Mixing
Backpropagation to calculate derivative
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 '18 at 0:10
MatKraviMatKravi
11
add a comment |
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 '18 at 0:10
MatKraviMatKravi
11
Also, please do not post questions that read like a text message: "i," "sth".
– amWhy
Nov 22 '18 at 0:18
Yes, you are supposed to use the chain rule.
– Matthew Towers
Nov 22 '18 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 '18 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 '18 at 0:54
add a comment |
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 '18 at 0:10
MatKraviMatKravi
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 '18 at 0:10
MatKraviMatKravi
11
asked Nov 22 '18 at 0:10
MatKraviMatKravi
11
edited Nov 22 '18 at 12:48
MatKravi
asked Nov 22 '18 at 0:10
MatKraviMatKravi
11
asked Nov 22 '18 at 0:10
MatKraviMatKravi
11
asked Nov 22 '18 at 0:10
MatKraviMatKravi
11
11
Also, please do not post questions that read like a text message: "i," "sth".
– amWhy
Nov 22 '18 at 0:18
Yes, you are supposed to use the chain rule.
– Matthew Towers
Nov 22 '18 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 '18 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 '18 at 0:54
add a comment |
Also, please do not post questions that read like a text message: "i," "sth".
– amWhy
Nov 22 '18 at 0:18
Yes, you are supposed to use the chain rule.
– Matthew Towers
Nov 22 '18 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 '18 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 '18 at 0:54
Also, please do not post questions that read like a text message: "i," "sth".
– amWhy
Nov 22 '18 at 0:18
Also, please do not post questions that read like a text message: "i," "sth".
– amWhy
Nov 22 '18 at 0:18
Yes, you are supposed to use the chain rule.
– Matthew Towers
Nov 22 '18 at 0:22
Yes, you are supposed to use the chain rule.
– Matthew Towers
Nov 22 '18 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 '18 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 '18 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 '18 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 '18 at 0:54
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008576%2fbackpropagation-to-calculate-derivative%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008576%2fbackpropagation-to-calculate-derivative%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Also, please do not post questions that read like a text message: "i," "sth".
– amWhy
Nov 22 '18 at 0:18
Yes, you are supposed to use the chain rule.
– Matthew Towers
Nov 22 '18 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 '18 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 '18 at 0:54