d\over d\theta}\int\int_{\Bbb{R}\times (0,\infty)}f_{\theta}(x,y)dxdy={d\over d\theta} \int_\Bbb{R} \int_{\tan \theta \cdot x'}^{\infty}f(x',y')dy'dx'$$
Using Leibniz integral rule, (note: $y=0$ gives $y'=\tan \theta \cdot x'$)
$${d\over d\theta} \int_\Bbb{R} \int_{\tan \theta \cdot x'}^{\infty}f(x',y')dy'dx'=-\int_\Bbb{R} f(x', \tan \theta x') x' \frac1{\cos^2 \theta} dx'$$
let $x=\frac{x'}{\cos \theta}$,
$$-\int_\Bbb{R} f(x', \tan \theta x') x' \frac1{\cos^2 \theta} dx'=-\int_\Bbb{R} f(x\cos \theta, x \sin \theta) x dx=-\int_\Bbb{R} f_{\theta}(x, 0) x dx$$
Thus, my answer is different a little. 
$${d\over d\theta}\int\int_{\Bbb{R}\times (0,\infty)}f_{\theta}(x,y)dxdy=-\int_\Bbb{R} f_{\theta}(x, 0) x dx$$
