# Tag Info

$\def\B{\Big}\def\L{\left}\def\R{\right}\def\o{{\tt1}}\def\p#1#2{\frac{\partial #1}{\partial #2}}$Define the all-ones vector $\o$ and a vector $v$ such that \eqalign{ v\odot v &= Ax\odot Ax + \delta^2\o \\ 2\,v\odot dv &= 2\,Ax\odot A\,dx \\ dv &= Ax\odot A\,dx\oslash v \\ } where $\odot$ denotes the elementwise/Hadamard product and $\... 2 Per your last equation, it's also equivalent to $$\frac{d}{dx}[\log(f_y)-\log(f)]=\frac{d}{dx}\log(f_y/f) =0.$$ Thus$\log(f_y/f)$is independent of$x$, i.e.,$f_y/f=\frac{\partial}{\partial y}\log f$is a function of$y$alone. This implies that$\log f(x,y)=G(x)+H(y)$for functions$G,H$, which is equivalent to$f(x,y)$being separable. A more symmetric ... 1 I think the notation is meant to be understood as $$g(x,y) \enspace = \enspace \frac{\partial f(x,y)}{\partial x} \; \Big|_{x = 2y}$$ So, your answer$g(x,y) = 1\$ would be correct. However, I would rather consider the context carefully - hard to say what was really meant here since the whole notation seems a bit off, e.g. the LHS still holds the dependency ...
Here is a hint: $$\frac{\partial}{\partial x}( \frac{\partial u}{\partial x} \frac{\partial f(u,v)}{\partial u}) = \frac{\partial^2 u}{\partial x^2} \frac{\partial f}{\partial u} +( \frac{\partial u}{\partial x} ) \frac{\partial}{\partial x}\left[ \frac{\partial f}{\partial u} \right]$$ Now, \frac{\partial }{\partial x} (\frac{\partial f}{\partial u}) = ... 1 Let \alpha(x,y)=xy and \beta(x,y)=x^2-2. Then, the function you're actually looking at is the composition F(x,y)=f(\alpha(x,y),\beta(x,y)). The chain rule tells you \begin{align} (\partial_1F)_{(x,y)} &=(\partial_1f)_{(\alpha(x,y),\beta(x,y))}\cdot (\partial_1\alpha)_{(x,y)} + (\partial_2f)_{(\alpha(x,y),\beta(x,y))}\cdot (\partial_1\beta)_{(x,y)}\\... 1 You start out wanting \frac{\partial}{\partial x}f(xy,x^2-2). There is this function f of two variables that I might call u and v: f=f(u,v). Here, u is composed with xy and v with x^2-2. The chain rule says:\begin{align} \frac{\partial}{\partial x}f(xy,x^2-2) &=\frac{\partial f}{\partial u}\cdot\frac{\partial u}{\partial x}+\frac{\...