Are these two equivalent
$begingroup$
When finding derivatives, a useful tool is $(a^x)' = (a^x)ln(a)$.
Now say I have $2^{3x}$. When trying to find the derivative, are these two equivalent?
$(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$
calculus
asked Dec 10 '18 at 7:33
mingming
3415
$endgroup$
add a comment |
$begingroup$
When finding derivatives, a useful tool is $(a^x)' = (a^x)ln(a)$.
Now say I have $2^{3x}$. When trying to find the derivative, are these two equivalent?
$(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$
calculus
asked Dec 10 '18 at 7:33
mingming
3415
$endgroup$
add a comment |
$begingroup$
When finding derivatives, a useful tool is $(a^x)' = (a^x)ln(a)$.
Now say I have $2^{3x}$. When trying to find the derivative, are these two equivalent?
$(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$
calculus
asked Dec 10 '18 at 7:33
mingming
3415
$endgroup$
When finding derivatives, a useful tool is $(a^x)' = (a^x)ln(a)$.
Now say I have $2^{3x}$. When trying to find the derivative, are these two equivalent?
$(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$
calculus
calculus
asked Dec 10 '18 at 7:33
mingming
3415
asked Dec 10 '18 at 7:33
mingming
3415
asked Dec 10 '18 at 7:33
mingming
3415
asked Dec 10 '18 at 7:33
mingming
3415
asked Dec 10 '18 at 7:33
mingming
3415
3415
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
They are not equivalent, the first derivative is given by $$(2^{3x})'=2^{3x}ln(2)cdot 3$$ by the chain rule.
answered Dec 10 '18 at 7:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
$endgroup$
$begingroup$
Why doesn't the power rule apply here as well?
$endgroup$
– ming
Dec 10 '18 at 17:25
$begingroup$
Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 17:29
add a comment |
$begingroup$
Both $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$ are wrong !
We have $2^{3x}=8^x$, hence $(2^{3x})'=8^x ln (8).$
answered Dec 10 '18 at 7:36
FredFred
45.7k1848
$endgroup$
$begingroup$
So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
$endgroup$
– ming
Dec 10 '18 at 17:58
add a comment |
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%2f3033586%2fare-these-two-equivalent%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
They are not equivalent, the first derivative is given by $$(2^{3x})'=2^{3x}ln(2)cdot 3$$ by the chain rule.
answered Dec 10 '18 at 7:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
$endgroup$
$begingroup$
Why doesn't the power rule apply here as well?
$endgroup$
– ming
Dec 10 '18 at 17:25
$begingroup$
Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 17:29
add a comment |
$begingroup$
They are not equivalent, the first derivative is given by $$(2^{3x})'=2^{3x}ln(2)cdot 3$$ by the chain rule.
answered Dec 10 '18 at 7:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
$endgroup$
$begingroup$
Why doesn't the power rule apply here as well?
$endgroup$
– ming
Dec 10 '18 at 17:25
$begingroup$
Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 17:29
add a comment |
$begingroup$
They are not equivalent, the first derivative is given by $$(2^{3x})'=2^{3x}ln(2)cdot 3$$ by the chain rule.
answered Dec 10 '18 at 7:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
$endgroup$
They are not equivalent, the first derivative is given by $$(2^{3x})'=2^{3x}ln(2)cdot 3$$ by the chain rule.
answered Dec 10 '18 at 7:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
answered Dec 10 '18 at 7:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
answered Dec 10 '18 at 7:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
answered Dec 10 '18 at 7:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
74.9k42865
$begingroup$
Why doesn't the power rule apply here as well?
$endgroup$
– ming
Dec 10 '18 at 17:25
$begingroup$
Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 17:29
add a comment |
$begingroup$
Why doesn't the power rule apply here as well?
$endgroup$
– ming
Dec 10 '18 at 17:25
$begingroup$
Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 17:29
$begingroup$
Why doesn't the power rule apply here as well?
$endgroup$
– ming
Dec 10 '18 at 17:25
$begingroup$
Why doesn't the power rule apply here as well?
$endgroup$
– ming
Dec 10 '18 at 17:25
$begingroup$
Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 17:29
$begingroup$
Since $$(a^x)'=a^xln(a)$$ and your case is $$a=2$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 17:29
add a comment |
$begingroup$
Both $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$ are wrong !
We have $2^{3x}=8^x$, hence $(2^{3x})'=8^x ln (8).$
answered Dec 10 '18 at 7:36
FredFred
45.7k1848
$endgroup$
$begingroup$
So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
$endgroup$
– ming
Dec 10 '18 at 17:58
add a comment |
$begingroup$
Both $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$ are wrong !
We have $2^{3x}=8^x$, hence $(2^{3x})'=8^x ln (8).$
answered Dec 10 '18 at 7:36
FredFred
45.7k1848
$endgroup$
$begingroup$
So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
$endgroup$
– ming
Dec 10 '18 at 17:58
add a comment |
$begingroup$
Both $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$ are wrong !
We have $2^{3x}=8^x$, hence $(2^{3x})'=8^x ln (8).$
answered Dec 10 '18 at 7:36
FredFred
45.7k1848
$endgroup$
Both $(2^x)ln(2)(3)$ and $(3x)2^{3x - 1}$ are wrong !
We have $2^{3x}=8^x$, hence $(2^{3x})'=8^x ln (8).$
answered Dec 10 '18 at 7:36
FredFred
45.7k1848
answered Dec 10 '18 at 7:36
FredFred
45.7k1848
answered Dec 10 '18 at 7:36
FredFred
45.7k1848
answered Dec 10 '18 at 7:36
FredFred
45.7k1848
45.7k1848
$begingroup$
So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
$endgroup$
– ming
Dec 10 '18 at 17:58
add a comment |
$begingroup$
So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
$endgroup$
– ming
Dec 10 '18 at 17:58
$begingroup$
So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
$endgroup$
– ming
Dec 10 '18 at 17:58
$begingroup$
So was the rule taught to me incorrect? $( a x ) ′ =( a x )ln(a)$
$endgroup$
– ming
Dec 10 '18 at 17:58
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%2f3033586%2fare-these-two-equivalent%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
