Why can't I differentiate $ x^{sin x} $ using the power rule?
$begingroup$
I'm trying to differentiate $ x^{sin x} $, with respect to $ x $ and $x > 0 $. My textbook initiates with $ y = x^{sin x} $, takes logarithms on both sides and arrives at the answer $$ x^{sin x - 1}.sin x + x^{sin x}.cos x log x $$
Why can't I use the power rule to proceed like this: $ y' = (sin x) x^{sin x-1} cos x $.
Here, I've first differentiated $ x $ with respect to $ sin x $ and then I've differentiated $ sin x $ with respect to $ x $ to get $ cos x $.
derivatives
asked Dec 25 '18 at 11:29
WorldGovWorldGov
324111
$endgroup$
add a comment |
$begingroup$
I'm trying to differentiate $ x^{sin x} $, with respect to $ x $ and $x > 0 $. My textbook initiates with $ y = x^{sin x} $, takes logarithms on both sides and arrives at the answer $$ x^{sin x - 1}.sin x + x^{sin x}.cos x log x $$
Why can't I use the power rule to proceed like this: $ y' = (sin x) x^{sin x-1} cos x $.
Here, I've first differentiated $ x $ with respect to $ sin x $ and then I've differentiated $ sin x $ with respect to $ x $ to get $ cos x $.
derivatives
asked Dec 25 '18 at 11:29
WorldGovWorldGov
324111
$endgroup$
add a comment |
$begingroup$
I'm trying to differentiate $ x^{sin x} $, with respect to $ x $ and $x > 0 $. My textbook initiates with $ y = x^{sin x} $, takes logarithms on both sides and arrives at the answer $$ x^{sin x - 1}.sin x + x^{sin x}.cos x log x $$
Why can't I use the power rule to proceed like this: $ y' = (sin x) x^{sin x-1} cos x $.
Here, I've first differentiated $ x $ with respect to $ sin x $ and then I've differentiated $ sin x $ with respect to $ x $ to get $ cos x $.
derivatives
asked Dec 25 '18 at 11:29
WorldGovWorldGov
324111
$endgroup$
I'm trying to differentiate $ x^{sin x} $, with respect to $ x $ and $x > 0 $. My textbook initiates with $ y = x^{sin x} $, takes logarithms on both sides and arrives at the answer $$ x^{sin x - 1}.sin x + x^{sin x}.cos x log x $$
Why can't I use the power rule to proceed like this: $ y' = (sin x) x^{sin x-1} cos x $.
Here, I've first differentiated $ x $ with respect to $ sin x $ and then I've differentiated $ sin x $ with respect to $ x $ to get $ cos x $.
derivatives
derivatives
asked Dec 25 '18 at 11:29
WorldGovWorldGov
324111
asked Dec 25 '18 at 11:29
WorldGovWorldGov
324111
asked Dec 25 '18 at 11:29
WorldGovWorldGov
324111
asked Dec 25 '18 at 11:29
WorldGovWorldGov
324111
asked Dec 25 '18 at 11:29
WorldGovWorldGov
324111
324111
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The power rule states that the derivative of a function of the form $x^k$ is $kx^{k-1}$ where $k$ is a real number. In your case, $sin x$ is variable, and can takes values from $-1$ to $+1$, so is not constant.
To prove the power rule, note that $y=x^kimplies ln y=kln ximplies frac{y'}y=frac kxcdot$ via the chain rule. Therefore, $y'=x^kfrac kx=kx^{k-1}$. When $K=sin x$, we need to do some more to manipulate the RHS, since $(sin xln x)'ne frac{sin x}x$.
answered Dec 25 '18 at 11:30
TheSimpliFireTheSimpliFire
12.7k62461
$endgroup$
$begingroup$
Does that proof hold for $k lt 0$?
$endgroup$
– DavidG
Jan 6 at 11:44
$begingroup$
Sorry, my apologies, I meant to type what if $x^k lt 0$ - can you still take the logarithm? If not, does this then only hold for $x gt 0$ ?
