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Calculating $u_t$, $u_x$, and $u_xx$ for $u(x, t) = -2 dfracpartialpartialxlog(phi(x,t))$
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I am trying to calculate $u_t$, $u_x$, and $u_xx$ for $u(x, t) = -2 dfracpartialpartialxlog(phi(x,t))$.
I've been trying for hours, but I've become so confused with the chain rule here that I don't know what to do.
Can someone please demonstrate how to correctly calculate these? Please show your calculations so that I can learn how I'm supposed to be applying the chain rule.
EDIT:
Thank you for all posting great answers. Here is my attempt without using simplification
$$u(x,t) = -2overbracefracpartialphi(x,t)partial x^;;textderivative of\textargument of logunderbracefrac1phi(x,t)_textderivative of log(x)\ textwith argument phi(x,t)$$
at the beginning.
$$u_x = -2 left[ fracpartialpartial x left( fracpartialpartial x (log(phi(x, t)) right) right] $$
$$= -2 left[ fracpartialpartial(log(phi)) left( fracpartialpartial x (log(phi(x, t)) right) right] times left[ fracpartialpartial(phi) (log(phi(x, t)) right] times fracpartialphipartialx$$
(By the chain rule.)
$$= -2 left[ fracpartialpartialx left[ left( frac1phi right) fracpartialphipartialx right] times frac1phi times fracpartialphipartialx right]$$
$$= -2 left[ fracpartialpartialx left[ left( frac1phi right) fracpartialphipartialx right] times frac1phi times fracpartialphipartialx right]$$
$$= -2 left[ left( -phi^-2(x, t) times fracpartialphipartialx + fracpartial^2phipartialx^2 times frac1phi right) times frac1phi times fracpartialphipartialx right]$$
Feedback on this attempt is appreciated.
real-analysis multivariable-calculus pde partial-derivative chain-rule
asked Aug 2 at 20:56
handler's handle
1277
add a comment |Â
up vote
1
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I am trying to calculate $u_t$, $u_x$, and $u_xx$ for $u(x, t) = -2 dfracpartialpartialxlog(phi(x,t))$.
I've been trying for hours, but I've become so confused with the chain rule here that I don't know what to do.
Can someone please demonstrate how to correctly calculate these? Please show your calculations so that I can learn how I'm supposed to be applying the chain rule.
EDIT:
Thank you for all posting great answers. Here is my attempt without using simplification
$$u(x,t) = -2overbracefracpartialphi(x,t)partial x^;;textderivative of\textargument of logunderbracefrac1phi(x,t)_textderivative of log(x)\ textwith argument phi(x,t)$$
at the beginning.
$$u_x = -2 left[ fracpartialpartial x left( fracpartialpartial x (log(phi(x, t)) right) right] $$
$$= -2 left[ fracpartialpartial(log(phi)) left( fracpartialpartial x (log(phi(x, t)) right) right] times left[ fracpartialpartial(phi) (log(phi(x, t)) right] times fracpartialphipartialx$$
(By the chain rule.)
$$= -2 left[ fracpartialpartialx left[ left( frac1phi right) fracpartialphipartialx right] times frac1phi times fracpartialphipartialx right]$$
$$= -2 left[ fracpartialpartialx left[ left( frac1phi right) fracpartialphipartialx right] times frac1phi times fracpartialphipartialx right]$$
$$= -2 left[ left( -phi^-2(x, t) times fracpartialphipartialx + fracpartial^2phipartialx^2 times frac1phi right) times frac1phi times fracpartialphipartialx right]$$
Feedback on this attempt is appreciated.
real-analysis multivariable-calculus pde partial-derivative chain-rule
asked Aug 2 at 20:56
handler's handle
1277
I don't quite understand why that $$fracpartialpartial x$$ became a $$fracpartialpartial log(phi)$$ if we use the chain rule we have that $$fracpartialpartial x = fracpartialpartial log(phi)fracpartiallog(phi)partial x$$ and this first equality doesn't touch the derivative that follows it. You could rewrite even that other partial derivative in the same way but it seems pretty unnecessary
– Davide Morgante
Aug 3 at 8:18
@DavideMorgante I apologise for doing it so badly. Can you please demonstrate this as an example with explanations? That way, I will get a better idea as to how to do it and what I am doing wrong. I will then try to do it for $u_xx$ and $u_t$ and post it.
– handler's handle
Aug 3 at 9:24
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to calculate $u_t$, $u_x$, and $u_xx$ for $u(x, t) = -2 dfracpartialpartialxlog(phi(x,t))$.
I've been trying for hours, but I've become so confused with the chain rule here that I don't know what to do.
Can someone please demonstrate how to correctly calculate these? Please show your calculations so that I can learn how I'm supposed to be applying the chain rule.
