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I'm deriving backpropagation step of training neural networks using vectorized equations. Following are two forward propagation equations between last hidden layer and output layer.

$$ Z^{[2]} = W^{[2]}A^{[1]} + b^{[2]} $$ $$ A^{[2]} = \hat{y} = g(Z^{[2]})$$

Where $g(z)$ is activation function.

Now in backpropagation, we calculate the change in cost function ($J$) w.r.t. all the parameters.

I've successfully calculated $\partial{J}/\partial{A^{[2]}}$ and $\partial{J}/\partial{Z^{[2]}}$ using chain rule. Now to calculate $\partial{J}/\partial{W^{[2]}}$, I've formed following chain

$$ \frac{\partial{J}}{\partial{W^{[2]}}} = \frac{\partial{J}}{\partial{A^{[2]}}} \frac{\partial{A^{[2]}}}{\partial{Z^{[2]}}} \frac{\partial{Z^{[2]}}}{\partial{W^{[2]}}}$$

Now to calculate $\frac{\partial{Z^{[2]}}}{\partial{W^{[2]}}}$, I used $ Z^{[2]} = W^{[2]}A^{[1]} + b^{[2]} $ equation which simply gives the derivation $ A^{[1]} $. But in literature, it's given as $ A^{[1]^{T}} $, which is transpose of my answer.

By checking dimensions of the answer, I could verify that the answer should be $ A^{[1]^{T}} $ instead if $ A^{[1]} $. But is there any general rule for such cases using which I could directly tell whether the derivative will be the transpose of a matrix or not without verifying dimensions?

I've also checked matrix cookbook but couldn't find any related thumb rules.

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Unfortunately, there are 2 common convention in matrix calculus: Jacobian (Numerator) and Gradient (Denominator).

For a function $f:\mathbb R^n \to \mathbb R^m$, then

  1. Jacobian convention: $\frac{\partial f}{\partial x}$ is represented by the $m\times n$ matrix with entries $\big(\tfrac{\partial f}{\partial x}\big)_{ij} = \tfrac{\partial f_i}{\partial x_j}$

  2. Gradiant convention: $\frac{\partial f}{\partial x}$ is represented by the $n\times m$ matrix with entries $\big(\tfrac{\partial f}{\partial x}\big)_{ij} = \tfrac{\partial f_j}{\partial x_i}$

If $f(x) = Ax$ then one can easily check that

  1. Under Jacobian convention $\frac{\partial f}{\partial x} = A$

  2. Under Gradient convention $\frac{\partial f}{\partial x} = A^T$

So it is a matter of which convention you are following.

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