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Need some assistance with this. I never took multivariable calculus in college, so all of this is new to me.

Anyways, let's suppose that $$ x = \left\{ x_1, x_2, \cdots, x_n \right\}^T $$ and $$ y = \left\{ y_1, y_2,\cdots, y_n \right\}^T $$ and $$ f\left(x, y\right) = x\log \left(y\right)^T + \left(1 - x\right)\log(1 - y)^T $$

The task is to find the partial derivative of $$ f\left(x, y\right) $$ with respect to $$ y $$ assuming that $$ x $$ is a constant.

The partial derivative can be easily ascertained by simplifying:

$$ f_y\left(x, y\right) = \frac{x}{y^T} + \frac{x - 1}{\left(1 - y\right)^T} $$

My question is, is the above correct? If yes, how am I supposed to interpret it? I'm having a difficult time switching between treating both $$ x $$ and $$ y $$ like I would regular vectors and treating them as arbitrary variables.

Thank you for reading.

AsukaMinato
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tuba09
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    This is definitely not mathematical notation. More like matlab notation. So it is not surprising that you are confused, the interpretation requires some arbitrary assumptions (Is scalar+vector component-wise? Are functions applied component-wise?). – orion Nov 05 '14 at 09:13
  • @wuyudi please stop editing posts to use \left and \right on every pair of parentheses. This is not helpful. (Apologies to the OP for this brief non-mathematical interlude.) – KReiser Feb 20 '20 at 05:44
  • @KReiser sorry about that, I wouldn't add \left \right anymore. – AsukaMinato Feb 20 '20 at 05:47

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If I understood your notation you have a function $$f:\mathbb R^n\times \mathbb R^n\rightarrow \mathbb R $$ with $(x,y)\mapsto f(x,y):=\langle x, \ln y\rangle + \langle (1-x),\ln (1-y)\rangle$, denoting by brackets the scalar product of vectors in $\mathbb R^n$. Equivalently, we can write

$$f(x,y)=\sum_{i=1}^nx_i\ln y_i + (1-x_i)\ln (1-y_i), (*) $$

as the vectors $1-x$ and $1-y$ are defined by

$$1-x := (1-x_1, 1-x_2,\dots, 1-x_n) $$ and $$1-y := (1-y_1, 1-y_2,\dots, 1-y_n) $$

Then, for all $j=1,\dots, n$

$$\frac{\partial f}{\partial y_j } = x_j \frac{d \ln y_j}{d y_j } + (1-x_j)\frac{d \ln (1-y_j)}{d y_j },$$

as all the terms in $(*)$ with $i\neq j$ do not depend on $y_j$. Computing the derivatives of the logarithms we arrive at

$$\frac{\partial f}{\partial y_j } = x_j \frac{1}{y_j } - (1-x_j)\frac{1}{1-y_j }.$$

Avitus
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