EDIT:
Thank you for all posting great answers. Here is my attempt without using simplification
$$u(x,t) = -2overbracefracpartialphi(x,t)partial x^;;textderivative of\textargument of logunderbracefrac1phi(x,t)_textderivative of log(x)\ textwith argument phi(x,t)$$
at the beginning.
$$u_x = -2 left[ fracpartialpartial x left( fracpartialpartial x (log(phi(x, t)) right) right] $$
$$= -2 left[ fracpartialpartial(log(phi)) left( fracpartialpartial x (log(phi(x, t)) right) right] times left[ fracpartialpartial(phi) (log(phi(x, t)) right] times fracpartialphipartialx$$
(By the chain rule.)
$$= -2 left[ fracpartialpartialx left[ left( frac1phi right) fracpartialphipartialx right] times frac1phi times fracpartialphipartialx right]$$
$$= -2 left[ fracpartialpartialx left[ left( frac1phi right) fracpartialphipartialx right] times frac1phi times fracpartialphipartialx right]$$
$$= -2 left[ left( -phi^-2(x, t) times fracpartialphipartialx + fracpartial^2phipartialx^2 times frac1phi right) times frac1phi times fracpartialphipartialx right]$$
Feedback on this attempt is appreciated.
real-analysis multivariable-calculus pde partial-derivative chain-rule
asked Aug 2 at 20:56
handler's handle
1277
I am trying to calculate $u_t$, $u_x$, and $u_xx$ for $u(x, t) = -2 dfracpartialpartialxlog(phi(x,t))$.
I've been trying for hours, but I've become so confused with the chain rule here that I don't know what to do.
Can someone please demonstrate how to correctly calculate these? Please show your calculations so that I can learn how I'm supposed to be applying the chain rule.
EDIT:
Thank you for all posting great answers. Here is my attempt without using simplification
$$u(x,t) = -2overbracefracpartialphi(x,t)partial x^;;textderivative of\textargument of logunderbracefrac1phi(x,t)_textderivative of log(x)\ textwith argument phi(x,t)$$
at the beginning.
$$u_x = -2 left[ fracpartialpartial x left( fracpartialpartial x (log(phi(x, t)) right) right] $$
$$= -2 left[ fracpartialpartial(log(phi)) left( fracpartialpartial x (log(phi(x, t)) right) right] times left[ fracpartialpartial(phi) (log(phi(x, t)) right] times fracpartialphipartialx$$
(By the chain rule.)
$$= -2 left[ fracpartialpartialx left[ left( frac1phi right) fracpartialphipartialx right] times frac1phi times fracpartialphipartialx right]$$
$$= -2 left[ fracpartialpartialx left[ left( frac1phi right) fracpartialphipartialx right] times frac1phi times fracpartialphipartialx right]$$
$$= -2 left[ left( -phi^-2(x, t) times fracpartialphipartialx + fracpartial^2phipartialx^2 times frac1phi right) times frac1phi times fracpartialphipartialx right]$$
Feedback on this attempt is appreciated.
real-analysis multivariable-calculus pde partial-derivative chain-rule
asked Aug 2 at 20:56
handler's handle
1277
edited Aug 3 at 0:15
asked Aug 2 at 20:56
handler's handle
1277
asked Aug 2 at 20:56
handler's handle
1277
asked Aug 2 at 20:56
handler's handle
1277
1277
I don't quite understand why that $$fracpartialpartial x$$ became a $$fracpartialpartial log(phi)$$ if we use the chain rule we have that $$fracpartialpartial x = fracpartialpartial log(phi)fracpartiallog(phi)partial x$$ and this first equality doesn't touch the derivative that follows it. You could rewrite even that other partial derivative in the same way but it seems pretty unnecessary
– Davide Morgante
Aug 3 at 8:18
@DavideMorgante I apologise for doing it so badly. Can you please demonstrate this as an example with explanations? That way, I will get a better idea as to how to do it and what I am doing wrong. I will then try to do it for $u_xx$ and $u_t$ and post it.
– handler's handle
Aug 3 at 9:24
add a comment |Â
I don't quite understand why that $$fracpartialpartial x$$ became a $$fracpartialpartial log(phi)$$ if we use the chain rule we have that $$fracpartialpartial x = fracpartialpartial log(phi)fracpartiallog(phi)partial x$$ and this first equality doesn't touch the derivative that follows it. You could rewrite even that other partial derivative in the same way but it seems pretty unnecessary
– Davide Morgante
Aug 3 at 8:18
@DavideMorgante I apologise for doing it so badly. Can you please demonstrate this as an example with explanations? That way, I will get a better idea as to how to do it and what I am doing wrong. I will then try to do it for $u_xx$ and $u_t$ and post it.
– handler's handle
Aug 3 at 9:24
I don't quite understand why that $$fracpartialpartial x$$ became a $$fracpartialpartial log(phi)$$ if we use the chain rule we have that $$fracpartialpartial x = fracpartialpartial log(phi)fracpartiallog(phi)partial x$$ and this first equality doesn't touch the derivative that follows it. You could rewrite even that other partial derivative in the same way but it seems pretty unnecessary
– Davide Morgante
Aug 3 at 8:18
I don't quite understand why that $$fracpartialpartial x$$ became a $$fracpartialpartial log(phi)$$ if we use the chain rule we have that $$fracpartialpartial x = fracpartialpartial log(phi)fracpartiallog(phi)partial x$$ and this first equality doesn't touch the derivative that follows it. You could rewrite even that other partial derivative in the same way but it seems pretty unnecessary
– Davide Morgante
Aug 3 at 8:18
@DavideMorgante I apologise for doing it so badly. Can you please demonstrate this as an example with explanations? That way, I will get a better idea as to how to do it and what I am doing wrong. I will then try to do it for $u_xx$ and $u_t$ and post it.
