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 $g(x,y) = \frac{\partial (3y)}{\partial x} =0.$


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

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



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.

  • $\begingroup$ 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. $\endgroup$ – bosco98 Mar 2 at 2:56
  • $\begingroup$ If updated my answer accordingly. $\endgroup$ – Octavius Mar 2 at 3:09
  • $\begingroup$ 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. $\endgroup$ – bosco98 Mar 2 at 21:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.