# How to evaluate partial derivative of $f$ with respect to $x$ as a function of $y$ only?

Hopefully this is just the notation confusing me, but how would I correctly evaluate the following function $$g$$ below?

$$g(x,y) = \frac{\partial f(2y, y)}{\partial x}$$ (or equivalently $$g(x,y) = f_{x}(2y, y)$$), where $$f(x,y) = x + y$$.

On one hand, $$\frac{\partial f(x, y)}{\partial x} = 1$$, so we could have $$g(x,y) = 1$$. On the other hand, $$f(2y, y) = 3y$$, suggesting $$g(x,y) = \frac{\partial (3y)}{\partial x} =0.$$

Edit:

To add some context, here is where this came from ($$\phi$$ below is a function of $$x$$ and $$t$$):

Define the mapping $$T$$ from $$C^{\infty}_{0}(\mathbb{R}^2)$$ to $$\mathbb{R}$$ by $$T: \phi \mapsto \int_{0}^\infty \phi(t, ct)dt$$, where $$c$$ is a real number. What is $$\partial_x T$$?

I think the notation is meant to be understood as

$$g(x,y) \enspace = \enspace \frac{\partial f(x,y)}{\partial x} \; \Big|_{x = 2y}$$

So, your answer $$g(x,y) = 1$$ would be correct. However, I would rather consider the context carefully - hard to say what was really meant here since the whole notation seems a bit off, e.g. the LHS still holds the dependency for $$x$$, whilst the RHS doesn't. So, I'd probably write it as

$$g(x = 2y, y ) \enspace = \enspace \frac{\partial f(x,y)}{\partial x} \; \Big|_{x = 2y}$$

$${}$$

EDIT:

With respect to your latest edit, I would understand your map as follows: The $$x$$-variable of $$\phi(t,x)$$ is a function of $$t$$, so one can rewrite it as $$\phi \big( t, x(t) \big)$$.

Therefore, your derivative can be solved as:

$$\partial_x T \enspace = \enspace \partial_x T \big( \phi \big) \enspace = \enspace \partial_x T \big( \phi(t,x(t)) \big)$$

$$= \enspace \int_0^{\infty} \partial_x \phi \big( t, x(t) \big) \, dt \enspace = \enspace \int_0^{\infty} \frac{\partial \phi \big( t, x(t) \big)}{\partial x} \, dt$$

So, you would have to derive $$\phi(t,ct)$$ as if $$ct$$ was $$x$$ and then - AFTER deriving it, but still BEFORE integrating it - fill in $$x = ct$$. However, I am not entirely sure.

• Thank you for the response. I agree the way I wrote it is ambiguous. I have added an edit with the context this came from. – bosco98 Mar 2 at 2:56
• If updated my answer accordingly. – Octavius Mar 2 at 3:09
• Thank you again. I have accepted your answer; after speaking with my professor today it seems this was how it was intended to be interpreted. – bosco98 Mar 2 at 21:33