– handler's handle
Aug 3 at 9:24
@DavideMorgante I apologise for doing it so badly. Can you please demonstrate this as an example with explanations? That way, I will get a better idea as to how to do it and what I am doing wrong. I will then try to do it for $u_xx$ and $u_t$ and post it.
– handler's handle
Aug 3 at 9:24
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
You could do all the calculations without evaluating the first derivative in the function, but we'll do it anyway $$u(x,t) = -2overbracefracpartialphi(x,t)partial x^;;textderivative of\textargument of logunderbracefrac1phi(x,t)_textderivative of log(x)\ textwith argument phi(x,t)$$ Now you can see that the function $u(x,t)$ is the product of two functions! We can evaluate the other derivatives using the derivative of the product of functions, mainly
$$u_x(x,t) = fracpartialpartial xleft(-2fracpartialphi(x,t)partial xfrac1phi(x,t)right) = underbrace-2fracpartial^2phipartial x^2frac1phi_D[1^textst]times2^textnd+underbrace2fracpartial phipartial xfracpartialphipartial xfrac1phi^2_D[2^textnd]times1^textst=2left(frac1phifracpartial phipartial xright)^2 -2fracpartial^2phipartial x^2frac1phi\
u_t(x,t) =-2fracpartial^2 phipartial x partial tfrac1phi+2fracpartial phipartial xfracpartial phipartial tfrac1phi^2\
u_xx(x,t) = fracpartialpartial xleft(colorred2left(frac1phifracpartial phipartial xright)^2 colorblue-2fracpartial^2phipartial x^2frac1phiright) = \=colorred-4left(frac1phifracpartial phipartial xright)left(frac1phi^2fracpartialphipartial x+frac1phifracpartial^2phipartial x^2right)colorblue-2left(frac1phifracpartial^3phipartial x^3-frac1phi^2fracpartial phipartial xfracpartial^2phipartial x^2right)$$
all done by using the chain rule. I used coloring to simplify the visualization of the derivatives
answered Aug 2 at 21:26
Davide Morgante
1,632220
Thank you. Your answer is very clear. Can you please also show how it's done without simplifying the first derivative? I'm trying to do that using the chain rule and find it very confusing.
– handler's handle
Aug 2 at 22:32
Without simplifying it you mean without collecting the squares together?
– Davide Morgante
Aug 2 at 22:33
I meant without this simplification: $u(x,t) = -2overbracefracpartialphi(x,t)partial x^;;textderivative of\textargument of logunderbracefrac1phi(x,t)_textderivative of log(x)\ textwith argument phi(x,t)$
– handler's handle
Aug 2 at 22:34
If I am not mistaken, this makes our use of the chain rule more complex.
– handler's handle
Aug 2 at 22:34
Oh ok, I don't think I can put it all in one comment, maybe I can do one just to clarify your doubts? Where are you stuck?
– Davide Morgante
Aug 2 at 22:35
 |Â
show 4 more comments
up vote
0
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First simplify the right side by finding the derivative.
$$fracpartialpartial xlogphi(x,t) = frac1phi(x,t)fracpartial phipartial x$$
so
$$u(x,t) = -2fracphi_x(x,t)phi(x,t)$$
Then the derivatives of $u$ are straightforward, each one follows the quotient rule.
$$fracpartial upartial t = -2fracpartialpartial tleft(fracphi_x(x,t)phi(x,t)right) = -2fracphi(x,t)phi_xt(x,t)-phi_x(x,t)phi_t(x,t)phi^2(x,t)$$
$$fracpartial upartial x = -2fracphi(x,t)phi_xx(x,t) - phi^2_x(x,t)phi^2(x,t)$$
The last is a bit of a doozy though. Dropping the $(x,t)$ as $phi$ is understood to be a fuction of both $x$ and $t$,
$$fracpartial^2upartial x^2=-2fracphi^2(phi_xphi_xx + phiphi_xxx - 2phi_xphi_xx) - 2phiphi_x(phiphi_xx-phi^2_x)phi^4$$
answered Aug 2 at 21:32
Skip
1,147212
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$$u(x,t)~=~-2fracpartialpartial xlog(phi(x,t))$$
The chain rule states that
$$fracpartial f(u(x,t))partial x~=~fracpartial f(u(x,t))partial u(x,t)fracpartial u(x,t)partial x$$
And so the original function becomes
$$-2fracpartialpartial xlog(phi(x,t))~=~-2frac1phi(x,t)phi_x(x,t)~=~-2fracphi_x(x,t)phi(x,t)$$
So for the first case $u_t$ we consider the function $phi(x,t)$ as $u$ and $fracpartialpartial xlog(u)$ as $f$ so we get
$$beginalign
fracpartialpartial tleft(-2fracphi_x(x,t)phi(x,t)right)~&=~-2fracphi_xt(x,t)phi(x,t)-phi_t(x,t)phi_x(x,t)phi^2(x,t)\
fracpartialpartial xleft(-2fracphi_x(x,t)phi(x,t)right)~&=~-2fracphi_xx(x,t)phi(x,t)-phi_x^2(x,t)phi^2(x,t)\
fracpartialpartial xleft(-2fracphi_xx(x,t)phi(x,t)-phi_x^2(x,t)phi^2(x,t)right)~&=~frac[phi_xxxphi+phi_xphi_xx-2phi_xphi_xx]phi^2-2phiphi_x[phi_xxphi-phi_x^2]phi^4\
&=~frac[phi_xxxphi-phi_xphi_xx]phi-2phi_x[phi_xxphi-phi_x^2]phi^3\
&=~fracphi_xxxphi^2-3phi_xxphi_xphi+2phi_x^2phiphi^3
endalign$$
This work is kind of messy so I hope I did not made a mistake somewhere. I guess the crucial point is the to get rid of the partial derivative with respect to $x$ in the definition of the funtion $u$. I hope I made clear how it works out.
