Backpropagation to calculate derivative
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I have got a task to calculate $frac{dv}{dx}$ using backpropagation, where
$v = z^2 + y$
$z = frac{y}{3}$
$y = arctan(x)$
Am I supposed to just use chain rule here or do something else? I'd really like to see an example of calculating this.
I did something like this
$frac{dv}{dz}$ = $2z$
$frac{dz}{dy}$ = $frac{1}{3}$
$frac{∂v}{∂y}$ = $1$
$frac{dv}{dy}$ = $frac{∂v}{∂z}$$*$$frac{dz}{dy}$ + $frac{∂v}{∂y}$ = $frac{2}{3}z$ $+1$
$frac{dv}{dx}$ = $frac{dv}{dy}$$*$$frac{dy}{dx}$ = $(frac{2}{3}z+1)*$ $frac{1}{(1+x^2)}$
calculus neural-networks
asked Nov 22 at 0:10
MatKravi
11
add a comment |
up vote
0
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I have got a task to calculate $frac{dv}{dx}$ using backpropagation, where
$v = z^2 + y$
$z = frac{y}{3}$
$y = arctan(x)$
Am I supposed to just use chain rule here or do something else? I'd really like to see an example of calculating this.
I did something like this
$frac{dv}{dz}$ = $2z$
$frac{dz}{dy}$ = $frac{1}{3}$
$frac{∂v}{∂y}$ = $1$
$frac{dv}{dy}$ = $frac{∂v}{∂z}$$*$$frac{dz}{dy}$ + $frac{∂v}{∂y}$ = $frac{2}{3}z$ $+1$
$frac{dv}{dx}$ = $frac{dv}{dy}$$*$$frac{dy}{dx}$ = $(frac{2}{3}z+1)*$ $frac{1}{(1+x^2)}$
calculus neural-networks
asked Nov 22 at 0:10
MatKravi
11
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 22 at 0:14
Also, please do not post questions that read like a text message: "i," "sth".
– amWhy
Nov 22 at 0:18
Yes, you are supposed to use the chain rule.
– Matthew Towers
Nov 22 at 0:22
Yes use the chain rule. If you like you can write this as $v=z^2+y=(y/3)^2+y=(arctan/3)^2+arctan(x)$ and calculate the derivative.
– Andres Mejia
Nov 22 at 0:40
My teacher wrote that i'm not supposed to use "normal" differentiation but the backpropagation. So am I supposed to simply calculate it like that?
– MatKravi
Nov 22 at 0:54
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have got a task to calculate $frac{dv}{dx}$ using backpropagation, where
$v = z^2 + y$
$z = frac{y}{3}$
$y = arctan(x)$
Am I supposed to just use chain rule here or do something else? I'd really like to see an example of calculating this.
I did something like this
$frac{dv}{dz}$ = $2z$
$frac{dz}{dy}$ = $frac{1}{3}$
$frac{∂v}{∂y}$ = $1$
$frac{dv}{dy}$ = $frac{∂v}{∂z}$$*$$frac{dz}{dy}$ + $frac{∂v}{∂y}$ = $frac{2}{3}z$ $+1$
$frac{dv}{dx}$ = $frac{dv}{dy}$$*$$frac{dy}{dx}$ = $(frac{2}{3}z+1)*$ $frac{1}{(1+x^2)}$
calculus neural-networks
asked Nov 22 at 0:10
MatKravi
11
I have got a task to calculate $frac{dv}{dx}$ using backpropagation, where
$v = z^2 + y$
$z = frac{y}{3}$
$y = arctan(x)$
Am I supposed to just use chain rule here or do something else? I'd really like to see an example of calculating this.
I did something like this
$frac{dv}{dz}$ = $2z$
$frac{dz}{dy}$ = $frac{1}{3}$
$frac{∂v}{∂y}$ = $1$
$frac{dv}{dy}$ = $frac{∂v}{∂z}$$*$$frac{dz}{dy}$ + $frac{∂v}{∂y}$ = $frac{2}{3}z$ $+1$
$frac{dv}{dx}$ = $frac{dv}{dy}$$*$$frac{dy}{dx}$ = $(frac{2}{3}z+1)*$ $frac{1}{(1+x^2)}$
calculus neural-networks
calculus neural-networks
asked Nov 22 at 0:10
MatKravi
11
asked Nov 22 at 0:10
MatKravi
11
edited Nov 22 at 12:48
asked Nov 22 at 0:10
MatKravi
11
asked Nov 22 at 0:10
MatKravi
11
asked Nov 22 at 0:10
MatKravi
11
11
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 22 at 0:14
Also, please do not post questions that read like a text message: "i," "sth".
– amWhy
Nov 22 at 0:18
Yes, you are supposed to use the chain rule.
– Matthew Towers
Nov 22 at 0:22
Yes use the chain rule. If you like you can write this as $v=z^2+y=(y/3)^2+y=(arctan/3)^2+arctan(x)$ and calculate the derivative.
– Andres Mejia
Nov 22 at 0:40
My teacher wrote that i'm not supposed to use "normal" differentiation but the backpropagation. So am I supposed to simply calculate it like that?
– MatKravi
Nov 22 at 0:54
add a comment |
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 22 at 0:14
Also, please do not post questions that read like a text message: "i," "sth".
– amWhy
Nov 22 at 0:18
Yes, you are supposed to use the chain rule.
– Matthew Towers
Nov 22 at 0:22
Yes use the chain rule. If you like you can write this as $v=z^2+y=(y/3)^2+y=(arctan/3)^2+arctan(x)$ and calculate the derivative.
– Andres Mejia
Nov 22 at 0:40
My teacher wrote that i'm not supposed to use "normal" differentiation but the backpropagation. So am I supposed to simply calculate it like that?
– MatKravi
Nov 22 at 0:54
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 22 at 0:14
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 22 at 0:14
Also, please do not post questions that read like a text message: "i," "sth".
– amWhy
Nov 22 at 0:18
Also, please do not post questions that read like a text message: "i," "sth".
– amWhy
Nov 22 at 0:18
Yes, you are supposed to use the chain rule.
– Matthew Towers
Nov 22 at 0:22
Yes, you are supposed to use the chain rule.
– Matthew Towers
Nov 22 at 0:22
Yes use the chain rule. If you like you can write this as $v=z^2+y=(y/3)^2+y=(arctan/3)^2+arctan(x)$ and calculate the derivative.
– Andres Mejia
Nov 22 at 0:40
Yes use the chain rule. If you like you can write this as $v=z^2+y=(y/3)^2+y=(arctan/3)^2+arctan(x)$ and calculate the derivative.
– Andres Mejia
Nov 22 at 0:40
My teacher wrote that i'm not supposed to use "normal" differentiation but the backpropagation. So am I supposed to simply calculate it like that?
– MatKravi
Nov 22 at 0:54
My teacher wrote that i'm not supposed to use "normal" differentiation but the backpropagation. So am I supposed to simply calculate it like that?
– MatKravi
Nov 22 at 0:54
add a comment |
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Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 22 at 0:14
Also, please do not post questions that read like a text message: "i," "sth".
– amWhy
Nov 22 at 0:18
Yes, you are supposed to use the chain rule.
– Matthew Towers
Nov 22 at 0:22
Yes use the chain rule. If you like you can write this as $v=z^2+y=(y/3)^2+y=(arctan/3)^2+arctan(x)$ and calculate the derivative.
– Andres Mejia
Nov 22 at 0:40
My teacher wrote that i'm not supposed to use "normal" differentiation but the backpropagation. So am I supposed to simply calculate it like that?
– MatKravi
Nov 22 at 0:54