$endgroup$
– DavidG
Jan 6 at 11:48
$begingroup$
No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
$endgroup$
– DavidG
Jan 6 at 11:50
$begingroup$
I agree, I'm not disputing that. Nor any I challenging that the derivative as you find holds for $x lt 0$. I'm just curious, in what you have presented in the solution - does it hold for all $x$? or is the reasoning (again as presented) only applicable for $x gt 0$?
$endgroup$
– DavidG
Jan 6 at 11:53
$begingroup$
@DavidG I said 'Yes' to your comment already (so it only holds for $x>0$) :P
$endgroup$
– TheSimpliFire
Jan 6 at 11:54
|
show 2 more comments
$begingroup$
$(x^k)'=kx^{k-1},$ where $k$ is constant.
In our case $sin{x}neq constant$ and we can not use this rule.
By the way,
$$left(x^{sin{x}}right)'=left(e^{sin{x}ln{x}}right)'=e^{sin{x}ln{x}}(sin{x}ln{x})'=...$$
Can you end it now?
answered Dec 25 '18 at 11:38
Michael RozenbergMichael Rozenberg
107k1895199
$endgroup$
add a comment |
$begingroup$
Because the power rule requires that the exponent be a constant. $sin x$ is most definitely not a constant.
answered Dec 25 '18 at 11:55
The_SympathizerThe_Sympathizer
7,7402246
$endgroup$
add a comment |
$begingroup$
You are trying to use the rule
$$frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$$
But this rule doesn't apply here, because $x^{sin x}$ is not just a function of $sin x$, it is a function of $x$ and $sin x$.
answered Dec 25 '18 at 11:39
TonyKTonyK
43k356135
$endgroup$
$begingroup$
Sometimes we can write $x=arcsinsin{x}$ and we'll obtain the function of $sin{x}.$ :)
$endgroup$
– Michael Rozenberg
Dec 25 '18 at 11:49
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The power rule states that the derivative of a function of the form $x^k$ is $kx^{k-1}$ where $k$ is a real number. In your case, $sin x$ is variable, and can takes values from $-1$ to $+1$, so is not constant.
To prove the power rule, note that $y=x^kimplies ln y=kln ximplies frac{y'}y=frac kxcdot$ via the chain rule. Therefore, $y'=x^kfrac kx=kx^{k-1}$. When $K=sin x$, we need to do some more to manipulate the RHS, since $(sin xln x)'ne frac{sin x}x$.
answered Dec 25 '18 at 11:30
TheSimpliFireTheSimpliFire
12.7k62461
$endgroup$
$begingroup$
Does that proof hold for $k lt 0$?
$endgroup$
– DavidG
Jan 6 at 11:44
$begingroup$
Sorry, my apologies, I meant to type what if $x^k lt 0$ - can you still take the logarithm? If not, does this then only hold for $x gt 0$ ?
$endgroup$
– DavidG
Jan 6 at 11:48
$begingroup$
No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
$endgroup$
– DavidG
Jan 6 at 11:50
$begingroup$
I agree, I'm not disputing that. Nor any I challenging that the derivative as you find holds for $x lt 0$. I'm just curious, in what you have presented in the solution - does it hold for all $x$? or is the reasoning (again as presented) only applicable for $x gt 0$?
$endgroup$
– DavidG
Jan 6 at 11:53
$begingroup$
@DavidG I said 'Yes' to your comment already (so it only holds for $x>0$) :P
$endgroup$
– TheSimpliFire
Jan 6 at 11:54
|
show 2 more comments
$begingroup$
The power rule states that the derivative of a function of the form $x^k$ is $kx^{k-1}$ where $k$ is a real number. In your case, $sin x$ is variable, and can takes values from $-1$ to $+1$, so is not constant.
To prove the power rule, note that $y=x^kimplies ln y=kln ximplies frac{y'}y=frac kxcdot$ via the chain rule. Therefore, $y'=x^kfrac kx=kx^{k-1}$. When $K=sin x$, we need to do some more to manipulate the RHS, since $(sin xln x)'ne frac{sin x}x$.
answered Dec 25 '18 at 11:30
TheSimpliFireTheSimpliFire
12.7k62461
$endgroup$
$begingroup$
Does that proof hold for $k lt 0$?
