Multi-variable chain rule with multi-variable functions as arguments
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What is the chain rule of a multi-variable function with arguments that are also multi-variable functions?
Suppose $x$, $y$, $z$ are independent variables. I mean changing $x$ won't change $y$ and $z$.
Is the general form of multi-variable chain rule similar to the following?
$$fracpartial w(u(x, y), v(y, z), t(y, z))partial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x $$
Thanks.
calculus linear-algebra multivariable-calculus
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up vote
1
down vote
favorite
What is the chain rule of a multi-variable function with arguments that are also multi-variable functions?
Suppose $x$, $y$, $z$ are independent variables. I mean changing $x$ won't change $y$ and $z$.
Is the general form of multi-variable chain rule similar to the following?
$$fracpartial w(u(x, y), v(y, z), t(y, z))partial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x $$
Thanks.
calculus linear-algebra multivariable-calculus
I am sure a gazillion people already know the answer... but all I find on the web is the $dx/dt$ thing and all have single variable function as argument of another function. Please tell me if you know this. Thanks.
â R zu
Aug 28 at 16:23
Thanks for the answers. You have my eternal gratitude.
â R zu
Aug 28 at 17:04
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
What is the chain rule of a multi-variable function with arguments that are also multi-variable functions?
Suppose $x$, $y$, $z$ are independent variables. I mean changing $x$ won't change $y$ and $z$.
Is the general form of multi-variable chain rule similar to the following?
$$fracpartial w(u(x, y), v(y, z), t(y, z))partial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x $$
Thanks.
calculus linear-algebra multivariable-calculus
What is the chain rule of a multi-variable function with arguments that are also multi-variable functions?
Suppose $x$, $y$, $z$ are independent variables. I mean changing $x$ won't change $y$ and $z$.
Is the general form of multi-variable chain rule similar to the following?
$$fracpartial w(u(x, y), v(y, z), t(y, z))partial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x $$
Thanks.
calculus linear-algebra multivariable-calculus
edited Aug 28 at 16:25
StubbornAtom
4,08511135
4,08511135
asked Aug 28 at 16:21
R zu
1587
1587
I am sure a gazillion people already know the answer... but all I find on the web is the $dx/dt$ thing and all have single variable function as argument of another function. Please tell me if you know this. Thanks.
â R zu
Aug 28 at 16:23
Thanks for the answers. You have my eternal gratitude.
â R zu
Aug 28 at 17:04
add a comment |Â
I am sure a gazillion people already know the answer... but all I find on the web is the $dx/dt$ thing and all have single variable function as argument of another function. Please tell me if you know this. Thanks.
â R zu
Aug 28 at 16:23
Thanks for the answers. You have my eternal gratitude.
â R zu
Aug 28 at 17:04
I am sure a gazillion people already know the answer... but all I find on the web is the $dx/dt$ thing and all have single variable function as argument of another function. Please tell me if you know this. Thanks.
â R zu
Aug 28 at 16:23
I am sure a gazillion people already know the answer... but all I find on the web is the $dx/dt$ thing and all have single variable function as argument of another function. Please tell me if you know this. Thanks.
â R zu
Aug 28 at 16:23
Thanks for the answers. You have my eternal gratitude.
â R zu
Aug 28 at 17:04
Thanks for the answers. You have my eternal gratitude.
â R zu
Aug 28 at 17:04
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
That's nearly right, but you left off the third term that accounts for $t$. So you should have
$$fracpartial wpartial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x + fracpartial wpartial tcdotfracpartial tpartial x$$
where $w$ is a function of $u,v,t$ and $u,v,t$ are functions of $x,y,z$.
One mnemonic device for remembering this is to think of it as summing over all products of "fractions" that will partially cancel and each give $partial w$ in the numerator and $partial x$ in the denominator. But be warned that that's not really what's happening, these aren't fractions and they aren't cancelling. Nonetheless, this helps you to remember it.
Addendum: Of course, if $x$ doesn't actually appear in the formula for $u$, $v$, or $t$, then that partial is zero. What I wrote is the most general formula, not just the formula for your specific case. In other words, even if $v=v(y,z)$ , you can still think of it as $v(x,y,z)$; but then $partial v/partial x = 0$ , and similarly for the others. Not sure if you really wrote what you meant, but my answer is applicable to the general case.
