Mixing
Show that $d/dx (a^x) = a^xln a$.
Show that
$$
frac{d}{dx} a^x = a^x ln a.
$$
How would I do a proof for this. I can't seem to get it to work anyway I try.
I know that
$$
frac{d}{dx} e^x = e^x.
$$
Does that help me here?
calculus
edited May 22 '13 at 3:48
kahen
13.2k32554
asked May 22 '13 at 3:25
1ftw1
3602615
add a comment |
Show that
$$
frac{d}{dx} a^x = a^x ln a.
$$
How would I do a proof for this. I can't seem to get it to work anyway I try.
I know that
$$
frac{d}{dx} e^x = e^x.
$$
Does that help me here?
calculus
edited May 22 '13 at 3:48
kahen
13.2k32554
asked May 22 '13 at 3:25
1ftw1
3602615
add a comment |
Show that
$$
frac{d}{dx} a^x = a^x ln a.
$$
How would I do a proof for this. I can't seem to get it to work anyway I try.
I know that
$$
frac{d}{dx} e^x = e^x.
$$
Does that help me here?
calculus
edited May 22 '13 at 3:48
kahen
13.2k32554
asked May 22 '13 at 3:25
1ftw1
3602615
Show that
$$
frac{d}{dx} a^x = a^x ln a.
$$
How would I do a proof for this. I can't seem to get it to work anyway I try.
I know that
$$
frac{d}{dx} e^x = e^x.
$$
Does that help me here?
calculus
calculus
edited May 22 '13 at 3:48
kahen
13.2k32554
asked May 22 '13 at 3:25
1ftw1
3602615
edited May 22 '13 at 3:48
kahen
13.2k32554
asked May 22 '13 at 3:25
1ftw1
3602615
edited May 22 '13 at 3:48
kahen
13.2k32554
edited May 22 '13 at 3:48
kahen
13.2k32554
edited May 22 '13 at 3:48
kahen
13.2k32554
13.2k32554
asked May 22 '13 at 3:25
1ftw1
3602615
asked May 22 '13 at 3:25
1ftw1
3602615
asked May 22 '13 at 3:25
1ftw1
3602615
3602615
add a comment |
add a comment |
7 Answers
7
active
oldest
votes
Hint: $a^x=e^{ln a^x}=e^{xln a}$.
answered May 22 '13 at 3:38
A. Chu
6,98583283
add a comment |
Let $f : mathbb{R} to mathbb{R}$ be given by $f(x)=a^x$ and consider the $ln$ function. We can take the composition so that we have:
$$(lncirc f)(x)=ln (a^x)=xln a$$
Now, if we take the derivative, on the left hand side we use the chain rule and on the right hand side we differentiate as usual so that we have:
$$frac{f'(x)}{f(x)}=ln a$$
Now solving for $f'(x)$ gives $f'(x) = f(x) ln a$ so that $f'(x) = a^x ln a$. This useful technique can be used to take derivatives of other functions: we compose the original function with the inverse and then differentiate on both sides and use the same idea we've used here, this technique can simplify many derivatives and save a lot of time in some situations.
answered May 22 '13 at 3:58
user1620696
11.5k341114
add a comment |
Hint: Write $y = a^x$ or equivalently $ln y = x ln a$ and use implicit differentiation.
answered May 22 '13 at 3:28
response
4,70121220
add a comment |
, the last term, 
, can be explained by L'Hopital's rule, just taking the limits of the numerator and denominator one of each time.
answered Feb 24 '14 at 0:22
user131054
111
add a comment |
Here are the steps
$$
frac{d}{dx} left[a^xright] = frac{d}{dx} left[e^{ln a^x}right]= e^{ln a^x} frac{d}{dx} left[ln a^xright]
$$
$$
= a^x frac{d}{dx} left[xln aright] = a^xleft(ln aright)frac{d}{dx} left[xright]= a^xln a
$$
answered Jan 5 '15 at 18:06
k170
7,45131540
add a comment |
$ frac{d}{dx} a^x=$
$=d/dx$ $e^{ln (a^x)}=$
$=d/dx e^{xln(a)}$ just as you said.
