Total differential in which arguments are a sum If $z=z(x+y)$, what is $dz$?
I see in one of my text books that it would be equal to $dz=z'*(dx+dy)$, but this confuses me.
If I am to treat $x+y$ as a single argument of the function, how can I differentiate them separately? It makes more sense to me to treat them as one argument and write the equation as $dz=z'*d(x+y)$ or is this completely wrong?
If the first method is correct, wouldn't this imply the following?
If $y=f(x)$ and $x=a+b$, then $dy=f'(x)dx$ and $dy=f'(x)(da+db)$ are equivalent?
Finally, quick sanity check: are $z'$ and $dz/d(x+y)$ the same? 
Thanks so much for your help!
 A: The notation is confusing. To be able to tell things apart, write instead $z(x,y)=f(x+y)$, where $f$ is a function of one variable. That is, you have $f(t)$ and form the two-variable function $z(x,y)$ by substituting $t=x+y$. Then, according to the chain rule,
$$
\frac{\partial z}{\partial x}(x,y) = \frac{df}{dt}(x+y) \cdot 1
,\qquad
\frac{\partial z}{\partial y}(x,y) = \frac{df}{dt}(x+y) \cdot 1
,
$$
where the ones are the derivatives of the inner function $t=x+y$:
$$\partial t/\partial x = 1, \qquad \partial t/\partial y = 1.$$
So
$$
\begin{aligned}
dz
&=
\frac{\partial z}{\partial x}(x,y) \, dx + \frac{\partial z}{\partial y}(x,y) \, dy
\\
&=
\frac{df}{dt}(x+y) \, dx
+ \frac{df}{dt}(x+y) \, dy
\\
&=
\frac{df}{dt}(x+y) \, (dx+dy)
.
\end{aligned}
$$
You see here that the notation $z'$ actually means $df/dt$ (evaluated at $t=x+y$).
A: Let $g(x,y) = x + y$. Then you are asking about $dz = dz(g(x,y))$. By definition,
$$
dz(g(x,y)) = \frac{dz}{dg}\frac{\partial g}{\partial x}\,dx + \frac{dz}{dg}\frac{\partial g}{\partial y}\,dy = \frac{dz}{dg}\left(\frac{\partial g}{\partial x}\,dx + \frac{\partial g}{\partial y}\,dy\right) = \frac{dz}{dg}(dx + dy)=z'\cdot(dx + dy).
$$
The correct interpretation of the symbol $dx$ is the differential 1-form $dx$. In your notation, yes $dz/d(x+y) = dz/dg = z'$.
