Prove that for a sufficiently differentiable $f$, if $f(x,y) = g(x+y)$ then its partial derivatives are equal Let $f$ be a sufficiently differentiable function. If there exists a function $g$ such that $f(x,y)  = g(x+y)$,  then $\frac {\delta f}{\delta x} = \frac {\delta f}{\delta y}$
I'm wondering if $f(x,y) = g(x+y)$ what are its partial derivatives equal to?
I feel like $\frac {\delta f}{\delta x}$ should be $g'(x+y) \cdot  \frac {\delta (x+y)}{\delta x} = g'(x+y)$, but I'm not sure whether this is correct or how to prove it.
If it is correct then the partial derivative with respect to $y$ is also $g'(x+y)$. But then this exercise that was part of a midterm would be too easy.
 A: (Note: I figured I might as well convert my comment into an answer.)
Your solution is perfectly correct. And it only uses the Chain Rule, so I don't think that it requires any further justification. Sometimes midterm exercises are that easy! :-)
Then you asked a follow-up question: why is the Chain Rule valid? I don't think you meant to ask how to prove it, but rather to justify how exactly it works here. You can find a very good general explanation of the multivariate Chain Rule here.
In this specific example, let's break our function apart as a composition of two functions. If we denote $z=f(x,y)$, then we have that
$$z=g(t),\; \text{where}\; t(x,y)=x+y, \quad \text{so}\; z=g(t(x,y))=g(x+y),$$
which does represent the given function $z=f(x,y)=g(x+y)$. Notice that the "outside" function $z=g(t)$ is a function of a single variable only, while the "inside" function $t(x,y)=x+y$ represents what we have inside $g$. In this case, the Chain Rule gives the following expressions:
$$\frac{\partial f}{\partial x}=\frac{dz}{dt}\cdot\frac{\partial t}{\partial x} \quad \text{and} \quad \frac{\partial f}{\partial y}=\frac{dz}{dt}\cdot\frac{\partial t}{\partial y}.$$
Observe that:


*

*since $z=g(t)$, then $\displaystyle \frac{dz}{dt}=g'(t)$;

*since $t=x+y$, then $\displaystyle \frac{\partial t}{\partial x}=1+0=1$;

*since $t=x+y$, then $\displaystyle \frac{\partial t}{\partial y}=0+1=1$;
and you have your answer.
A: Let $F$ denote the 'flip map' $(x,y) \mapsto (y,x)$. Clearly $f \circ F$ is just $f$ again. This means that for the total derivative, we have 
$(\partial_x f, \partial_y f)(x,y) = Df(x,y)=D(f\circ F)(x,y)=Df(x,y)\circ DF(x,y) =  (\partial_x f, \partial_y f)(x,y) \circ M =  (\partial_y f, \partial_x f)(x,y)$
where the $2 \times 2$ matrix $M$ just has $1$ on the off diagonal, and $0$ on the diagonal. 
