Find $\frac{\partial}{\partial p}$ and $\frac{\partial}{\partial q}$ of $U(f(p) + q^2, g(q)^2)$

Question

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.

Answer 1

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 :

$$\frac{\partial G}{\partial q} = 2q \frac{\partial U}{\partial x} + 2g'(q)g(q) \frac{\partial U}{\partial y}$$

Answer 2

Hope everything is well with you, What can make your job easier is to consider it like this, $$U(u,v) : u = f(p)+q^2 , v = g(q)^2$$ Then now apply derivative as usual, $$dU/dp = \frac{dU}{du} \frac{du}{dp} + \frac{dU}{dv} \frac{dv}{dp} $$ $$dU/dp = \frac{dU}{du} f'(p) $$ Same happens with q. Good Luck