answered Aug 2 at 21:25
mrtaurho
607117
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
You could do all the calculations without evaluating the first derivative in the function, but we'll do it anyway $$u(x,t) = -2overbracefracpartialphi(x,t)partial x^;;textderivative of\textargument of logunderbracefrac1phi(x,t)_textderivative of log(x)\ textwith argument phi(x,t)$$ Now you can see that the function $u(x,t)$ is the product of two functions! We can evaluate the other derivatives using the derivative of the product of functions, mainly
$$u_x(x,t) = fracpartialpartial xleft(-2fracpartialphi(x,t)partial xfrac1phi(x,t)right) = underbrace-2fracpartial^2phipartial x^2frac1phi_D[1^textst]times2^textnd+underbrace2fracpartial phipartial xfracpartialphipartial xfrac1phi^2_D[2^textnd]times1^textst=2left(frac1phifracpartial phipartial xright)^2 -2fracpartial^2phipartial x^2frac1phi\
u_t(x,t) =-2fracpartial^2 phipartial x partial tfrac1phi+2fracpartial phipartial xfracpartial phipartial tfrac1phi^2\
u_xx(x,t) = fracpartialpartial xleft(colorred2left(frac1phifracpartial phipartial xright)^2 colorblue-2fracpartial^2phipartial x^2frac1phiright) = \=colorred-4left(frac1phifracpartial phipartial xright)left(frac1phi^2fracpartialphipartial x+frac1phifracpartial^2phipartial x^2right)colorblue-2left(frac1phifracpartial^3phipartial x^3-frac1phi^2fracpartial phipartial xfracpartial^2phipartial x^2right)$$
all done by using the chain rule. I used coloring to simplify the visualization of the derivatives
answered Aug 2 at 21:26
Davide Morgante
1,632220
Thank you. Your answer is very clear. Can you please also show how it's done without simplifying the first derivative? I'm trying to do that using the chain rule and find it very confusing.
– handler's handle
Aug 2 at 22:32
Without simplifying it you mean without collecting the squares together?
– Davide Morgante
Aug 2 at 22:33
I meant without this simplification: $u(x,t) = -2overbracefracpartialphi(x,t)partial x^;;textderivative of\textargument of logunderbracefrac1phi(x,t)_textderivative of log(x)\ textwith argument phi(x,t)$
– handler's handle
Aug 2 at 22:34
If I am not mistaken, this makes our use of the chain rule more complex.
– handler's handle
Aug 2 at 22:34
Oh ok, I don't think I can put it all in one comment, maybe I can do one just to clarify your doubts? Where are you stuck?
– Davide Morgante
Aug 2 at 22:35
 |Â
show 4 more comments
up vote
4
down vote
accepted
You could do all the calculations without evaluating the first derivative in the function, but we'll do it anyway $$u(x,t) = -2overbracefracpartialphi(x,t)partial x^;;textderivative of\textargument of logunderbracefrac1phi(x,t)_textderivative of log(x)\ textwith argument phi(x,t)$$ Now you can see that the function $u(x,t)$ is the product of two functions! We can evaluate the other derivatives using the derivative of the product of functions, mainly
$$u_x(x,t) = fracpartialpartial xleft(-2fracpartialphi(x,t)partial xfrac1phi(x,t)right) = underbrace-2fracpartial^2phipartial x^2frac1phi_D[1^textst]times2^textnd+underbrace2fracpartial phipartial xfracpartialphipartial xfrac1phi^2_D[2^textnd]times1^textst=2left(frac1phifracpartial phipartial xright)^2 -2fracpartial^2phipartial x^2frac1phi\
u_t(x,t) =-2fracpartial^2 phipartial x partial tfrac1phi+2fracpartial phipartial xfracpartial phipartial tfrac1phi^2\
u_xx(x,t) = fracpartialpartial xleft(colorred2left(frac1phifracpartial phipartial xright)^2 colorblue-2fracpartial^2phipartial x^2frac1phiright) = \=colorred-4left(frac1phifracpartial phipartial xright)left(frac1phi^2fracpartialphipartial x+frac1phifracpartial^2phipartial x^2right)colorblue-2left(frac1phifracpartial^3phipartial x^3-frac1phi^2fracpartial phipartial xfracpartial^2phipartial x^2right)$$
all done by using the chain rule. I used coloring to simplify the visualization of the derivatives
answered Aug 2 at 21:26
Davide Morgante
1,632220
Thank you. Your answer is very clear. Can you please also show how it's done without simplifying the first derivative? I'm trying to do that using the chain rule and find it very confusing.