$endgroup$
– DavidG
Jan 6 at 11:44
$begingroup$
Sorry, my apologies, I meant to type what if $x^k lt 0$ - can you still take the logarithm? If not, does this then only hold for $x gt 0$ ?
$endgroup$
– DavidG
Jan 6 at 11:48
$begingroup$
No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
$endgroup$
– DavidG
Jan 6 at 11:50
$begingroup$
I agree, I'm not disputing that. Nor any I challenging that the derivative as you find holds for $x lt 0$. I'm just curious, in what you have presented in the solution - does it hold for all $x$? or is the reasoning (again as presented) only applicable for $x gt 0$?
$endgroup$
– DavidG
Jan 6 at 11:53
$begingroup$
@DavidG I said 'Yes' to your comment already (so it only holds for $x>0$) :P
$endgroup$
– TheSimpliFire
Jan 6 at 11:54
|
show 2 more comments
$begingroup$
The power rule states that the derivative of a function of the form $x^k$ is $kx^{k-1}$ where $k$ is a real number. In your case, $sin x$ is variable, and can takes values from $-1$ to $+1$, so is not constant.
To prove the power rule, note that $y=x^kimplies ln y=kln ximplies frac{y'}y=frac kxcdot$ via the chain rule. Therefore, $y'=x^kfrac kx=kx^{k-1}$. When $K=sin x$, we need to do some more to manipulate the RHS, since $(sin xln x)'ne frac{sin x}x$.
answered Dec 25 '18 at 11:30
TheSimpliFireTheSimpliFire
12.7k62461
$endgroup$
The power rule states that the derivative of a function of the form $x^k$ is $kx^{k-1}$ where $k$ is a real number. In your case, $sin x$ is variable, and can takes values from $-1$ to $+1$, so is not constant.
To prove the power rule, note that $y=x^kimplies ln y=kln ximplies frac{y'}y=frac kxcdot$ via the chain rule. Therefore, $y'=x^kfrac kx=kx^{k-1}$. When $K=sin x$, we need to do some more to manipulate the RHS, since $(sin xln x)'ne frac{sin x}x$.
answered Dec 25 '18 at 11:30
TheSimpliFireTheSimpliFire
12.7k62461
answered Dec 25 '18 at 11:30
TheSimpliFireTheSimpliFire
12.7k62461
answered Dec 25 '18 at 11:30
TheSimpliFireTheSimpliFire
12.7k62461
answered Dec 25 '18 at 11:30
TheSimpliFireTheSimpliFire
12.7k62461
12.7k62461
$begingroup$
Does that proof hold for $k lt 0$?
$endgroup$
– DavidG
Jan 6 at 11:44
$begingroup$
Sorry, my apologies, I meant to type what if $x^k lt 0$ - can you still take the logarithm? If not, does this then only hold for $x gt 0$ ?
$endgroup$
– DavidG
Jan 6 at 11:48
$begingroup$
No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
$endgroup$
– DavidG
Jan 6 at 11:50
$begingroup$
I agree, I'm not disputing that. Nor any I challenging that the derivative as you find holds for $x lt 0$. I'm just curious, in what you have presented in the solution - does it hold for all $x$? or is the reasoning (again as presented) only applicable for $x gt 0$?
$endgroup$
– DavidG
Jan 6 at 11:53
$begingroup$
@DavidG I said 'Yes' to your comment already (so it only holds for $x>0$) :P
$endgroup$
– TheSimpliFire
Jan 6 at 11:54
|
show 2 more comments
$begingroup$
Does that proof hold for $k lt 0$?
$endgroup$
– DavidG
Jan 6 at 11:44
$begingroup$
Sorry, my apologies, I meant to type what if $x^k lt 0$ - can you still take the logarithm? If not, does this then only hold for $x gt 0$ ?
$endgroup$
– DavidG
Jan 6 at 11:48
$begingroup$
No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
$endgroup$
– DavidG
Jan 6 at 11:50
$begingroup$
I agree, I'm not disputing that. Nor any I challenging that the derivative as you find holds for $x lt 0$. I'm just curious, in what you have presented in the solution - does it hold for all $x$? or is the reasoning (again as presented) only applicable for $x gt 0$?