Now it's time for chain rule.
$f(x)=e^x$
$g(x)=x*ln(a)$
$a^x= f(g(x))$
last comment before I get solving
$a=e^{ln(a)}$
now let's get it done
$d/dx e^{x ln(a)}=$
$=e^{x*ln(a)} d/dx (x ln(a))=$ (by chain rule)
$=e^{x ln(a)} * ln(a)$
and that is the solution. Wait it doesnt look correct --> we should do some algebra!
Show for yourself that
$e^{x*ln(a)}=( e^{ln(a)} )^x$ but as we said that $a = e^{ln(a)}$ so $( e^{ln(a)} )^x$ actually equals $a^x$ and of course $e^{x*ln(a)}$ equals a^x
With what we just said you can see that $e^{x * ln(a)} ln(a)$ equals $a^x ln(a)$ or
$$d/dx a^x = a^x ln(a)$$
edited Jan 9 '15 at 20:35
Community♦
1
answered Jan 5 '15 at 17:29
grekiki
1
1
Use TeX while typing math symbols
– supremum
Jan 5 '15 at 17:59
For help, see this FAQ entry about $LaTeX$ on Math.SE.
– Ruslan
Jan 5 '15 at 18:43
add a comment |
Let $y = a^x$
Then taking log on both side to the base $e$
We have;
$ ln y = ln a^x$
$ ln y = xcdot ln a$
Taking derivative with respect to $x$;
$frac d{dx} ln y = frac d{dx} xcdot ln a$
(I have applied chain rule here>>)
$ frac 1y frac{dy}{dx} = ln a + 0$
$ frac {dy}{dx} = ycdot ln a$
We know $y= a^x$
So, $frac {dy}{dx} = a^x ln a$
Hope it was easier than other methodologies used to derive it!
edited Nov 21 '18 at 9:46
idea
2,15441025
answered Nov 21 '18 at 9:17
Vichitra Attri
11
The question is five years old and your answer adds nothing new to the already existing answers.
– José Carlos Santos
Nov 21 '18 at 9:39
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%2f398826%2fshow-that-d-dx-ax-ax-ln-a%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint: $a^x=e^{ln a^x}=e^{xln a}$.
answered May 22 '13 at 3:38
A. Chu
6,98583283
add a comment |
Hint: $a^x=e^{ln a^x}=e^{xln a}$.
answered May 22 '13 at 3:38
A. Chu
6,98583283
add a comment |
Hint: $a^x=e^{ln a^x}=e^{xln a}$.
answered May 22 '13 at 3:38
A. Chu
6,98583283
Hint: $a^x=e^{ln a^x}=e^{xln a}$.
answered May 22 '13 at 3:38
A. Chu
6,98583283
answered May 22 '13 at 3:38
A. Chu
6,98583283
answered May 22 '13 at 3:38
A. Chu
6,98583283
answered May 22 '13 at 3:38
A. Chu
6,98583283
6,98583283
add a comment |
add a comment |
Let $f : mathbb{R} to mathbb{R}$ be given by $f(x)=a^x$ and consider the $ln$ function. We can take the composition so that we have:
$$(lncirc f)(x)=ln (a^x)=xln a$$
Now, if we take the derivative, on the left hand side we use the chain rule and on the right hand side we differentiate as usual so that we have:
$$frac{f'(x)}{f(x)}=ln a$$
Now solving for $f'(x)$ gives $f'(x) = f(x) ln a$ so that $f'(x) = a^x ln a$. This useful technique can be used to take derivatives of other functions: we compose the original function with the inverse and then differentiate on both sides and use the same idea we've used here, this technique can simplify many derivatives and save a lot of time in some situations.
answered May 22 '13 at 3:58
user1620696
11.5k341114
add a comment |
Let $f : mathbb{R} to mathbb{R}$ be given by $f(x)=a^x$ and consider the $ln$ function. We can take the composition so that we have:
$$(lncirc f)(x)=ln (a^x)=xln a$$
Now, if we take the derivative, on the left hand side we use the chain rule and on the right hand side we differentiate as usual so that we have:
$$frac{f'(x)}{f(x)}=ln a$$
Now solving for $f'(x)$ gives $f'(x) = f(x) ln a$ so that $f'(x) = a^x ln a$. This useful technique can be used to take derivatives of other functions: we compose the original function with the inverse and then differentiate on both sides and use the same idea we've used here, this technique can simplify many derivatives and save a lot of time in some situations.