– handler's handle
Aug 2 at 22:32
Without simplifying it you mean without collecting the squares together?
– Davide Morgante
Aug 2 at 22:33
I meant without this simplification: $u(x,t) = -2overbracefracpartialphi(x,t)partial x^;;textderivative of\textargument of logunderbracefrac1phi(x,t)_textderivative of log(x)\ textwith argument phi(x,t)$
– handler's handle
Aug 2 at 22:34
If I am not mistaken, this makes our use of the chain rule more complex.
– handler's handle
Aug 2 at 22:34
Oh ok, I don't think I can put it all in one comment, maybe I can do one just to clarify your doubts? Where are you stuck?
– Davide Morgante
Aug 2 at 22:35
 |Â
show 4 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
You could do all the calculations without evaluating the first derivative in the function, but we'll do it anyway $$u(x,t) = -2overbracefracpartialphi(x,t)partial x^;;textderivative of\textargument of logunderbracefrac1phi(x,t)_textderivative of log(x)\ textwith argument phi(x,t)$$ Now you can see that the function $u(x,t)$ is the product of two functions! We can evaluate the other derivatives using the derivative of the product of functions, mainly
$$u_x(x,t) = fracpartialpartial xleft(-2fracpartialphi(x,t)partial xfrac1phi(x,t)right) = underbrace-2fracpartial^2phipartial x^2frac1phi_D[1^textst]times2^textnd+underbrace2fracpartial phipartial xfracpartialphipartial xfrac1phi^2_D[2^textnd]times1^textst=2left(frac1phifracpartial phipartial xright)^2 -2fracpartial^2phipartial x^2frac1phi\
u_t(x,t) =-2fracpartial^2 phipartial x partial tfrac1phi+2fracpartial phipartial xfracpartial phipartial tfrac1phi^2\
u_xx(x,t) = fracpartialpartial xleft(colorred2left(frac1phifracpartial phipartial xright)^2 colorblue-2fracpartial^2phipartial x^2frac1phiright) = \=colorred-4left(frac1phifracpartial phipartial xright)left(frac1phi^2fracpartialphipartial x+frac1phifracpartial^2phipartial x^2right)colorblue-2left(frac1phifracpartial^3phipartial x^3-frac1phi^2fracpartial phipartial xfracpartial^2phipartial x^2right)$$
all done by using the chain rule. I used coloring to simplify the visualization of the derivatives
answered Aug 2 at 21:26
Davide Morgante
1,632220
You could do all the calculations without evaluating the first derivative in the function, but we'll do it anyway $$u(x,t) = -2overbracefracpartialphi(x,t)partial x^;;textderivative of\textargument of logunderbracefrac1phi(x,t)_textderivative of log(x)\ textwith argument phi(x,t)$$ Now you can see that the function $u(x,t)$ is the product of two functions! We can evaluate the other derivatives using the derivative of the product of functions, mainly
$$u_x(x,t) = fracpartialpartial xleft(-2fracpartialphi(x,t)partial xfrac1phi(x,t)right) = underbrace-2fracpartial^2phipartial x^2frac1phi_D[1^textst]times2^textnd+underbrace2fracpartial phipartial xfracpartialphipartial xfrac1phi^2_D[2^textnd]times1^textst=2left(frac1phifracpartial phipartial xright)^2 -2fracpartial^2phipartial x^2frac1phi\
u_t(x,t) =-2fracpartial^2 phipartial x partial tfrac1phi+2fracpartial phipartial xfracpartial phipartial tfrac1phi^2\
u_xx(x,t) = fracpartialpartial xleft(colorred2left(frac1phifracpartial phipartial xright)^2 colorblue-2fracpartial^2phipartial x^2frac1phiright) = \=colorred-4left(frac1phifracpartial phipartial xright)left(frac1phi^2fracpartialphipartial x+frac1phifracpartial^2phipartial x^2right)colorblue-2left(frac1phifracpartial^3phipartial x^3-frac1phi^2fracpartial phipartial xfracpartial^2phipartial x^2right)$$
all done by using the chain rule. I used coloring to simplify the visualization of the derivatives
answered Aug 2 at 21:26
Davide Morgante
1,632220
answered Aug 2 at 21:26
Davide Morgante
1,632220
answered Aug 2 at 21:26
Davide Morgante
1,632220
answered Aug 2 at 21:26
Davide Morgante
1,632220
1,632220
Thank you. Your answer is very clear. Can you please also show how it's done without simplifying the first derivative? I'm trying to do that using the chain rule and find it very confusing.
– handler's handle
Aug 2 at 22:32
Without simplifying it you mean without collecting the squares together?
– Davide Morgante
Aug 2 at 22:33
I meant without this simplification: $u(x,t) = -2overbracefracpartialphi(x,t)partial x^;;textderivative of\textargument of logunderbracefrac1phi(x,t)_textderivative of log(x)\ textwith argument phi(x,t)$
– handler's handle
Aug 2 at 22:34
If I am not mistaken, this makes our use of the chain rule more complex.
– handler's handle
Aug 2 at 22:34
Oh ok, I don't think I can put it all in one comment, maybe I can do one just to clarify your doubts? Where are you stuck?
