Derivative of x^(2x)
$begingroup$
Maybe a simple question but I can't find a good example of it on the internet.
What is the derivative of:
$x^{2x}$
What is the simplest method of determining the derivative and what kind of rules are involved?
derivatives
asked Oct 28 '14 at 20:47
BasBas
1233
$endgroup$
add a comment |
$begingroup$
Maybe a simple question but I can't find a good example of it on the internet.
What is the derivative of:
$x^{2x}$
What is the simplest method of determining the derivative and what kind of rules are involved?
derivatives
asked Oct 28 '14 at 20:47
BasBas
1233
$endgroup$
add a comment |
$begingroup$
Maybe a simple question but I can't find a good example of it on the internet.
What is the derivative of:
$x^{2x}$
What is the simplest method of determining the derivative and what kind of rules are involved?
derivatives
asked Oct 28 '14 at 20:47
BasBas
1233
$endgroup$
Maybe a simple question but I can't find a good example of it on the internet.
What is the derivative of:
$x^{2x}$
What is the simplest method of determining the derivative and what kind of rules are involved?
derivatives
derivatives
asked Oct 28 '14 at 20:47
BasBas
1233
asked Oct 28 '14 at 20:47
BasBas
1233
asked Oct 28 '14 at 20:47
BasBas
1233
asked Oct 28 '14 at 20:47
BasBas
1233
asked Oct 28 '14 at 20:47
BasBas
1233
1233
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: $x^{2x} = e^{2x ln(x)}$. Use the chain rule.
answered Oct 28 '14 at 20:48
TomTom
9,0571218
$endgroup$
add a comment |
$begingroup$
HINT: Consider $y=x^{2x}$. Take $log$ both side. So, we get $log y=2x cdot log x$. Then apply chain rule..
$endgroup$
add a comment |
$begingroup$
As mentioned, first rewrite $x^{2x} = e^{2xln(x)}$. Then, by the chain rule,
$$frac{d}{dx}big(e^{2xln(x)}big)=(2ln(x)+2)e^{2xln(x)}=(2(ln(x)+1))e^{2xln(x)}=2x^{2x}(ln(x)+1)$$.
answered Jan 2 at 3:09
Axion004Axion004
430413
$endgroup$
add a comment |
Your Answer
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%2f995604%2fderivative-of-x2x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: $x^{2x} = e^{2x ln(x)}$. Use the chain rule.
answered Oct 28 '14 at 20:48
TomTom
9,0571218
$endgroup$
add a comment |
$begingroup$
Hint: $x^{2x} = e^{2x ln(x)}$. Use the chain rule.
answered Oct 28 '14 at 20:48
TomTom
9,0571218
$endgroup$
add a comment |
$begingroup$
Hint: $x^{2x} = e^{2x ln(x)}$. Use the chain rule.
answered Oct 28 '14 at 20:48
TomTom
9,0571218
$endgroup$
Hint: $x^{2x} = e^{2x ln(x)}$. Use the chain rule.
answered Oct 28 '14 at 20:48
TomTom
9,0571218
answered Oct 28 '14 at 20:48
TomTom
9,0571218
answered Oct 28 '14 at 20:48
TomTom
9,0571218
answered Oct 28 '14 at 20:48
TomTom
9,0571218
9,0571218
add a comment |
add a comment |
$begingroup$
HINT: Consider $y=x^{2x}$. Take $log$ both side. So, we get $log y=2x cdot log x$. Then apply chain rule..
$endgroup$
add a comment |
$begingroup$
HINT: Consider $y=x^{2x}$. Take $log$ both side. So, we get $log y=2x cdot log x$. Then apply chain rule..
$endgroup$
add a comment |
$begingroup$
HINT: Consider $y=x^{2x}$. Take $log$ both side. So, we get $log y=2x cdot log x$. Then apply chain rule..
$endgroup$
HINT: Consider $y=x^{2x}$. Take $log$ both side. So, we get $log y=2x cdot log x$. Then apply chain rule..
answered Oct 28 '14 at 20:58
community wiki
Argha
add a comment |
add a comment |
$begingroup$
As mentioned, first rewrite $x^{2x} = e^{2xln(x)}$. Then, by the chain rule,
$$frac{d}{dx}big(e^{2xln(x)}big)=(2ln(x)+2)e^{2xln(x)}=(2(ln(x)+1))e^{2xln(x)}=2x^{2x}(ln(x)+1)$$.
answered Jan 2 at 3:09
Axion004Axion004
430413
$endgroup$
add a comment |
$begingroup$
As mentioned, first rewrite $x^{2x} = e^{2xln(x)}$. Then, by the chain rule,
$$frac{d}{dx}big(e^{2xln(x)}big)=(2ln(x)+2)e^{2xln(x)}=(2(ln(x)+1))e^{2xln(x)}=2x^{2x}(ln(x)+1)$$.
answered Jan 2 at 3:09
Axion004Axion004
430413
$endgroup$
add a comment |
$begingroup$
As mentioned, first rewrite $x^{2x} = e^{2xln(x)}$. Then, by the chain rule,
$$frac{d}{dx}big(e^{2xln(x)}big)=(2ln(x)+2)e^{2xln(x)}=(2(ln(x)+1))e^{2xln(x)}=2x^{2x}(ln(x)+1)$$.
answered Jan 2 at 3:09
Axion004Axion004
430413
$endgroup$
As mentioned, first rewrite $x^{2x} = e^{2xln(x)}$. Then, by the chain rule,
$$frac{d}{dx}big(e^{2xln(x)}big)=(2ln(x)+2)e^{2xln(x)}=(2(ln(x)+1))e^{2xln(x)}=2x^{2x}(ln(x)+1)$$.
answered Jan 2 at 3:09
Axion004Axion004
430413
answered Jan 2 at 3:09
Axion004Axion004
430413
answered Jan 2 at 3:09
Axion004Axion004
430413
answered Jan 2 at 3:09
Axion004Axion004
430413
430413
add a comment |
add a comment |
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.
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%2f995604%2fderivative-of-x2x%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
