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$\def\B{\Big}\def\L{\left}\def\R{\right}\def\o{{\tt1}}\def\p#1#2{\frac{\partial #1}{\partial #2}}$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 $\...


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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 ...


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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 ...


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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}) = ...


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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)}\\...


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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{\...


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