– Davide Morgante
Aug 2 at 22:35
 |Â
show 4 more comments
Thank you. Your answer is very clear. Can you please also show how it's done without simplifying the first derivative? I'm trying to do that using the chain rule and find it very confusing.
– handler's handle
Aug 2 at 22:32
Without simplifying it you mean without collecting the squares together?
– Davide Morgante
Aug 2 at 22:33
I meant without this simplification: $u(x,t) = -2overbracefracpartialphi(x,t)partial x^;;textderivative of\textargument of logunderbracefrac1phi(x,t)_textderivative of log(x)\ textwith argument phi(x,t)$
– handler's handle
Aug 2 at 22:34
If I am not mistaken, this makes our use of the chain rule more complex.
– handler's handle
Aug 2 at 22:34
Oh ok, I don't think I can put it all in one comment, maybe I can do one just to clarify your doubts? Where are you stuck?
– Davide Morgante
Aug 2 at 22:35
Thank you. Your answer is very clear. Can you please also show how it's done without simplifying the first derivative? I'm trying to do that using the chain rule and find it very confusing.
– handler's handle
Aug 2 at 22:32
Thank you. Your answer is very clear. Can you please also show how it's done without simplifying the first derivative? I'm trying to do that using the chain rule and find it very confusing.
– handler's handle
Aug 2 at 22:32
Without simplifying it you mean without collecting the squares together?
– Davide Morgante
Aug 2 at 22:33
Without simplifying it you mean without collecting the squares together?
– Davide Morgante
Aug 2 at 22:33
I meant without this simplification: $u(x,t) = -2overbracefracpartialphi(x,t)partial x^;;textderivative of\textargument of logunderbracefrac1phi(x,t)_textderivative of log(x)\ textwith argument phi(x,t)$
– handler's handle
Aug 2 at 22:34
I meant without this simplification: $u(x,t) = -2overbracefracpartialphi(x,t)partial x^;;textderivative of\textargument of logunderbracefrac1phi(x,t)_textderivative of log(x)\ textwith argument phi(x,t)$
– handler's handle
Aug 2 at 22:34
If I am not mistaken, this makes our use of the chain rule more complex.
– handler's handle
Aug 2 at 22:34
If I am not mistaken, this makes our use of the chain rule more complex.
– handler's handle
Aug 2 at 22:34
Oh ok, I don't think I can put it all in one comment, maybe I can do one just to clarify your doubts? Where are you stuck?
– Davide Morgante
Aug 2 at 22:35
Oh ok, I don't think I can put it all in one comment, maybe I can do one just to clarify your doubts? Where are you stuck?
– Davide Morgante
Aug 2 at 22:35
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First simplify the right side by finding the derivative.
$$fracpartialpartial xlogphi(x,t) = frac1phi(x,t)fracpartial phipartial x$$
so
$$u(x,t) = -2fracphi_x(x,t)phi(x,t)$$
Then the derivatives of $u$ are straightforward, each one follows the quotient rule.
$$fracpartial upartial t = -2fracpartialpartial tleft(fracphi_x(x,t)phi(x,t)right) = -2fracphi(x,t)phi_xt(x,t)-phi_x(x,t)phi_t(x,t)phi^2(x,t)$$
$$fracpartial upartial x = -2fracphi(x,t)phi_xx(x,t) - phi^2_x(x,t)phi^2(x,t)$$
The last is a bit of a doozy though. Dropping the $(x,t)$ as $phi$ is understood to be a fuction of both $x$ and $t$,
$$fracpartial^2upartial x^2=-2fracphi^2(phi_xphi_xx + phiphi_xxx - 2phi_xphi_xx) - 2phiphi_x(phiphi_xx-phi^2_x)phi^4$$
answered Aug 2 at 21:32
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First simplify the right side by finding the derivative.
$$fracpartialpartial xlogphi(x,t) = frac1phi(x,t)fracpartial phipartial x$$
so
$$u(x,t) = -2fracphi_x(x,t)phi(x,t)$$
Then the derivatives of $u$ are straightforward, each one follows the quotient rule.
$$fracpartial upartial t = -2fracpartialpartial tleft(fracphi_x(x,t)phi(x,t)right) = -2fracphi(x,t)phi_xt(x,t)-phi_x(x,t)phi_t(x,t)phi^2(x,t)$$
$$fracpartial upartial x = -2fracphi(x,t)phi_xx(x,t) - phi^2_x(x,t)phi^2(x,t)$$
The last is a bit of a doozy though. Dropping the $(x,t)$ as $phi$ is understood to be a fuction of both $x$ and $t$,
$$fracpartial^2upartial x^2=-2fracphi^2(phi_xphi_xx + phiphi_xxx - 2phi_xphi_xx) - 2phiphi_x(phiphi_xx-phi^2_x)phi^4$$
answered Aug 2 at 21:32
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First simplify the right side by finding the derivative.
$$fracpartialpartial xlogphi(x,t) = frac1phi(x,t)fracpartial phipartial x$$
so
$$u(x,t) = -2fracphi_x(x,t)phi(x,t)$$
Then the derivatives of $u$ are straightforward, each one follows the quotient rule.