answered May 22 '13 at 3:58
user1620696
11.5k341114
add a comment |
Let $f : mathbb{R} to mathbb{R}$ be given by $f(x)=a^x$ and consider the $ln$ function. We can take the composition so that we have:
$$(lncirc f)(x)=ln (a^x)=xln a$$
Now, if we take the derivative, on the left hand side we use the chain rule and on the right hand side we differentiate as usual so that we have:
$$frac{f'(x)}{f(x)}=ln a$$
Now solving for $f'(x)$ gives $f'(x) = f(x) ln a$ so that $f'(x) = a^x ln a$. This useful technique can be used to take derivatives of other functions: we compose the original function with the inverse and then differentiate on both sides and use the same idea we've used here, this technique can simplify many derivatives and save a lot of time in some situations.
answered May 22 '13 at 3:58
user1620696
11.5k341114
Let $f : mathbb{R} to mathbb{R}$ be given by $f(x)=a^x$ and consider the $ln$ function. We can take the composition so that we have:
$$(lncirc f)(x)=ln (a^x)=xln a$$
Now, if we take the derivative, on the left hand side we use the chain rule and on the right hand side we differentiate as usual so that we have:
$$frac{f'(x)}{f(x)}=ln a$$
Now solving for $f'(x)$ gives $f'(x) = f(x) ln a$ so that $f'(x) = a^x ln a$. This useful technique can be used to take derivatives of other functions: we compose the original function with the inverse and then differentiate on both sides and use the same idea we've used here, this technique can simplify many derivatives and save a lot of time in some situations.
answered May 22 '13 at 3:58
user1620696
11.5k341114
edited Jan 5 '15 at 23:28
answered May 22 '13 at 3:58
user1620696
11.5k341114
answered May 22 '13 at 3:58
user1620696
11.5k341114
answered May 22 '13 at 3:58
user1620696
11.5k341114
11.5k341114
add a comment |
add a comment |
Hint: Write $y = a^x$ or equivalently $ln y = x ln a$ and use implicit differentiation.
answered May 22 '13 at 3:28
response
4,70121220
add a comment |
Hint: Write $y = a^x$ or equivalently $ln y = x ln a$ and use implicit differentiation.
answered May 22 '13 at 3:28
response
4,70121220
add a comment |
Hint: Write $y = a^x$ or equivalently $ln y = x ln a$ and use implicit differentiation.
answered May 22 '13 at 3:28
response
4,70121220
Hint: Write $y = a^x$ or equivalently $ln y = x ln a$ and use implicit differentiation.
answered May 22 '13 at 3:28
response
4,70121220
answered May 22 '13 at 3:28
response
4,70121220
answered May 22 '13 at 3:28
response
4,70121220
answered May 22 '13 at 3:28
response
4,70121220
4,70121220
add a comment |
add a comment |
, the last term, , can be explained by L'Hopital's rule, just taking the limits of the numerator and denominator one of each time.
answered Feb 24 '14 at 0:22
user131054
111
add a comment |
, the last term, , can be explained by L'Hopital's rule, just taking the limits of the numerator and denominator one of each time.
answered Feb 24 '14 at 0:22
user131054
111
add a comment |
, the last term, , can be explained by L'Hopital's rule, just taking the limits of the numerator and denominator one of each time.
answered Feb 24 '14 at 0:22
user131054
111
, the last term, , can be explained by L'Hopital's rule, just taking the limits of the numerator and denominator one of each time.
