Please help with my chain rule confusion I was trying to follow an earlier post that involved differentiating 
$f(x+y)=f(x)+f(y)+g(xy)$
with respect to $x$
The correct answer is:
$f'(x+y)=f'(x)+yg'(xy)$
However I get $f'(x+y)=f'(x)+\frac{dg}{du}y$ 
Is this the same thing?
I got this using $g(xy)=g(u)$ where $u=xy\implies\frac{dg}{dx}=\frac{dg}{du}\frac{du}{dx}=\frac{dg}{du}y$
I asked the answerer (who kindly tried to explain) but I'm struggling.
My main problem is that $g'(x)$ seems to indicate a derivative with respect to $x$ whilst my notation $\frac{dg}{du}$ is with respect to $u$.  If it is where is the gap in my knowledge?  
I would be extremely grateful for any help.  For my life I've only ever used Liebenitz notation as it is easy to remember.  I'm trying to iron out a few basics in preparation for teaching at a higher level and so really want to be more rigorous with myself first. 
Thanks in advance.
 A: We have $$f(x+y) = f(x) + f(y) + g(xy)$$
The tricky part here is $g(xy)$.
If we differentiate $g(xy)$ with respect to $x$ then $y$ appears to be constant.
The definition of the chain rule is: $$\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)$$
In this case, we would have $$\frac{d}{dx}g(h(x)) = g'(h(x))h'(x)$$
Where $h(x) = xy)$.
So, $$\frac{d}{dx} g(xy) = g'(xy)(\frac{d}{dx}xy) = yg'(xy)$$
You have done the rest of the work yourself. So we get $$f'(x+y) = f'(x) + yg'(xy)$$

It is harder to interpet using the other $\frac{dg}{dx}$ notation as we are using an unknown function in which case you can use the definition.
Anyway, here you go:
Let $$h = g(xy) \text{ and } u = xy$$
Then $$\frac{dh}{dx} = g'(xy)$$
So we want to find $\frac{dh}{dx}$.
With $u = xy$, then $$\frac{du}{dx} = y$$
We also have that $$h = g(xy) \text{ but } u = xy \implies h = g(u)$$
So therefore, $$\frac{dh}{du} = g'(u)$$Just like if $y = f(x)$, then $\frac{dh}{dx} = f'(x)$.
So, now, using $$\frac{dh}{dx} = \frac{dh}{du}\frac{du}{dx}$$
Then $$\frac{d}{dx}g(xy) = yg'(xy)$$
It is a bit easier when you use $h = g(xy)$ and treat it as a different function.
A: See $g(xy)$ is a function . now lets make a similar example whats $y'$ of $ln(sin(x))$ what we do is we first differentiate $dln(sin(x)))/dx=1/sin(x)$ then we differentiate inner function ie sin so we get $cos(x)$ hence the derivative is $cot(x)$ now we do it in similar way with $g(xy)$ we first differentiate the whole outer function so we get $g'(xy)$ and then differentiate $xy$ which is y thus $d(g(xy))/dx=g'(xy)y$