$$fracpartial upartial t = -2fracpartialpartial tleft(fracphi_x(x,t)phi(x,t)right) = -2fracphi(x,t)phi_xt(x,t)-phi_x(x,t)phi_t(x,t)phi^2(x,t)$$
$$fracpartial upartial x = -2fracphi(x,t)phi_xx(x,t) - phi^2_x(x,t)phi^2(x,t)$$
The last is a bit of a doozy though. Dropping the $(x,t)$ as $phi$ is understood to be a fuction of both $x$ and $t$,
$$fracpartial^2upartial x^2=-2fracphi^2(phi_xphi_xx + phiphi_xxx - 2phi_xphi_xx) - 2phiphi_x(phiphi_xx-phi^2_x)phi^4$$
answered Aug 2 at 21:32
Skip
1,147212
First simplify the right side by finding the derivative.
$$fracpartialpartial xlogphi(x,t) = frac1phi(x,t)fracpartial phipartial x$$
so
$$u(x,t) = -2fracphi_x(x,t)phi(x,t)$$
Then the derivatives of $u$ are straightforward, each one follows the quotient rule.
$$fracpartial upartial t = -2fracpartialpartial tleft(fracphi_x(x,t)phi(x,t)right) = -2fracphi(x,t)phi_xt(x,t)-phi_x(x,t)phi_t(x,t)phi^2(x,t)$$
$$fracpartial upartial x = -2fracphi(x,t)phi_xx(x,t) - phi^2_x(x,t)phi^2(x,t)$$
The last is a bit of a doozy though. Dropping the $(x,t)$ as $phi$ is understood to be a fuction of both $x$ and $t$,
$$fracpartial^2upartial x^2=-2fracphi^2(phi_xphi_xx + phiphi_xxx - 2phi_xphi_xx) - 2phiphi_x(phiphi_xx-phi^2_x)phi^4$$
answered Aug 2 at 21:32
Skip
1,147212
answered Aug 2 at 21:32
Skip
1,147212
answered Aug 2 at 21:32
Skip
1,147212
answered Aug 2 at 21:32
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1,147212
1,147212
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$$u(x,t)~=~-2fracpartialpartial xlog(phi(x,t))$$
The chain rule states that
$$fracpartial f(u(x,t))partial x~=~fracpartial f(u(x,t))partial u(x,t)fracpartial u(x,t)partial x$$
And so the original function becomes
$$-2fracpartialpartial xlog(phi(x,t))~=~-2frac1phi(x,t)phi_x(x,t)~=~-2fracphi_x(x,t)phi(x,t)$$
So for the first case $u_t$ we consider the function $phi(x,t)$ as $u$ and $fracpartialpartial xlog(u)$ as $f$ so we get
$$beginalign
fracpartialpartial tleft(-2fracphi_x(x,t)phi(x,t)right)~&=~-2fracphi_xt(x,t)phi(x,t)-phi_t(x,t)phi_x(x,t)phi^2(x,t)\
fracpartialpartial xleft(-2fracphi_x(x,t)phi(x,t)right)~&=~-2fracphi_xx(x,t)phi(x,t)-phi_x^2(x,t)phi^2(x,t)\
fracpartialpartial xleft(-2fracphi_xx(x,t)phi(x,t)-phi_x^2(x,t)phi^2(x,t)right)~&=~frac[phi_xxxphi+phi_xphi_xx-2phi_xphi_xx]phi^2-2phiphi_x[phi_xxphi-phi_x^2]phi^4\
&=~frac[phi_xxxphi-phi_xphi_xx]phi-2phi_x[phi_xxphi-phi_x^2]phi^3\
&=~fracphi_xxxphi^2-3phi_xxphi_xphi+2phi_x^2phiphi^3
endalign$$
This work is kind of messy so I hope I did not made a mistake somewhere. I guess the crucial point is the to get rid of the partial derivative with respect to $x$ in the definition of the funtion $u$. I hope I made clear how it works out.
answered Aug 2 at 21:25
mrtaurho
607117
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$$u(x,t)~=~-2fracpartialpartial xlog(phi(x,t))$$
The chain rule states that
$$fracpartial f(u(x,t))partial x~=~fracpartial f(u(x,t))partial u(x,t)fracpartial u(x,t)partial x$$
And so the original function becomes
$$-2fracpartialpartial xlog(phi(x,t))~=~-2frac1phi(x,t)phi_x(x,t)~=~-2fracphi_x(x,t)phi(x,t)$$
So for the first case $u_t$ we consider the function $phi(x,t)$ as $u$ and $fracpartialpartial xlog(u)$ as $f$ so we get
$$beginalign
fracpartialpartial tleft(-2fracphi_x(x,t)phi(x,t)right)~&=~-2fracphi_xt(x,t)phi(x,t)-phi_t(x,t)phi_x(x,t)phi^2(x,t)\
fracpartialpartial xleft(-2fracphi_x(x,t)phi(x,t)right)~&=~-2fracphi_xx(x,t)phi(x,t)-phi_x^2(x,t)phi^2(x,t)\
fracpartialpartial xleft(-2fracphi_xx(x,t)phi(x,t)-phi_x^2(x,t)phi^2(x,t)right)~&=~frac[phi_xxxphi+phi_xphi_xx-2phi_xphi_xx]phi^2-2phiphi_x[phi_xxphi-phi_x^2]phi^4\
&=~frac[phi_xxxphi-phi_xphi_xx]phi-2phi_x[phi_xxphi-phi_x^2]phi^3\
&=~fracphi_xxxphi^2-3phi_xxphi_xphi+2phi_x^2phiphi^3
endalign$$
This work is kind of messy so I hope I did not made a mistake somewhere. I guess the crucial point is the to get rid of the partial derivative with respect to $x$ in the definition of the funtion $u$. I hope I made clear how it works out.