answered Feb 24 '14 at 0:22
user131054
111
answered Feb 24 '14 at 0:22
user131054
111
answered Feb 24 '14 at 0:22
user131054
111
answered Feb 24 '14 at 0:22
user131054
111
111
add a comment |
add a comment |
Here are the steps
$$
frac{d}{dx} left[a^xright] = frac{d}{dx} left[e^{ln a^x}right]= e^{ln a^x} frac{d}{dx} left[ln a^xright]
$$
$$
= a^x frac{d}{dx} left[xln aright] = a^xleft(ln aright)frac{d}{dx} left[xright]= a^xln a
$$
answered Jan 5 '15 at 18:06
k170
7,45131540
add a comment |
Here are the steps
$$
frac{d}{dx} left[a^xright] = frac{d}{dx} left[e^{ln a^x}right]= e^{ln a^x} frac{d}{dx} left[ln a^xright]
$$
$$
= a^x frac{d}{dx} left[xln aright] = a^xleft(ln aright)frac{d}{dx} left[xright]= a^xln a
$$
answered Jan 5 '15 at 18:06
k170
7,45131540
add a comment |
Here are the steps
$$
frac{d}{dx} left[a^xright] = frac{d}{dx} left[e^{ln a^x}right]= e^{ln a^x} frac{d}{dx} left[ln a^xright]
$$
$$
= a^x frac{d}{dx} left[xln aright] = a^xleft(ln aright)frac{d}{dx} left[xright]= a^xln a
$$
answered Jan 5 '15 at 18:06
k170
7,45131540
Here are the steps
$$
frac{d}{dx} left[a^xright] = frac{d}{dx} left[e^{ln a^x}right]= e^{ln a^x} frac{d}{dx} left[ln a^xright]
$$
$$
= a^x frac{d}{dx} left[xln aright] = a^xleft(ln aright)frac{d}{dx} left[xright]= a^xln a
$$
answered Jan 5 '15 at 18:06
k170
7,45131540
answered Jan 5 '15 at 18:06
k170
7,45131540
answered Jan 5 '15 at 18:06
k170
7,45131540
answered Jan 5 '15 at 18:06
k170
7,45131540
7,45131540
add a comment |
add a comment |
$ frac{d}{dx} a^x=$
$=d/dx$ $e^{ln (a^x)}=$
$=d/dx e^{xln(a)}$ just as you said.
Now it's time for chain rule.
$f(x)=e^x$
$g(x)=x*ln(a)$
$a^x= f(g(x))$
last comment before I get solving
$a=e^{ln(a)}$
now let's get it done
$d/dx e^{x ln(a)}=$
$=e^{x*ln(a)} d/dx (x ln(a))=$ (by chain rule)
$=e^{x ln(a)} * ln(a)$
and that is the solution. Wait it doesnt look correct --> we should do some algebra!
Show for yourself that
$e^{x*ln(a)}=( e^{ln(a)} )^x$ but as we said that $a = e^{ln(a)}$ so $( e^{ln(a)} )^x$ actually equals $a^x$ and of course $e^{x*ln(a)}$ equals a^x
With what we just said you can see that $e^{x * ln(a)} ln(a)$ equals $a^x ln(a)$ or
$$d/dx a^x = a^x ln(a)$$
edited Jan 9 '15 at 20:35
Community♦
1
answered Jan 5 '15 at 17:29
grekiki
1
1
Use TeX while typing math symbols
– supremum
Jan 5 '15 at 17:59
For help, see this FAQ entry about $LaTeX$ on Math.SE.
– Ruslan
Jan 5 '15 at 18:43
add a comment |
$ frac{d}{dx} a^x=$
$=d/dx$ $e^{ln (a^x)}=$
$=d/dx e^{xln(a)}$ just as you said.
Now it's time for chain rule.
$f(x)=e^x$
$g(x)=x*ln(a)$
$a^x= f(g(x))$
last comment before I get solving
$a=e^{ln(a)}$
now let's get it done
$d/dx e^{x ln(a)}=$
$=e^{x*ln(a)} d/dx (x ln(a))=$ (by chain rule)
$=e^{x ln(a)} * ln(a)$
and that is the solution. Wait it doesnt look correct --> we should do some algebra!
