How to express the partial derivatives of a function?

Question

My calculus is a bit rusty and I would need a refresher on this:

Consider 2 functions $F$ and $G$. Both functions are $R \rightarrow R$. Consider a third function $y(u,v) = F(u)+G(v)$.

How can you show that $\frac{\partial^2{y}}{\partial{u}\partial{v}}=0$.

I would need a hint on how to tackle this. Thanks!

I took the words "of a composite function" out because $y(u, v)$ is the sum of two function, not a composition of those functions. — amWhy, Jun 08 '17 at 18:52
$\frac{\partial y}{\partial v}(u, v) = G'(v)$. Does not depend on $u$. Hence $\frac{\partial^2 y}{\partial u \partial v}(u, v) = 0$. Do you need something deeper than this? — user66081, Jun 08 '17 at 18:53
Indeed as simple as that. Nearly feeling embarrassed to having asked the question :). — BigONotation, Jun 08 '17 at 18:56
@user66081 Please do post an answer (transcribe your comment into an answer field, and post). You were the first to note the issue. If you post your comment as an answer, I'll upvote it. — amWhy, Jun 08 '17 at 19:02

score 1 · Accepted Answer · answered Jun 08 '17 at 18:55

$y(u,v)=F(u)+G(v)\\ \frac{\partial y}{\partial u} = \frac {df}{du}\\ \frac{\partial y}{\partial v} = \frac {dg}{dv}$

$\frac{\partial y}{\partial u}$ is a function of $u$ and $\frac{\partial y}{\partial v}$ is a function of $v$

$\frac{\partial^2 y}{\partial u\partial v} = \frac{\partial }{\partial u}\frac{\partial y}{\partial v} =\frac{\partial }{\partial v}\frac{\partial y}{\partial u} = 0$

How to express the partial derivatives of a function?

1 Answers1