$endgroup$
– DavidG
Jan 6 at 11:53
$begingroup$
@DavidG I said 'Yes' to your comment already (so it only holds for $x>0$) :P
$endgroup$
– TheSimpliFire
Jan 6 at 11:54
$begingroup$
Does that proof hold for $k lt 0$?
$endgroup$
– DavidG
Jan 6 at 11:44
$begingroup$
Does that proof hold for $k lt 0$?
$endgroup$
– DavidG
Jan 6 at 11:44
$begingroup$
Sorry, my apologies, I meant to type what if $x^k lt 0$ - can you still take the logarithm? If not, does this then only hold for $x gt 0$ ?
$endgroup$
– DavidG
Jan 6 at 11:48
$begingroup$
Sorry, my apologies, I meant to type what if $x^k lt 0$ - can you still take the logarithm? If not, does this then only hold for $x gt 0$ ?
$endgroup$
– DavidG
Jan 6 at 11:48
$begingroup$
No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
$endgroup$
– DavidG
Jan 6 at 11:50
$begingroup$
No, I appreciate that - but just for clarity - this method as presented only holds for $x gt 0$?
$endgroup$
– DavidG
Jan 6 at 11:50
$begingroup$
I agree, I'm not disputing that. Nor any I challenging that the derivative as you find holds for $x lt 0$. I'm just curious, in what you have presented in the solution - does it hold for all $x$? or is the reasoning (again as presented) only applicable for $x gt 0$?
$endgroup$
– DavidG
Jan 6 at 11:53
$begingroup$
I agree, I'm not disputing that. Nor any I challenging that the derivative as you find holds for $x lt 0$. I'm just curious, in what you have presented in the solution - does it hold for all $x$? or is the reasoning (again as presented) only applicable for $x gt 0$?
$endgroup$
– DavidG
Jan 6 at 11:53
$begingroup$
@DavidG I said 'Yes' to your comment already (so it only holds for $x>0$) :P
$endgroup$
– TheSimpliFire
Jan 6 at 11:54
$begingroup$
@DavidG I said 'Yes' to your comment already (so it only holds for $x>0$) :P
$endgroup$
– TheSimpliFire
Jan 6 at 11:54
|
show 2 more comments
$begingroup$
$(x^k)'=kx^{k-1},$ where $k$ is constant.
In our case $sin{x}neq constant$ and we can not use this rule.
By the way,
$$left(x^{sin{x}}right)'=left(e^{sin{x}ln{x}}right)'=e^{sin{x}ln{x}}(sin{x}ln{x})'=...$$
Can you end it now?
answered Dec 25 '18 at 11:38
Michael RozenbergMichael Rozenberg
107k1895199
$endgroup$
add a comment |
$begingroup$
$(x^k)'=kx^{k-1},$ where $k$ is constant.
In our case $sin{x}neq constant$ and we can not use this rule.
By the way,
$$left(x^{sin{x}}right)'=left(e^{sin{x}ln{x}}right)'=e^{sin{x}ln{x}}(sin{x}ln{x})'=...$$
Can you end it now?
answered Dec 25 '18 at 11:38
Michael RozenbergMichael Rozenberg
107k1895199
$endgroup$
add a comment |
$begingroup$
$(x^k)'=kx^{k-1},$ where $k$ is constant.
In our case $sin{x}neq constant$ and we can not use this rule.
By the way,
$$left(x^{sin{x}}right)'=left(e^{sin{x}ln{x}}right)'=e^{sin{x}ln{x}}(sin{x}ln{x})'=...$$
Can you end it now?
answered Dec 25 '18 at 11:38
Michael RozenbergMichael Rozenberg
107k1895199
$endgroup$
$(x^k)'=kx^{k-1},$ where $k$ is constant.
In our case $sin{x}neq constant$ and we can not use this rule.