Show for yourself that
$e^{x*ln(a)}=( e^{ln(a)} )^x$ but as we said that $a = e^{ln(a)}$ so $( e^{ln(a)} )^x$ actually equals $a^x$ and of course $e^{x*ln(a)}$ equals a^x
With what we just said you can see that $e^{x * ln(a)} ln(a)$ equals $a^x ln(a)$ or
$$d/dx a^x = a^x ln(a)$$
edited Jan 9 '15 at 20:35
Community♦
1
answered Jan 5 '15 at 17:29
grekiki
1
1
Use TeX while typing math symbols
– supremum
Jan 5 '15 at 17:59
For help, see this FAQ entry about $LaTeX$ on Math.SE.
– Ruslan
Jan 5 '15 at 18:43
add a comment |
$ frac{d}{dx} a^x=$
$=d/dx$ $e^{ln (a^x)}=$
$=d/dx e^{xln(a)}$ just as you said.
Now it's time for chain rule.
$f(x)=e^x$
$g(x)=x*ln(a)$
$a^x= f(g(x))$
last comment before I get solving
$a=e^{ln(a)}$
now let's get it done
$d/dx e^{x ln(a)}=$
$=e^{x*ln(a)} d/dx (x ln(a))=$ (by chain rule)
$=e^{x ln(a)} * ln(a)$
and that is the solution. Wait it doesnt look correct --> we should do some algebra!
Show for yourself that
$e^{x*ln(a)}=( e^{ln(a)} )^x$ but as we said that $a = e^{ln(a)}$ so $( e^{ln(a)} )^x$ actually equals $a^x$ and of course $e^{x*ln(a)}$ equals a^x
With what we just said you can see that $e^{x * ln(a)} ln(a)$ equals $a^x ln(a)$ or
$$d/dx a^x = a^x ln(a)$$
edited Jan 9 '15 at 20:35
Community♦
1
answered Jan 5 '15 at 17:29
grekiki
1
$ frac{d}{dx} a^x=$
$=d/dx$ $e^{ln (a^x)}=$
$=d/dx e^{xln(a)}$ just as you said.
Now it's time for chain rule.
$f(x)=e^x$
$g(x)=x*ln(a)$
$a^x= f(g(x))$
last comment before I get solving
$a=e^{ln(a)}$
now let's get it done
$d/dx e^{x ln(a)}=$
$=e^{x*ln(a)} d/dx (x ln(a))=$ (by chain rule)
$=e^{x ln(a)} * ln(a)$
and that is the solution. Wait it doesnt look correct --> we should do some algebra!
Show for yourself that
$e^{x*ln(a)}=( e^{ln(a)} )^x$ but as we said that $a = e^{ln(a)}$ so $( e^{ln(a)} )^x$ actually equals $a^x$ and of course $e^{x*ln(a)}$ equals a^x
With what we just said you can see that $e^{x * ln(a)} ln(a)$ equals $a^x ln(a)$ or
$$d/dx a^x = a^x ln(a)$$
edited Jan 9 '15 at 20:35
Community♦
1
answered Jan 5 '15 at 17:29
grekiki
1
edited Jan 9 '15 at 20:35
Community♦
1
edited Jan 9 '15 at 20:35
Community♦
1
edited Jan 9 '15 at 20:35
Community♦
1
1
answered Jan 5 '15 at 17:29
grekiki
1
answered Jan 5 '15 at 17:29
grekiki
1
answered Jan 5 '15 at 17:29
grekiki
1
1
1
Use TeX while typing math symbols
– supremum
Jan 5 '15 at 17:59
For help, see this FAQ entry about $LaTeX$ on Math.SE.
– Ruslan
Jan 5 '15 at 18:43
add a comment |
1
Use TeX while typing math symbols
– supremum
Jan 5 '15 at 17:59
For help, see this FAQ entry about $LaTeX$ on Math.SE.
– Ruslan
Jan 5 '15 at 18:43
1
1
Use TeX while typing math symbols
– supremum
Jan 5 '15 at 17:59
Use TeX while typing math symbols
– supremum
Jan 5 '15 at 17:59
For help, see this FAQ entry about $LaTeX$ on Math.SE.
– Ruslan
Jan 5 '15 at 18:43
For help, see this FAQ entry about $LaTeX$ on Math.SE.