I saw a linear algebra proof of the simple chain rule: proofwiki.org/wiki/Chain_Rule_for_Real-Valued_Functions But I haven't studied that yet. Thanks a lot for the answer.
â R zu
Aug 28 at 17:05
add a comment |Â
up vote
1
down vote
beginalign
fracpartial w(u(x, y), v(y, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotoverbracefracpartial vpartial x^0+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
&=fracpartial wpartial ucdotfracpartial upartial x,
endalign
or what I think you meant to write
beginalign
fracpartial w(u(x, y), v(colorredx, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
&=fracpartial wpartial ucdotfracpartial upartial x+ fracpartial wpartial vcdotfracpartial vpartial x.
endalign
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
That's nearly right, but you left off the third term that accounts for $t$. So you should have
$$fracpartial wpartial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x + fracpartial wpartial tcdotfracpartial tpartial x$$
where $w$ is a function of $u,v,t$ and $u,v,t$ are functions of $x,y,z$.
One mnemonic device for remembering this is to think of it as summing over all products of "fractions" that will partially cancel and each give $partial w$ in the numerator and $partial x$ in the denominator. But be warned that that's not really what's happening, these aren't fractions and they aren't cancelling. Nonetheless, this helps you to remember it.
Addendum: Of course, if $x$ doesn't actually appear in the formula for $u$, $v$, or $t$, then that partial is zero. What I wrote is the most general formula, not just the formula for your specific case. In other words, even if $v=v(y,z)$ , you can still think of it as $v(x,y,z)$; but then $partial v/partial x = 0$ , and similarly for the others. Not sure if you really wrote what you meant, but my answer is applicable to the general case.
I saw a linear algebra proof of the simple chain rule: proofwiki.org/wiki/Chain_Rule_for_Real-Valued_Functions But I haven't studied that yet. Thanks a lot for the answer.
â R zu
Aug 28 at 17:05
add a comment |Â
up vote
3
down vote
accepted
That's nearly right, but you left off the third term that accounts for $t$. So you should have
$$fracpartial wpartial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x + fracpartial wpartial tcdotfracpartial tpartial x$$
where $w$ is a function of $u,v,t$ and $u,v,t$ are functions of $x,y,z$.
One mnemonic device for remembering this is to think of it as summing over all products of "fractions" that will partially cancel and each give $partial w$ in the numerator and $partial x$ in the denominator. But be warned that that's not really what's happening, these aren't fractions and they aren't cancelling. Nonetheless, this helps you to remember it.
Addendum: Of course, if $x$ doesn't actually appear in the formula for $u$, $v$, or $t$, then that partial is zero. What I wrote is the most general formula, not just the formula for your specific case. In other words, even if $v=v(y,z)$ , you can still think of it as $v(x,y,z)$; but then $partial v/partial x = 0$ , and similarly for the others. Not sure if you really wrote what you meant, but my answer is applicable to the general case.
I saw a linear algebra proof of the simple chain rule: proofwiki.org/wiki/Chain_Rule_for_Real-Valued_Functions But I haven't studied that yet. Thanks a lot for the answer.
â R zu
Aug 28 at 17:05
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
That's nearly right, but you left off the third term that accounts for $t$. So you should have
$$fracpartial wpartial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x + fracpartial wpartial tcdotfracpartial tpartial x$$
where $w$ is a function of $u,v,t$ and $u,v,t$ are functions of $x,y,z$.
One mnemonic device for remembering this is to think of it as summing over all products of "fractions" that will partially cancel and each give $partial w$ in the numerator and $partial x$ in the denominator. But be warned that that's not really what's happening, these aren't fractions and they aren't cancelling. Nonetheless, this helps you to remember it.
Addendum: Of course, if $x$ doesn't actually appear in the formula for $u$, $v$, or $t$, then that partial is zero. What I wrote is the most general formula, not just the formula for your specific case. In other words, even if $v=v(y,z)$ , you can still think of it as $v(x,y,z)$; but then $partial v/partial x = 0$ , and similarly for the others. Not sure if you really wrote what you meant, but my answer is applicable to the general case.
