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Around the phrase in the book of http://neuralnetworksanddeeplearning.com/chap2.html

which obviously is easily computable.

There is $C = \frac{1}{2}\sum_{j}(y_j - a_{j}^{L}) ^2$

Then why is: $\frac{\partial C}{\partial a_{j}^{L}} = (a_{j}^{L} - y_j)$?

I thought it would be: $\frac{\partial C}{\partial a_{j}^{L}} = (y_j - a_{j}^{L} )$

The answer in the book would flip the sign, wouldn't it?

Is the flip between $a_{j}^{L}$ and $y_j$ a typo or intentional?

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  • $\begingroup$ $(f(x)^2)'=2f'(x)f(x).$ $\endgroup$ – Surb Feb 5 '19 at 16:02
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The book is correct. When you do the chain rule, you need to multiply by the derivative of $y-a$ with respect to $a$ which is $-1$. So the derivative is $-(y-a) = a-y$

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  • $\begingroup$ Oh! I didn't see that. Thanks! I just now realized it's the chain rule since it is $\frac{1}{2}(...)^2$ and then whatever is inside of it which is $-a$. $\endgroup$ – Melvin Roest Feb 5 '19 at 16:40

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