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I am not sure how to prove this partial derivative. It seems to me that they are both vectors of integers. So when taking a derivative of the values we get 0 but then the total would be zero overall. Instead the value at one position seems to be 1 and it removes the transpose form the non derived value. I am not quite grasping this can anyone provide an explanation?

https://imgur.com/a/v2wRKYE

\begin{equation} \frac{\partial}{\partial Vc} = Uo^T Vc \end{equation}

Thank you

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  • $\begingroup$ Got it, sorry about that. $\endgroup$ – Jacob B Jul 27 '18 at 22:09
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$u_0$ and $v_c$ are vectors of the same dimension. The expression $\frac{\partial}{\partial v_c}(u_0^\top v_c)$ is not really a partial derivative, but rather the gradient of the function $f(v_c) = u_0^\top v_c = \sum_i (u_0)_i (v_c)_i$. In other words, $\frac{\partial}{\partial v_c}(u_0^\top v_c)$ is a vector of the partial derivatives $\frac{\partial}{\partial (v_c)_i} (u_0^\top v_c)$ for each $i$. Noting that $\frac{\partial}{\partial (v_c)_i} (u_0^\top v_c) = (u_0)_i$ shows that the gradient is indeed the vector $u_0$.

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  • $\begingroup$ Hello, thank you for your help! I fail to see how $\frac{\partial}{\partial (v_c)_i} (u_0^\top v_c) = (u_0)_i$ Why is the answer not transposed and how does $(V_c)$ go away leaving one element of $(u_0)$? $\endgroup$ – Jacob B Jul 27 '18 at 22:26
  • $\begingroup$ @JacobB Write $u_0^\top v_c$ as $\sum_i (u_0)_i (v_c)_i$ and take the partial derivative with respect to $(v_c)_i$. This is a simple calculus problem that does not involve vectors. $\endgroup$ – angryavian Jul 27 '18 at 22:43
  • $\begingroup$ That part makes sense to a point, but to me it looks like it would always be 0? The values are constants in my mind. $\endgroup$ – Jacob B Jul 28 '18 at 0:14

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