Derivative of a function of vector parameter. Problem with notation. I have an error function $$err=\frac{1}{N}\left[\textbf{y}^T\ln{\textbf{x}}+(\textbf{1}-\textbf{y})^T\ln{(\textbf{1}-\textbf{x})}\right]$$ I need to find the gradient $\bigtriangledown_x{}err$, such that find derivation over vector $\textbf{x}$. I am learning matrix calculus and multivariable calculus and somehow I cannot figure out how to do it. I know what the derivative of $\ln{(x)}$ is but struggle to write it in vector notation e.g. $\frac{1}{\textbf{x}}$ ??? Which I believe is not right. I will be grateful for some help. Cheers!
 A: The issue I think you are having is in thinking about what the log of a vector is. For me this is simply another vector function, i.e. for vector $\textbf{x} = [x_{1},\ldots,x_{n}]^{T}$
$\ln(1-\textbf{x}) = [\ln(1-x_{1}),\ldots,\ln(1-x_{n})]^{T}$
and so
$\displaystyle\nabla_{x}\ln(1-\textbf{x}) = \left[\frac{\text{d}}{\text{d}x_{1}}\ln(1-x_{1}), \ldots, \frac{\text{d}}{\text{d}x_{n}}\ln(1-x_{n})\right]^{T}$
Then we use the notation $\nabla_{x}\ln(\textbf{x})$ for the derivative of $\ln(\textbf{x})$ wrt the vector $\textbf{x}$. I.e. 
$\nabla_{x}\text{err} = \displaystyle\frac{1}{N} [\textbf{y}^{T}\nabla_{x}\ln(\textbf{x}) + (1-\textbf{y})^{T}\nabla_{x}\ln(1-\textbf{x})]$.
This makes it clear what is being done, and the notation is not too verbose. As you mention $1/\textbf{x}$ does not make sense for a vector, so this notation would be wrong. Alternatively you could define a dummy variable $\textbf{z} = \nabla_{x}\ln(\textbf{x}) = [1/x_{1},\ldots,1/x_{n}]$ if you want the notation to be more compact.
