Find $\frac{\partial}{\partial p}$ and $\frac{\partial}{\partial q}$ of $U(f(p) + q^2, g(q)^2)$ Here's my question:
Suppose $U(x_1, x_2)$ is a differentiable function. Suppose $f(p)$ and $g(q)$ are  differentiable functions that depend only on $p$ and $q$ respectively.
Find $\frac{\partial}{\partial p}$ and $\frac{\partial}{\partial q}$ of $U(f(p) + q^2, g(q)^2).$
Be sure to include the value at which each partial derivative is evaluated.
I know the chain rule for multivariable functions has the general look of:
$\frac{\partial f}{\partial t_j} = \frac{\partial f}{\partial x_1}\frac{\partial x_1}{\partial t_j} + \frac{\partial f}{\partial x_2}\frac{\partial x_2}{\partial t_j} + ... + \frac{\partial f}{\partial x_n}\frac{\partial x_n}{\partial t_j}$
I understand that $p$ and $q$ are supposed to be my $t_1$ and $t_2$, and that $x_1$ and $x_2$ are the same. I'm not sure how to apply the chain rule to this though, since $x_1$ is a function of two variables but $x_2$ is a function of just one. 
Any ideas of what I'm supposed to do?
Thanks ahead of time.

First note that you can defined the functions : $x_1 : (p,q) \mapsto f(p)+q^2$ and $x_2 :(p,q) \mapsto g(q)^2$. Then we can defined : $G : (p,q) \mapsto U(x_1(p,q), x_2(p,q))$.
So we want to calculate : $\frac{\partial G}{\partial p}$. To do so let's use the chain rule : 
$$\frac{\partial G}{\partial p}(p,q) = f'(p) \frac{\partial U}{\partial x}(x_1(p,q), x_2(p,q))$$
Note that since $f$ is a function of one variable we have : $f'(p) = \frac{\partial f}{\partial x}(p)$..
We also get using once again the chain rule : 
$$\frac{\partial G}{\partial q} = 2q \frac{\partial U}{\partial x} + 2g'(q)g(q) \frac{\partial U}{\partial y}$$