answered Aug 2 at 21:25
mrtaurho
607117
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up vote
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up vote
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$$u(x,t)~=~-2fracpartialpartial xlog(phi(x,t))$$
The chain rule states that
$$fracpartial f(u(x,t))partial x~=~fracpartial f(u(x,t))partial u(x,t)fracpartial u(x,t)partial x$$
And so the original function becomes
$$-2fracpartialpartial xlog(phi(x,t))~=~-2frac1phi(x,t)phi_x(x,t)~=~-2fracphi_x(x,t)phi(x,t)$$
So for the first case $u_t$ we consider the function $phi(x,t)$ as $u$ and $fracpartialpartial xlog(u)$ as $f$ so we get
$$beginalign
fracpartialpartial tleft(-2fracphi_x(x,t)phi(x,t)right)~&=~-2fracphi_xt(x,t)phi(x,t)-phi_t(x,t)phi_x(x,t)phi^2(x,t)\
fracpartialpartial xleft(-2fracphi_x(x,t)phi(x,t)right)~&=~-2fracphi_xx(x,t)phi(x,t)-phi_x^2(x,t)phi^2(x,t)\
fracpartialpartial xleft(-2fracphi_xx(x,t)phi(x,t)-phi_x^2(x,t)phi^2(x,t)right)~&=~frac[phi_xxxphi+phi_xphi_xx-2phi_xphi_xx]phi^2-2phiphi_x[phi_xxphi-phi_x^2]phi^4\
&=~frac[phi_xxxphi-phi_xphi_xx]phi-2phi_x[phi_xxphi-phi_x^2]phi^3\
&=~fracphi_xxxphi^2-3phi_xxphi_xphi+2phi_x^2phiphi^3
endalign$$
This work is kind of messy so I hope I did not made a mistake somewhere. I guess the crucial point is the to get rid of the partial derivative with respect to $x$ in the definition of the funtion $u$. I hope I made clear how it works out.
answered Aug 2 at 21:25
mrtaurho
607117
$$u(x,t)~=~-2fracpartialpartial xlog(phi(x,t))$$
The chain rule states that
$$fracpartial f(u(x,t))partial x~=~fracpartial f(u(x,t))partial u(x,t)fracpartial u(x,t)partial x$$
And so the original function becomes
$$-2fracpartialpartial xlog(phi(x,t))~=~-2frac1phi(x,t)phi_x(x,t)~=~-2fracphi_x(x,t)phi(x,t)$$
So for the first case $u_t$ we consider the function $phi(x,t)$ as $u$ and $fracpartialpartial xlog(u)$ as $f$ so we get
$$beginalign
fracpartialpartial tleft(-2fracphi_x(x,t)phi(x,t)right)~&=~-2fracphi_xt(x,t)phi(x,t)-phi_t(x,t)phi_x(x,t)phi^2(x,t)\
fracpartialpartial xleft(-2fracphi_x(x,t)phi(x,t)right)~&=~-2fracphi_xx(x,t)phi(x,t)-phi_x^2(x,t)phi^2(x,t)\
fracpartialpartial xleft(-2fracphi_xx(x,t)phi(x,t)-phi_x^2(x,t)phi^2(x,t)right)~&=~frac[phi_xxxphi+phi_xphi_xx-2phi_xphi_xx]phi^2-2phiphi_x[phi_xxphi-phi_x^2]phi^4\
&=~frac[phi_xxxphi-phi_xphi_xx]phi-2phi_x[phi_xxphi-phi_x^2]phi^3\
&=~fracphi_xxxphi^2-3phi_xxphi_xphi+2phi_x^2phiphi^3
endalign$$
This work is kind of messy so I hope I did not made a mistake somewhere. I guess the crucial point is the to get rid of the partial derivative with respect to $x$ in the definition of the funtion $u$. I hope I made clear how it works out.
answered Aug 2 at 21:25
mrtaurho
607117
edited Aug 2 at 21:34
answered Aug 2 at 21:25
mrtaurho
607117
answered Aug 2 at 21:25
mrtaurho
607117
answered Aug 2 at 21:25
mrtaurho
607117
607117
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I don't quite understand why that $$fracpartialpartial x$$ became a $$fracpartialpartial log(phi)$$ if we use the chain rule we have that $$fracpartialpartial x = fracpartialpartial log(phi)fracpartiallog(phi)partial x$$ and this first equality doesn't touch the derivative that follows it. You could rewrite even that other partial derivative in the same way but it seems pretty unnecessary
– Davide Morgante
Aug 3 at 8:18
@DavideMorgante I apologise for doing it so badly. Can you please demonstrate this as an example with explanations? That way, I will get a better idea as to how to do it and what I am doing wrong. I will then try to do it for $u_xx$ and $u_t$ and post it.
– handler's handle
Aug 3 at 9:24