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Given the real-valued functions $f_1$ and $f_2$ with $x\in\mathbb{R}$, then $$ \frac{df_2(f_1(x))}{dx} = \frac{df_2(f_1(x))}{d f_1(x)}\frac{df_1(x)}{dx}$$

Is it then the case that if we also have a real-valued function $f_3$, then $$ \frac{df_3(f_2(f_1(x)))}{dx} = \frac{df_3(f_2(f_1(x)))}{df_2(f_1(x))}\frac{df_2(f_1(x))}{df_1(x)}\frac{df_1(x)}{dx}$$

Is this correct?

If so, does the chain rule generalize to the following statement: $$ \frac{df_{N}(f_{N-1}(...f_2(f_1(x))))}{dx} $$ $$= \frac{df_{N}(f_{N-1}(...f_2(f_1(x))))}{df_{N-1}(...f_2(f_1(x)))} \frac{df_{N-1}(...f_2(f_1(x)))}{df_{N-2}(...f_2(f_1(x)))}...\frac{df_2(f_1(x))}{df_1(x)}\frac{df_1(x)}{dx}$$

My apologies about this looking like a straight calculus question. I am asking to figure out the backpropagation algorithm in its general form.

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  • $\begingroup$ that's true.... $\endgroup$
    – Fat32
    Commented Jun 18, 2017 at 1:59
  • $\begingroup$ it's fine with me to post it here, but there is a nice mathematics SE which would be an ideal place for this post. $\endgroup$ Commented Jun 18, 2017 at 3:58

1 Answer 1

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I'm not a mathematician but if the multiple composition $ f_1(f_2(f_3 \dots)))$ is equivalent to a sequence of single compositions $g_2(g_1)$, and another $g_3(g_2)$ and so forth.

The generalization looks reasonable for $x, f_{1}, f_{2} \dots f_N \in \cal{R}$ and $f_i$ differentiable with no discontinuities.

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  • $\begingroup$ I'm far from a mathematician myself, and I think the three responses saying it is correct is good enough for me. :) $\endgroup$
    – Josh
    Commented Jun 18, 2017 at 4:33

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