That's nearly right, but you left off the third term that accounts for $t$. So you should have
$$fracpartial wpartial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x + fracpartial wpartial tcdotfracpartial tpartial x$$
where $w$ is a function of $u,v,t$ and $u,v,t$ are functions of $x,y,z$.
One mnemonic device for remembering this is to think of it as summing over all products of "fractions" that will partially cancel and each give $partial w$ in the numerator and $partial x$ in the denominator. But be warned that that's not really what's happening, these aren't fractions and they aren't cancelling. Nonetheless, this helps you to remember it.
Addendum: Of course, if $x$ doesn't actually appear in the formula for $u$, $v$, or $t$, then that partial is zero. What I wrote is the most general formula, not just the formula for your specific case. In other words, even if $v=v(y,z)$ , you can still think of it as $v(x,y,z)$; but then $partial v/partial x = 0$ , and similarly for the others. Not sure if you really wrote what you meant, but my answer is applicable to the general case.
edited Aug 28 at 17:19
answered Aug 28 at 16:46
MPW
28.6k11853
28.6k11853
I saw a linear algebra proof of the simple chain rule: proofwiki.org/wiki/Chain_Rule_for_Real-Valued_Functions But I haven't studied that yet. Thanks a lot for the answer.
â R zu
Aug 28 at 17:05
add a comment |Â
I saw a linear algebra proof of the simple chain rule: proofwiki.org/wiki/Chain_Rule_for_Real-Valued_Functions But I haven't studied that yet. Thanks a lot for the answer.
â R zu
Aug 28 at 17:05
I saw a linear algebra proof of the simple chain rule: proofwiki.org/wiki/Chain_Rule_for_Real-Valued_Functions But I haven't studied that yet. Thanks a lot for the answer.
â R zu
Aug 28 at 17:05
I saw a linear algebra proof of the simple chain rule: proofwiki.org/wiki/Chain_Rule_for_Real-Valued_Functions But I haven't studied that yet. Thanks a lot for the answer.
â R zu
Aug 28 at 17:05
add a comment |Â
up vote
1
down vote
beginalign
fracpartial w(u(x, y), v(y, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotoverbracefracpartial vpartial x^0+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
&=fracpartial wpartial ucdotfracpartial upartial x,
endalign
or what I think you meant to write
beginalign
fracpartial w(u(x, y), v(colorredx, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
&=fracpartial wpartial ucdotfracpartial upartial x+ fracpartial wpartial vcdotfracpartial vpartial x.
endalign
add a comment |Â
up vote
1
down vote
beginalign
fracpartial w(u(x, y), v(y, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotoverbracefracpartial vpartial x^0+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
&=fracpartial wpartial ucdotfracpartial upartial x,
endalign
or what I think you meant to write
beginalign
fracpartial w(u(x, y), v(colorredx, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
&=fracpartial wpartial ucdotfracpartial upartial x+ fracpartial wpartial vcdotfracpartial vpartial x.
endalign
add a comment |Â
up vote
1
down vote
up vote
1
down vote
beginalign
fracpartial w(u(x, y), v(y, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotoverbracefracpartial vpartial x^0+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
&=fracpartial wpartial ucdotfracpartial upartial x,
endalign
or what I think you meant to write
beginalign
fracpartial w(u(x, y), v(colorredx, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
&=fracpartial wpartial ucdotfracpartial upartial x+ fracpartial wpartial vcdotfracpartial vpartial x.
endalign
beginalign
fracpartial w(u(x, y), v(y, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotoverbracefracpartial vpartial x^0+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
&=fracpartial wpartial ucdotfracpartial upartial x,
endalign
or what I think you meant to write
beginalign
fracpartial w(u(x, y), v(colorredx, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
&=fracpartial wpartial ucdotfracpartial upartial x+ fracpartial wpartial vcdotfracpartial vpartial x.
endalign
answered Aug 28 at 16:53
mzp
1,16421134
1,16421134
add a comment |Â
add a comment |Â
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I am sure a gazillion people already know the answer... but all I find on the web is the $dx/dt$ thing and all have single variable function as argument of another function. Please tell me if you know this. Thanks.
â R zu
Aug 28 at 16:23
Thanks for the answers. You have my eternal gratitude.
â R zu
Aug 28 at 17:04