Suppose there is a DNN like this:

$h = t(w_1x+b_1)$
$g = t(w_2h+b_2)$
$\hat y = t(w_3g+b_3)$
$Loss = \frac 12\sum (y - \hat y)^2$

Just one input $x$
$t$ is activation function.

Could you explain with equations how the gradients of $w_1$, $w_2$, $w_3$ is calculated?

By the way, when combining partial derivative and chain rule,

$\frac {dk}{dg}\frac {dg}{dh}\frac {dh}{dx}$ -> $\frac {\partial k}{\partial g}\frac {\partial g}{\partial h}\frac {\partial h}{\partial w_1}$

is the above fine?

  • $\begingroup$ how is y related to h,g and k? $\endgroup$
    – gunes
    Dec 28, 2019 at 9:09
  • $\begingroup$ I edited my question. sorry. $\endgroup$
    – Crispy13
    Dec 28, 2019 at 9:14

1 Answer 1


We just execute the chain rule: $$\frac{\partial L}{\partial w_1}=\frac{\partial L}{\partial \hat y}\frac{\partial \hat y}{\partial g}\frac{\partial g}{\partial h}\frac{\partial h}{\partial w_1}$$

Then, we calculate each term separately, e.g. $$\frac{\partial h}{\partial w_1}=t'(w_1x+b_1)x,\frac{\partial g}{\partial h}=t'(w_2h+b_2)w_2,\cdots$$

Your chain rule is not fine, because we don't ever calculate $\frac{\partial h}{\partial x}$ since $x$ is not a parameter nor an outcome.

  • $\begingroup$ What about calculating w_2? Is it just to substitue w_1 into w_2 and calculate the equation which you wrote? $\endgroup$
    – Crispy13
    Dec 28, 2019 at 9:29
  • $\begingroup$ it’s similar, but upto the term h, not g because it’s not in the first layer. $\endgroup$
    – gunes
    Dec 28, 2019 at 9:32
  • $\begingroup$ so $\frac {\partial L}{\partial w_2} = \frac {\partial L}{\partial \hat y}\frac {\partial \hat y}{\partial g}\frac {\partial g}{\partial w_2}$ is right? $\endgroup$
    – Crispy13
    Dec 28, 2019 at 9:36
  • $\begingroup$ Yes, that’s right $\endgroup$
    – gunes
    Dec 28, 2019 at 9:58

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