Partial differential of $f(x,y)=x+y$ given $x+y=1$

Question

Suppose a function $f(x,y)=x+y$ and We have to find $\partial f/\partial x$ given $x+y=1$. There two ways I can do this but I'm confused about which one is right and why?

$$f(x,y)=x+y=1$$ $$\Rightarrow \frac{\partial f}{\partial x}=0$$

$$\frac{\partial f}{\partial x}=\frac{\partial }{\partial x}(x+y)=1$$

which one is correct?

The first one. Given the constraint, the second one should have $\dfrac\partial{\partial x}y=-1$ — J. W. Tanner, Apr 06 '21 at 05:16
$\dfrac\partial{\partial x}(x+y)=\dfrac\partial{\partial x}x+\dfrac\partial{\partial x}y=1+(-1)=0$ — J. W. Tanner, Apr 06 '21 at 05:23
I get it! Thanks a lot! You have written $-1$ on the first comment by mistake, See if you can edit it. — Young Kindaichi, Apr 06 '21 at 05:26
I wrote $-1$ for $\dfrac\partial{\partial x}y$, whereas $\dfrac\partial{\partial x}(x+y)=0$ — J. W. Tanner, Apr 06 '21 at 05:29

score 1 · Answer 1 · answered Apr 06 '21 at 05:22

1

It's not clear what you mean by "given $x+y=1$".

If you have to find $\frac{\partial f}{\partial x}$ at a location $(x,y)$ where $x+y=1$, then it's just the general derivative, $1$, evaluated at that $(x,y)$, but the result is $1$ since that is a constant function. This is your second proposal.
If you are saying the domain of $f$ is restricted to the line where $x+y=1$, then a change in $x$ induces a change in $y$ (so that you remain within the domain) and the function is a constant $1$ on its domain. And then this derivative is $0$. This is your first proposal.

answered Apr 06 '21 at 05:22

2'5 9'2

54,717

$x+y=1$ is a constraint on a function. – Young Kindaichi Apr 06 '21 at 05:24
@YoungKindaichi Then this is just a constant function, and any derivative that has meaning is $0$. – 2'5 9'2 Apr 06 '21 at 06:01

Partial differential of $f(x,y)=x+y$ given $x+y=1$

1 Answers1

Linked