Let:
$f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable.
$u:\mathbb{R}^2\rightarrow\mathbb{R}$ and $u(x,y)=e^{-y}f(x+y^2)$
Show that: $$2y\frac{\partial{u}}{\partial{x}}-\frac{\partial{u}}{\partial{y}}=u$$
My attempt has been to first work out the two partial derivatives and plug them in, however I am not very confident in my notation, for example:
$e^{-y}\frac{\partial{f(x+y^2)}}{\partial{y}}$ is the second part of my working of partial u/partial y, first * derivative of second
The next step is: $2ye^{-y}\frac{\partial{f(x+y^2)}}{\partial{y}}$ but that looks the same as before, my notation doesn't show f(x+y^2) and the partial derivative of f wrt a variable at a point as being different. How do I overcome this?
I've never been formally taught or had the need for the vertical bar notation, but I sense this is what the notation solves, how do I write it?
Addendum: I've not LaTeXed a lot of it, I find it a laggy choir at present, please don't take this as laziness, I'm also going to try introducing a new variable used for x+y^2 but on paper. If I answer it I shall report back, sorry for mixing up my mu and u, it's one of those prints of a scan of an old text-book, the toner is shiny on the cheap paper and it's black and white, not gray-scale a choir to read!
Addendum 2: Here, it works, but I'm not sure if it's errors canceling or not:
Let: $\lambda:\mathbb{R}^2\rightarrow\mathbb{R}$, $\lambda(x,y)=x+y^2$
Now:
$u(x,y) = e^{-y}f(\lambda)$ (I feel I need not write $\lambda(x,y)$ all the time)
$$\frac{\partial{u}}{\partial{x}}=e^{-y}\frac{df(\lambda)}{\partial{x}}=e^{-y}\frac{df}{d\lambda}.\frac{\partial{\lambda}}{\partial{x}}=e^{-y}\frac{df}{d\lambda}$$
and
$$\frac{\partial{u}}{\partial{y}}=e^{-y}\frac{df(\lambda)}{\partial{y}}+\frac{\partial}{\partial{y}}[e^{-y}]f(\lambda)$$ Now there I've written partial by partial-y, this is why I am not that happy with my notation.... but d/partial y doesn't look right either...
Anyway: $$\frac{\partial{u}}{\partial{y}}=-e^{-y}f(\lambda)+e^{-y}\frac{df}{d\lambda}.\frac{\partial{\lambda}}{\partial{y}}=e^{-y}[2y\frac{df}{d\lambda}-f(\lambda)]$$
To show that is just to bung those in.