By the way,
$$left(x^{sin{x}}right)'=left(e^{sin{x}ln{x}}right)'=e^{sin{x}ln{x}}(sin{x}ln{x})'=...$$
Can you end it now?
answered Dec 25 '18 at 11:38
Michael RozenbergMichael Rozenberg
107k1895199
edited Dec 25 '18 at 11:45
answered Dec 25 '18 at 11:38
Michael RozenbergMichael Rozenberg
107k1895199
answered Dec 25 '18 at 11:38
Michael RozenbergMichael Rozenberg
107k1895199
answered Dec 25 '18 at 11:38
Michael RozenbergMichael Rozenberg
107k1895199
107k1895199
add a comment |
add a comment |
$begingroup$
Because the power rule requires that the exponent be a constant. $sin x$ is most definitely not a constant.
answered Dec 25 '18 at 11:55
The_SympathizerThe_Sympathizer
7,7402246
$endgroup$
add a comment |
$begingroup$
Because the power rule requires that the exponent be a constant. $sin x$ is most definitely not a constant.
answered Dec 25 '18 at 11:55
The_SympathizerThe_Sympathizer
7,7402246
$endgroup$
add a comment |
$begingroup$
Because the power rule requires that the exponent be a constant. $sin x$ is most definitely not a constant.
answered Dec 25 '18 at 11:55
The_SympathizerThe_Sympathizer
7,7402246
$endgroup$
Because the power rule requires that the exponent be a constant. $sin x$ is most definitely not a constant.
answered Dec 25 '18 at 11:55
The_SympathizerThe_Sympathizer
7,7402246
answered Dec 25 '18 at 11:55
The_SympathizerThe_Sympathizer
7,7402246
answered Dec 25 '18 at 11:55
The_SympathizerThe_Sympathizer
7,7402246
answered Dec 25 '18 at 11:55
The_SympathizerThe_Sympathizer
7,7402246
7,7402246
add a comment |
add a comment |
$begingroup$
You are trying to use the rule
$$frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$$
But this rule doesn't apply here, because $x^{sin x}$ is not just a function of $sin x$, it is a function of $x$ and $sin x$.
answered Dec 25 '18 at 11:39
TonyKTonyK
43k356135
$endgroup$
$begingroup$
Sometimes we can write $x=arcsinsin{x}$ and we'll obtain the function of $sin{x}.$ :)
$endgroup$
– Michael Rozenberg
Dec 25 '18 at 11:49
add a comment |
$begingroup$
You are trying to use the rule
$$frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$$
But this rule doesn't apply here, because $x^{sin x}$ is not just a function of $sin x$, it is a function of $x$ and $sin x$.
answered Dec 25 '18 at 11:39
TonyKTonyK
43k356135
$endgroup$
$begingroup$
Sometimes we can write $x=arcsinsin{x}$ and we'll obtain the function of $sin{x}.$ :)
$endgroup$
– Michael Rozenberg
Dec 25 '18 at 11:49
add a comment |
$begingroup$
You are trying to use the rule
$$frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$$
But this rule doesn't apply here, because $x^{sin x}$ is not just a function of $sin x$, it is a function of $x$ and $sin x$.
answered Dec 25 '18 at 11:39
TonyKTonyK
43k356135
$endgroup$
You are trying to use the rule
$$frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$$
But this rule doesn't apply here, because $x^{sin x}$ is not just a function of $sin x$, it is a function of $x$ and $sin x$.
answered Dec 25 '18 at 11:39
TonyKTonyK
43k356135
answered Dec 25 '18 at 11:39
TonyKTonyK
43k356135
answered Dec 25 '18 at 11:39
TonyKTonyK
43k356135
answered Dec 25 '18 at 11:39
TonyKTonyK
43k356135
43k356135
$begingroup$
Sometimes we can write $x=arcsinsin{x}$ and we'll obtain the function of $sin{x}.$ :)
$endgroup$
– Michael Rozenberg
Dec 25 '18 at 11:49
add a comment |
$begingroup$
Sometimes we can write $x=arcsinsin{x}$ and we'll obtain the function of $sin{x}.$ :)
$endgroup$
– Michael Rozenberg
Dec 25 '18 at 11:49
$begingroup$
Sometimes we can write $x=arcsinsin{x}$ and we'll obtain the function of $sin{x}.$ :)
$endgroup$
– Michael Rozenberg
Dec 25 '18 at 11:49
$begingroup$
Sometimes we can write $x=arcsinsin{x}$ and we'll obtain the function of $sin{x}.$ :)
$endgroup$
– Michael Rozenberg
Dec 25 '18 at 11:49
add a comment |
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