– Ruslan
Jan 5 '15 at 18:43
add a comment |
Let $y = a^x$
Then taking log on both side to the base $e$
We have;
$ ln y = ln a^x$
$ ln y = xcdot ln a$
Taking derivative with respect to $x$;
$frac d{dx} ln y = frac d{dx} xcdot ln a$
(I have applied chain rule here>>)
$ frac 1y frac{dy}{dx} = ln a + 0$
$ frac {dy}{dx} = ycdot ln a$
We know $y= a^x$
So, $frac {dy}{dx} = a^x ln a$
Hope it was easier than other methodologies used to derive it!
edited Nov 21 '18 at 9:46
idea
2,15441025
answered Nov 21 '18 at 9:17
Vichitra Attri
11
The question is five years old and your answer adds nothing new to the already existing answers.
– José Carlos Santos
Nov 21 '18 at 9:39
add a comment |
Let $y = a^x$
Then taking log on both side to the base $e$
We have;
$ ln y = ln a^x$
$ ln y = xcdot ln a$
Taking derivative with respect to $x$;
$frac d{dx} ln y = frac d{dx} xcdot ln a$
(I have applied chain rule here>>)
$ frac 1y frac{dy}{dx} = ln a + 0$
$ frac {dy}{dx} = ycdot ln a$
We know $y= a^x$
So, $frac {dy}{dx} = a^x ln a$
Hope it was easier than other methodologies used to derive it!
edited Nov 21 '18 at 9:46
idea
2,15441025
answered Nov 21 '18 at 9:17
Vichitra Attri
11
The question is five years old and your answer adds nothing new to the already existing answers.
– José Carlos Santos
Nov 21 '18 at 9:39
add a comment |
Let $y = a^x$
Then taking log on both side to the base $e$
We have;
$ ln y = ln a^x$
$ ln y = xcdot ln a$
Taking derivative with respect to $x$;
$frac d{dx} ln y = frac d{dx} xcdot ln a$
(I have applied chain rule here>>)
$ frac 1y frac{dy}{dx} = ln a + 0$
$ frac {dy}{dx} = ycdot ln a$
We know $y= a^x$
So, $frac {dy}{dx} = a^x ln a$
Hope it was easier than other methodologies used to derive it!
edited Nov 21 '18 at 9:46
idea
2,15441025
answered Nov 21 '18 at 9:17
Vichitra Attri
11
Let $y = a^x$
Then taking log on both side to the base $e$
We have;
$ ln y = ln a^x$
$ ln y = xcdot ln a$
Taking derivative with respect to $x$;
$frac d{dx} ln y = frac d{dx} xcdot ln a$
(I have applied chain rule here>>)
$ frac 1y frac{dy}{dx} = ln a + 0$
$ frac {dy}{dx} = ycdot ln a$
We know $y= a^x$
So, $frac {dy}{dx} = a^x ln a$
Hope it was easier than other methodologies used to derive it!
edited Nov 21 '18 at 9:46
idea
2,15441025
answered Nov 21 '18 at 9:17
Vichitra Attri
11
edited Nov 21 '18 at 9:46
idea
2,15441025
edited Nov 21 '18 at 9:46
idea
2,15441025
edited Nov 21 '18 at 9:46
idea
2,15441025
2,15441025
answered Nov 21 '18 at 9:17
Vichitra Attri
11
answered Nov 21 '18 at 9:17
Vichitra Attri
11
answered Nov 21 '18 at 9:17
Vichitra Attri
11
11
The question is five years old and your answer adds nothing new to the already existing answers.
– José Carlos Santos
Nov 21 '18 at 9:39
add a comment |
The question is five years old and your answer adds nothing new to the already existing answers.
– José Carlos Santos
Nov 21 '18 at 9:39
The question is five years old and your answer adds nothing new to the already existing answers.
– José Carlos Santos
Nov 21 '18 at 9:39
The question is five years old and your answer adds nothing new to the already existing answers.
– José Carlos Santos
Nov 21 '18 at 9:39
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.
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%2f398826%2fshow-that-d-dx-ax-ax-ln-a%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
